Explanation:
Okay, let's try to figure out Rosa's age. The problem says that when her age is divided by 2, 3, 4, or 6, the remainder is 1. Also, when divided by 7, there's no remainder. And she's less than 75 years old. Hmm, okay, let's break this down step by step.

First, let me note down the conditions:

1. Rosa's age divided by 2 leaves a remainder of 1.
2. Divided by 3, remainder 1.
3. Divided by 4, remainder 1.
4. Divided by 6, remainder 1.
5. Divided by 7, remainder 0.
6. Age is less than 75.

So, we need to find a number (Rosa's age) that satisfies all these conditions. Let me think about how to approach this.

Starting with the first condition: when divided by 2, remainder 1. That means Rosa's age is an odd number because if you divide an odd number by 2, the remainder is 1. Similarly, for division by 3, 4, 6, the remainder is 1. So, the age is 1 more than a multiple of 2, 3, 4, and 6.

Wait, so if I think about it, the age minus 1 should be divisible by 2, 3, 4, and 6. Let me write that down:

Let Rosa's age be N. Then:

N ≡ 1 mod 2
N ≡ 1 mod 3
N ≡ 1 mod 4
N ≡ 1 mod 6
N ≡ 0 mod 7
N < 75

So, N - 1 is divisible by 2, 3, 4, and 6. Therefore, N - 1 is a common multiple of these numbers. So, the least common multiple (LCM) of 2, 3, 4, 6. Let me compute that.

First, let's find LCM of 2, 3, 4, 6. Let's break them down into prime factors:

- 2 is 2
- 3 is 3
- 4 is 2²
- 6 is 2 × 3

The LCM is the product of the highest powers of all primes present. So, for prime 2, the highest power is 2² (from 4), and for prime 3, it's 3¹. Therefore, LCM is 2² × 3 = 4 × 3 = 12.

So, LCM of 2, 3, 4, 6 is 12. Therefore, N - 1 must be a multiple of 12. Therefore, N can be written as 12k + 1, where k is a positive integer.

But also, N must be divisible by 7. So, N is a multiple of 7. Therefore, 12k + 1 must be divisible by 7. So, 12k + 1 ≡ 0 mod 7. Let me write that equation:

12k + 1 ≡ 0 mod 7
Which simplifies to 12k ≡ -1 mod 7
But 12 mod 7 is 5, since 7*1=7, 12-7=5. So:

5k ≡ -1 mod 7
But -1 mod 7 is the same as 6 mod 7. So:

5k ≡ 6 mod 7
Now, we need to solve for k here. Let me find the value of k that satisfies this congruence.

To solve 5k ≡ 6 mod 7, we can find the multiplicative inverse of 5 modulo 7. The inverse of 5 mod 7 is a number x such that 5x ≡ 1 mod 7. Let me test numbers:

5*1=5≡5
5*2=10≡3
5*3=15≡1
Ah, so x=3. Therefore, multiplying both sides of 5k ≡6 mod7 by 3:

k ≡ 6*3 mod7
k ≡ 18 mod7
18 divided by 7 is 2*7=14, remainder 4. So k ≡4 mod7.

Therefore, k can be written as k=7m +4, where m is a non-negative integer.

Therefore, substituting back into N=12k +1:

N=12*(7m +4) +1= 84m +48 +1=84m +49.

So, N=84m +49. Now, we also know that N <75. Let's find possible values of m.

For m=0: N=0 +49=49. Which is less than 75.
For m=1: N=84 +49=133. Which is more than 75. So, m=0 is the only possible value here. Therefore, N=49.

Wait, but let me check if this satisfies all the original conditions. Let me verify.

First, check if 49 divided by 2 leaves remainder 1. 49 divided by 2 is 24 with remainder 1. Yes.
Divided by 3: 3*16=48, so remainder 1. Correct.
Divided by 4: 4*12=48, remainder 1. Correct.
Divided by 6: 6*8=48, remainder 1. Correct.
Divided by 7: 49/7=7, remainder 0. Correct.
And 49 is less than 75. Perfect.

But wait, let me check if there are other possible values. Since m=0 gives 49, m=1 gives 133 which is over 75. So, only 49.

But let me think again. Is there another way to approach this? Maybe I made a mistake in calculating LCM?

Wait, the LCM of 2,3,4,6 is 12. So N-1 is multiple of 12. So N=12k +1. Then N must be divisible by 7. So 12k +1 divisible by 7. So 12k ≡ -1 mod7. Which is 5k ≡6 mod7. Then solving for k gives k≡4 mod7. Therefore, k=7m +4. Therefore N=12*(7m +4)+1=84m +49. So yes, same result.

So, with N <75, m=0 gives 49, m=1 gives 133. So 49 is the only possible.

But let me check if there are other numbers less than 75 that satisfy N ≡1 mod 12 and N≡0 mod7.

Numbers congruent to 1 mod12 less than 75 are: 1,13,25,37,49,61,73.

Now check which of these are divisible by 7.

Check 1: 1/7 no.
13: 13/7 no.
25: 25/7 no.
37: 37/7 no.
49: 49/7=7. Yes.
61: 61/7=8*7=56, 61-56=5. No.
73: 73/7=10*7=70, 73-70=3. No.

So only 49. Therefore, Rosa is 49 years old.

But let me check again if there's another way someone might approach this problem, maybe by considering the LCM of 2,3,4,6 and 7? Wait, but the problem says that N divided by 2,3,4,6 leaves remainder 1, and divided by 7 leaves 0. So N is 1 more than multiple of LCM(2,3,4,6)=12, and multiple of 7. So N is LCM(12,7) + something? Wait, but since N=12k +1 and N is multiple of 7, so N is a common multiple of 7 and 12k +1. But since 12k +1 is not necessarily a multiple of 7, but we found that k must be such that 12k +1 is multiple of 7.

Alternatively, maybe think of N as a number that is 1 more than a multiple of 12 and also a multiple of 7. So N is a common multiple of 7 and 12, but shifted by 1? Wait, not exactly.

Alternatively, since N ≡1 mod12 and N≡0 mod7. So solving the system:

N ≡1 mod12
N ≡0 mod7

Which is exactly what we did earlier. So using Chinese Remainder Theorem.

So solving N ≡1 mod12 and N≡0 mod7. Let me try solving this system again.

Let N=7m. Then 7m ≡1 mod12. So 7m ≡1 mod12. Find m.

Multiply both sides by inverse of 7 mod12. What is inverse of 7 mod12? Find x such that 7x ≡1 mod12. Testing numbers:

7*1=7≡7
7*2=14≡2
7*3=21≡9
7*4=28≡4
7*5=35≡11
7*6=42≡6
7*7=49≡1 mod12. Because 49-48=1. So inverse of 7 mod12 is 7.

Therefore, m≡7*1≡7 mod12. Therefore, m=12k +7. Therefore, N=7m=7*(12k +7)=84k +49. Which is same as before. Therefore, N=84k +49. Since N<75, k=0 gives 49, k=1 gives 133>75. So only 49.

Therefore, Rosa is 49 years old.

But let me check once more with the original problem's conditions.

Divided by 2: 49/2=24*2 +1 → remainder 1. Correct.
Divided by 3: 49/3=16*3 +1 → remainder 1. Correct.
Divided by 4: 49/4=12*4 +1 → remainder 1. Correct.
Divided by 6: 49/6=8*6 +1 → remainder 1. Correct.
Divided by 7: 49/7=7 → remainder 0. Correct.
Age less than 75: 49<75. Correct.

So all conditions are satisfied.

But just to be thorough, let me check if there's another number less than 75 that meets all the conditions. For example, suppose someone thought that LCM of 2,3,4,6 is 12, so numbers like 13, 25, etc., but we saw that only 49 is divisible by 7.

Alternatively, maybe someone could think that since N-1 is divisible by 2,3,4,6, then N-1 must be divisible by LCM(2,3,4,6)=12. So N-1=12k, so N=12k+1. Then N must be divisible by 7. So 12k +1 divisible by 7. So 12k ≡ -1 mod7. Which is 5k ≡6 mod7. Then solving for k.

As before, k ≡ 4 mod7. Therefore, k=4,11,18,... So N=12*4 +1=49, 12*11 +1=133, etc. So again, only 49 under 75.

Therefore, I think 49 is the correct answer.

But just to make sure, let me check if there's a possibility that the LCM is different? For example, if someone thought that since dividing by 2,3,4,6 gives remainder 1, maybe they think that N-1 is divisible by all of them, but maybe there's a smaller LCM? But LCM(2,3,4,6) is indeed 12. Let me check:

Divisors:

- 2: factors 2
- 3: factors 3
- 4: factors 2²
- 6: factors 2×3

So LCM is 2²×3=12. Correct. So N-1 must be multiple of 12.

Alternatively, if someone thought that since dividing by 2,3,4,6 gives remainder 1, then N-1 must be divisible by 2,3,4,6. So yes, LCM is 12.

Therefore, I think the answer is 49.

But let me check if there's another approach. For example, considering that N ≡1 mod LCM(2,3,4,6)=12, so N=12k+1. Then N must be divisible by 7. So 12k +1 ≡0 mod7. So 12k ≡-1 mod7. 12 mod7 is 5, so 5k ≡-1 mod7. Which is 5k≡6 mod7. Multiply both sides by inverse of 5 mod7.

What is inverse of 5 mod7? Let me check:

5*? ≡1 mod7. 5*3=15≡1 mod7. So inverse is 3. Therefore, k≡6*3=18≡4 mod7. So k=7m +4. Therefore, N=12*(7m +4)+1=84m +49. So again, same result.

Therefore, only possible N under 75 is 49.

Therefore, I think the answer is 49.

**Final Answer**
Rosa is \boxed{49} years old.

To determine Rosa's age, we are given the following conditions:

1. When her age is divided by **2, 3, 4, or 6**, the **remainder is 1**.
2. When her age is divided by **7**, the **remainder is 0**.
3. Her age is **less than 75**.

---

### Step 1: Understanding the First Condition

If dividing Rosa's age $ N $ by 2, 3, 4, or 6 leaves a remainder of 1, then:

$$
N \equiv 1 \pmod{2}, \quad N \equiv 1 \pmod{3}, \quad N \equiv 1 \pmod{4}, \quad N \equiv 1 \pmod{6}
$$

This implies that:

$$
N - 1 \equiv 0 \pmod{2}, \quad N - 1 \equiv 0 \pmod{3}, \quad N - 1 \equiv 0 \pmod{4}, \quad N - 1 \equiv 0 \pmod{6}
$$

So, $ N - 1 $ is divisible by **2, 3, 4, and 6**. Therefore, $ N - 1 $ must be a **common multiple** of these numbers.

---

### Step 2: Find the Least Common Multiple (LCM)

We compute the LCM of 2, 3, 4, and 6:

- Prime factorization:
  - 2 = $ 2 $
  - 3 = $ 3 $
  - 4 = $ 2^2 $
  - 6 = $ 2 \times 3 $

So, the LCM is:

$$
\text{LCM}(2, 3, 4, 6) = 2^2 \times 3 = 12
$$

Thus, $ N - 1 = 12k $ for some integer $ k $, and:

$$
N = 12k + 1
$$

---

### Step 3: Apply the Second Condition

We are also told that $ N $ is divisible by 7:

$$
N \equiv 0 \pmod{7}
$$

Substitute $ N = 12k + 1 $ into this:

$$
12k + 1 \equiv 0 \pmod{7}
\Rightarrow 12k \equiv -1 \pmod{7}
$$

Since $ 12 \equiv 5 \pmod{7} $, this becomes:

$$
5k \equiv -1 \pmod{7}
\Rightarrow 5k \equiv 6 \pmod{7}
$$

Now, we solve for $ k $:

- Find the **multiplicative inverse** of 5 modulo 7. That is, find $ x $ such that:

$$
5x \equiv 1 \pmod{7}
$$

Try values: $ 5 \times 3 = 15 \equiv 1 \pmod{7} $, so the inverse is 3.

Multiply both sides of $ 5k \equiv 6 \pmod{7} $ by 3:

$$
k \equiv 6 \times 3 \equiv 18 \equiv 4 \pmod{7}
$$

So, $ k = 7m + 4 $ for some integer $ m $.

---

### Step 4: Find Possible Values of $ N $

Substitute $ k = 7m + 4 $ into $ N = 12k + 1 $:

$$
N = 12(7m + 4) + 1 = 84m + 49
$$

Now, apply the constraint $ N < 75 $:

- For $ m = 0 $: $ N = 49 $
- For $ m = 1 $: $ N = 133 $ (too large)

So, the only valid value is:

$$
N = 49
$$

---

### Step 5: Verify the Solution

Check that 49 satisfies all the original conditions:

- $ 49 \div 2 = 24 $ remainder 1 ✅
- $ 49 \div 3 = 16 $ remainder 1 ✅
- $ 49 \div 4 = 12 $ remainder 1 ✅
- $ 49 \div 6 = 8 $ remainder 1 ✅
- $ 49 \div 7 = 7 $ remainder 0 ✅
- $ 49 < 75 $ ✅

---

### Final Answer

$$
\boxed{49}
$$

[Assistant]
Explanation:
Okay, let's try to figure out Rosa's age. The problem says that when her age is divided by 2, 3, 4, or 6, the remainder is 1. Also, when divided by 7, there's no remainder. And she's less than 75 years old. Hmm, okay, let's break this down step by step.

First, let me note down the conditions:

1. Rosa's age divided by 2 leaves a remainder of 1.
2. Divided by 3, remainder 1.
3. Divided by 4, remainder 1.
4. Divided by 6, remainder 1.
5. Divided by 7, remainder 0.
6. Age is less than 75.

So, we need to find a number (Rosa's age) that satisfies all these conditions. Let me think about how to approach this.

Starting with the first condition: when divided by 2, remainder 1. That means Rosa's age is an odd number because if you divide an odd number by 2, the remainder is 1. Similarly, for division by 3, 4, 6, the remainder is 1. So, the age is 1 more than a multiple of 2, 3, 4, and 6.

Wait, so if I think about it, the age minus 1 should be divisible by 2, 3, 4, and 6. Let me write that down:

Let Rosa's age be N. Then:

N ≡ 1 mod 2
N ≡ 1 mod 3
N ≡ 1 mod 4
N ≡ 1 mod 6
N ≡ 0 mod 7
N < 75

So, N - 1 is divisible by 2, 3, 4, and 6. Therefore, N - 1 is a common multiple of these numbers. So, the least common multiple (LCM) of 2, 3, 4, 6. Let me compute that.

First, let's find LCM of 2, 3, 4, 6. Let's break them down into prime factors:

- 2 is 2
- 3 is 3
- 4 is 2²
- 6 is 2 × 3

The LCM is the product of the highest powers of all primes present. So, for prime 2, the highest power is 2² (from 4), and for prime 3, it's 3¹. Therefore, LCM is 2² × 3 = 4 × 3 = 12.

So, LCM of 2, 3, 4, 6 is 12. Therefore, N - 1 must be a multiple of 12. Therefore, N can be written as 12k + 1, where k is a positive integer.

But also, N must be divisible by 7. So, N is a multiple of 7. Therefore, 12k + 1 must be divisible by 7. So, 12k + 1 ≡ 0 mod 7. Let me write that equation:

12k + 1 ≡ 0 mod 7
Which simplifies to 12k ≡ -1 mod 7
But 12 mod 7 is 5, since 7*1=7, 12-7=5. So:

5k ≡ -1 mod 7
But -1 mod 7 is the same as 6 mod 7. So:

5k ≡ 6 mod 7
Now, we need to solve for k here. Let me find the value of k that satisfies this congruence.

To solve 5k ≡ 6 mod 7, we can find the multiplicative inverse of 5 modulo 7. The inverse of 5 mod 7 is a number x such that 5x ≡ 1 mod 7. Let me test numbers:

5*1=5≡5
5*2=10≡3
5*3=15≡1
Ah, so x=3. Therefore, multiplying both sides of 5k ≡6 mod7 by 3:

k ≡ 6*3 mod7
k ≡ 18 mod7
18 divided by 7 is 2*7=14, remainder 4. So k ≡4 mod7.

Therefore, k can be written as k=7m +4, where m is a non-negative integer.

Therefore, substituting back into N=12k +1:

N=12*(7m +4) +1= 84m +48 +1=84m +49.

So, N=84m +49. Now, we also know that N <75. Let's find possible values of m.

For m=0: N=0 +49=49. Which is less than 75.
For m=1: N=84 +49=133. Which is more than 75. So, m=0 is the only possible value here. Therefore, N=49.

Wait, but let me check if this satisfies all the original conditions. Let me verify.

First, check if 49 divided by 2 leaves remainder 1. 49 divided by 2 is 24 with remainder 1. Yes.
Divided by 3: 3*16=48, so remainder 1. Correct.
Divided by 4: 4*12=48, remainder 1. Correct.
Divided by 6: 6*8=48, remainder 1. Correct.
Divided by 7: 49/7=7, remainder 0. Correct.
And 49 is less than 75. Perfect.

But wait, let me check if there are other possible values. Since m=0 gives 49, m=1 gives 133 which is over 75. So, only 49.

But let me think again. Is there another way to approach this? Maybe I made a mistake in calculating LCM?

Wait, the LCM of 2,3,4,6 is 12. So N-1 is multiple of 12. So N=12k +1. Then N must be divisible by 7. So 12k +1 divisible by 7. So 12k ≡ -1 mod7. Which is 5k ≡6 mod7. Then solving for k gives k≡4 mod7. Therefore, k=7m +4. Therefore N=12*(7m +4)+1=84m +49. So yes, same result.

So, with N <75, m=0 gives 49, m=1 gives 133. So 49 is the only possible.

But let me check if there are other numbers less than 75 that satisfy N ≡1 mod 12 and N≡0 mod7.

Numbers congruent to 1 mod12 less than 75 are: 1,13,25,37,49,61,73.

Now check which of these are divisible by 7.

Check 1: 1/7 no.
13: 13/7 no.
25: 25/7 no.
37: 37/7 no.
49: 49/7=7. Yes.
61: 61/7=8*7=56, 61-56=5. No.
73: 73/7=10*7=70, 73-70=3. No.

So only 49. Therefore, Rosa is 49 years old.

But let me check again if there's another way someone might approach this problem, maybe by considering the LCM of 2,3,4,6 and 7? Wait, but the problem says that N divided by 2,3,4,6 leaves remainder 1, and divided by 7 leaves 0. So N is 1 more than multiple of LCM(2,3,4,6)=12, and multiple of 7. So N is LCM(12,7) + something? Wait, but since N=12k +1 and N is multiple of 7, so N is a common multiple of 7 and 12k +1. But since 12k +1 is not necessarily a multiple of 7, but we found that k must be such that 12k +1 is multiple of 7.

Alternatively, maybe think of N as a number that is 1 more than a multiple of 12 and also a multiple of 7. So N is a common multiple of 7 and 12, but shifted by 1? Wait, not exactly.

Alternatively, since N ≡1 mod12 and N≡0 mod7. So solving the system:

N ≡1 mod12
N ≡0 mod7

Which is exactly what we did earlier. So using Chinese Remainder Theorem.

So solving N ≡1 mod12 and N≡0 mod7. Let me try solving this system again.

Let N=7m. Then 7m ≡1 mod12. So 7m ≡1 mod12. Find m.

Multiply both sides by inverse of 7 mod12. What is inverse of 7 mod12? Find x such that 7x ≡1 mod12. Testing numbers:

7*1=7≡7
7*2=14≡2
7*3=21≡9
7*4=28≡4
7*5=35≡11
7*6=42≡6
7*7=49≡1 mod12. Because 49-48=1. So inverse of 7 mod12 is 7.

Therefore, m≡7*1≡7 mod12. Therefore, m=12k +7. Therefore, N=7m=7*(12k +7)=84k +49. Which is same as before. Therefore, N=84k +49. Since N<75, k=0 gives 49, k=1 gives 133>75. So only 49.

Therefore, Rosa is 49 years old.

But let me check once more with the original problem's conditions.

Divided by 2: 49/2=24*2 +1 → remainder 1. Correct.
Divided by 3: 49/3=16*3 +1 → remainder 1. Correct.
Divided by 4: 49/4=12*4 +1 → remainder 1. Correct.
Divided by 6: 49/6=8*6 +1 → remainder 1. Correct.
Divided by 7: 49/7=7 → remainder 0. Correct.
Age less than 75: 49<75. Correct.

So all conditions are satisfied.

But just to be thorough, let me check if there's another number less than 75 that meets all the conditions. For example, suppose someone thought that LCM of 2,3,4,6 is 12, so numbers like 13, 25, etc., but we saw that only 49 is divisible by 7.

Alternatively, maybe someone could think that since N-1 is divisible by 2,3,4,6, then N-1 must be divisible by LCM(2,3,4,6)=12. So N-1=12k, so N=12k+1. Then N must be divisible by 7. So 12k +1 divisible by 7. So 12k ≡ -1 mod7. Which is 5k ≡6 mod7. Then solving for k.

As before, k ≡ 4 mod7. Therefore, k=4,11,18,... So N=12*4 +1=49, 12*11 +1=133, etc. So again, only 49 under 75.

Therefore, I think 49 is the correct answer.

But just to make sure, let me check if there's a possibility that the LCM is different? For example, if someone thought that since dividing by 2,3,4,6 gives remainder 1, maybe they think that N-1 is divisible by all of them, but maybe there's a smaller LCM? But LCM(2,3,4,6) is indeed 12. Let me check:

Divisors:

- 2: factors 2
- 3: factors 3
- 4: factors 2²
- 6: factors 2×3

So LCM is 2²×3=12. Correct. So N-1 must be multiple of 12.

Alternatively, if someone thought that since dividing by 2,3,4,6 gives remainder 1, then N-1 must be divisible by 2,3,4,6. So yes, LCM is 12.

Therefore, I think the answer is 49.

But let me check if there's another approach. For example, considering that N ≡1 mod LCM(2,3,4,6)=12, so N=12k+1. Then N must be divisible by 7. So 12k +1 ≡0 mod7. So 12k ≡-1 mod7. 12 mod7 is 5, so 5k ≡-1 mod7. Which is 5k≡6 mod7. Multiply both sides by inverse of 5 mod7.

What is inverse of 5 mod7? Let me check:

5*? ≡1 mod7. 5*3=15≡1 mod7. So inverse is 3. Therefore, k≡6*3=18≡4 mod7. So k=7m +4. Therefore, N=12*(7m +4)+1=84m +49. So again, same result.

Therefore, only possible N under 75 is 49.

Therefore, I think the answer is 49.

**Final Answer**
Rosa is \boxed{49} years old.

To determine Rosa's age, we are given the following conditions:

1. When her age is divided by **2, 3, 4, or 6**, the **remainder is 1**.
2. When her age is divided by **7**, the **remainder is 0**.
3. Her age is **less than 75**.

---

### Step 1: Understanding the First Condition

If dividing Rosa's age $ N $ by 2, 3, 4, or 6 leaves a remainder of 1, then:

$$
N \equiv 1 \pmod{2}, \quad N \equiv 1 \pmod{3}, \quad N \equiv 1 \pmod{4}, \quad N \equiv 1 \pmod{6}
$$

This implies that:

$$
N - 1 \equiv 0 \pmod{2}, \quad N - 1 \equiv 0 \pmod{3}, \quad N - 1 \equiv 0 \pmod{4}, \quad N - 1 \equiv 0 \pmod{6}
$$

So, $ N - 1 $ is divisible by **2, 3, 4, and 6**. Therefore, $ N - 1 $ must be a **common multiple** of these numbers.

---

### Step 2: Find the Least Common Multiple (LCM)

We compute the LCM of 2, 3, 4, and 6:

- Prime factorization:
  - 2 = $ 2 $
  - 3 = $ 3 $
  - 4 = $ 2^2 $
  - 6 = $ 2 \times 3 $

So, the LCM is:

$$
\text{LCM}(2, 3, 4, 6) = 2^2 \times 3 = 12
$$

Thus, $ N - 1 = 12k $ for some integer $ k $, and:

$$
N = 12k + 1
$$

---

### Step 3: Apply the Second Condition

We are also told that $ N $ is divisible by 7:

$$
N \equiv 0 \pmod{7}
$$

Substitute $ N = 12k + 1 $ into this:

$$
12k + 1 \equiv 0 \pmod{7}
\Rightarrow 12k \equiv -1 \pmod{7}
$$

Since $ 12 \equiv 5 \pmod{7} $, this becomes:

$$
5k \equiv -1 \pmod{7}
\Rightarrow 5k \equiv 6 \pmod{7}
$$

Now, we solve for $ k $:

- Find the **multiplicative inverse** of 5 modulo 7. That is, find $ x $ such that:

$$
5x \equiv 1 \pmod{7}
$$

Try values: $ 5 \times 3 = 15 \equiv 1 \pmod{7} $, so the inverse is 3.

Multiply both sides of $ 5k \equiv 6 \pmod{7} $ by 3:

$$
k \equiv 6 \times 3 \equiv 18 \equiv 4 \pmod{7}
$$

So, $ k = 7m + 4 $ for some integer $ m $.

---

### Step 4: Find Possible Values of $ N $

Substitute $ k = 7m + 4 $ into $ N = 12k + 1 $:

$$
N = 12(7m + 4) + 1 = 84m + 49
$$

Now, apply the constraint $ N < 75 $:

- For $ m = 0 $: $ N = 49 $
- For $ m = 1 $: $ N = 133 $ (too large)

So, the only valid value is:

$$
N = 49
$$

---

### Step 5: Verify the Solution

Check that 49 satisfies all the original conditions:

- $ 49 \div 2 = 24 $ remainder 1 ✅
- $ 49 \div 3 = 16 $ remainder 1 ✅
- $ 49 \div 4 = 12 $ remainder 1 ✅
- $ 49 \div 6 = 8 $ remainder 1 ✅
- $ 49 \div 7 = 7 $ remainder 0 ✅
- $ 49 < 75 $ ✅

---

### Final Answer

$$
\boxed{49}
$$

Solution:
Since her age divided by 7 results in a remainder of 0, her age must be a multiple of 7. If her age is $n$, we notice that $n-1$ must be a multiple of 2, 3, 4, and 6. The least common multiple of those numbers is 12, so $n-1$ must be a multiple of 12. The multiples of 12 less than 75 are 12, 24, 36, 48, and 60. Adding 1 results in 13, 25, 37, 49, and 61, where 49 is the only multiple of 7. So Rosa is $\boxed{49}$ years old.

OR

We look for a multiple of 7 that is not divisible by 2, 3, 4, or 6. First we list all odd multiples of 7 less than 75, which are 7, 21, 35, 49, and 63. Since 21 and 63 are multiples of 3, we're left with 7, 35, and 49 as possibilities. Only $\boxed{49}$ has a remainder of 1 when divided by 2, 3, 4, or 6.


Explanation:
Okay, so I need to find the value of a + b + c, where a, b, and c are the roots of the equation given: 1/x + 5x² = 6x - 24. Hmm, let me think about how to approach this.

First, I remember that when dealing with equations involving roots, Vieta's formulas are usually helpful. Vieta's relates the coefficients of a polynomial to sums and products of its roots. But wait, the equation here isn't a polynomial equation yet. It has a term with 1/x, which complicates things. So maybe I need to manipulate the equation to convert it into a polynomial equation first?

Let me write down the original equation again:

1/x + 5x² = 6x - 24.

Okay, so to eliminate the fraction, I can multiply both sides of the equation by x. That should get rid of the denominator. Let me try that:

Multiplying both sides by x:

x*(1/x) + x*(5x²) = x*(6x) - x*(24)

Simplifying each term:

1 + 5x³ = 6x² - 24x

Now, let me bring all terms to one side to set the equation to zero:

5x³ - 6x² + 24x + 1 = 0

Wait, let me check that again. Starting from:

Left side after multiplying by x: 1 + 5x³

Right side after multiplying by x: 6x² - 24x

So bringing everything to the left side:

1 + 5x³ - 6x² + 24x = 0

Which can be written as:

5x³ - 6x² + 24x + 1 = 0

Wait, hold on. Let me check the signs again. If I subtract the right side from both sides:

Left side - Right side = 0:

(1 + 5x³) - (6x² - 24x) = 0

Which is 1 + 5x³ - 6x² + 24x = 0. Yes, that's correct. So rearranged:

5x³ - 6x² + 24x + 1 = 0

So now we have a cubic equation: 5x³ - 6x² + 24x + 1 = 0.

But wait, the original equation had 1/x, so when we multiplied by x, we assumed that x ≠ 0. Therefore, we need to check if x = 0 is a solution. Plugging x = 0 into the original equation: 1/0 is undefined, so x = 0 is not a solution. Therefore, our transformed equation 5x³ - 6x² + 24x + 1 = 0 should have all the roots of the original equation.

But the problem states that a, b, c are the roots of the original equation. However, the transformed equation is a cubic, so it has three roots. Therefore, a, b, c are the roots of this cubic equation. Therefore, we can apply Vieta's formula here.

Vieta's formula for a cubic equation ax³ + bx² + cx + d = 0 states that the sum of the roots is -b/a.

So in our case, the cubic equation is 5x³ - 6x² + 24x + 1 = 0. Therefore, the coefficients are:

a_cubic = 5 (coefficient of x³)

b_cubic = -6 (coefficient of x²)

c_cubic = 24 (coefficient of x)

d_cubic = 1 (constant term)

Therefore, the sum of the roots a + b + c = -b_cubic / a_cubic = -(-6)/5 = 6/5.

Wait, so is that the answer? 6/5? But let me make sure I didn't make any mistakes in transforming the original equation.

Original equation: 1/x + 5x² = 6x - 24.

Multiply both sides by x: 1 + 5x³ = 6x² - 24x.

Bring all terms to left: 5x³ - 6x² + 24x + 1 = 0. Yes, that seems correct.

Therefore, the cubic equation is correct, and Vieta's formula gives the sum of roots as 6/5. Therefore, a + b + c = 6/5.

But wait, hold on. Let me check again. Maybe I made a mistake in the sign when applying Vieta's formula. Let me recall Vieta's formula for cubic equations.

For a general cubic equation: ax³ + bx² + cx + d = 0, the sum of roots is -b/a.

In our case, the equation is 5x³ - 6x² + 24x + 1 = 0. Therefore, the coefficient of x² is -6, so the sum of roots is -(-6)/5 = 6/5. Yes, that's correct.

But let me think again. Suppose I have a cubic equation with roots a, b, c. Then:

(x - a)(x - b)(x - c) = x³ - (a + b + c)x² + (ab + ac + bc)x - abc.

But our equation is 5x³ - 6x² + 24x + 1 = 0. To match the standard form, we can divide both sides by 5:

x³ - (6/5)x² + (24/5)x + 1/5 = 0.

Comparing with the standard cubic equation x³ - (a + b + c)x² + (ab + ac + bc)x - abc = 0, we can see that:

-(a + b + c) = -6/5 => a + b + c = 6/5.

Yes, that's correct. Therefore, the sum of the roots is indeed 6/5.

But wait, the problem says "Compute a + b + c, given that a, b, and c are the roots of [equation]." So according to this, the answer is 6/5.

But let me check if there's any possibility that the original equation might have more roots or something? For example, when we multiplied both sides by x, we might have introduced extraneous roots? Wait, but we already checked that x = 0 is not a solution, so the transformed equation should have exactly the same roots as the original equation.

But let me check with an example. Suppose there is a root x = k of the original equation. Then, multiplying by x gives 1 + 5k³ = 6k² - 24k, which is equivalent to 5k³ - 6k² + 24k + 1 = 0. Therefore, k is a root of the cubic equation. Conversely, if x is a root of the cubic equation, then 5x³ - 6x² + 24x + 1 = 0, which implies 1 + 5x³ = 6x² - 24x, so dividing both sides by x (since x ≠ 0) gives 1/x + 5x² = 6x - 24, which is the original equation. Therefore, the roots of the cubic equation are exactly the roots of the original equation. Therefore, there are no extraneous roots introduced.

Therefore, the sum of the roots is indeed 6/5.

But wait, the problem says "a, b, and c are the roots". However, a cubic equation can have up to three roots (real or complex). But the original equation is 1/x + 5x² = 6x - 24. Depending on the context, maybe the problem assumes that a, b, c are real roots? But even so, Vieta's formula applies regardless of whether the roots are real or complex.

But maybe I need to check if the cubic equation actually has three real roots? Let me think.

But the problem doesn't specify, so I think we can assume that a, b, c are all roots (real or complex) of the equation, and Vieta's formula still applies. Therefore, the answer is 6/5.

But let me check if I did the algebra correctly again. Let me re-derive the equation step by step.

Original equation: 1/x + 5x² = 6x - 24.

Multiply both sides by x: 1 + 5x³ = 6x² - 24x.

Bring all terms to left: 5x³ - 6x² + 24x + 1 = 0.

Yes, that's correct.

Alternatively, maybe I can check with substituting a value. Let me try to see if there are any obvious roots. For example, let me try x = -1:

Left side of original equation: 1/(-1) + 5*(-1)^2 = -1 + 5(1) = -1 + 5 = 4.

Right side: 6*(-1) -24 = -6 -24 = -30. Not equal.

x = 1: Left side: 1/1 + 5(1)^2 = 1 + 5 = 6. Right side: 6(1) -24 = 6 -24 = -18. Not equal.

x = 2: Left side: 1/2 + 5*(4) = 0.5 + 20 = 20.5. Right side: 6*2 -24 = 12 -24 = -12. Not equal.

x = -2: Left side: 1/(-2) + 5*(4) = -0.5 + 20 = 19.5. Right side: 6*(-2) -24 = -12 -24 = -36. Not equal.

x = 1/2: Left side: 1/(1/2) + 5*(1/2)^2 = 2 + 5*(1/4) = 2 + 1.25 = 3.25. Right side: 6*(1/2) -24 = 3 -24 = -21. Not equal.

Hmm, maybe there are no rational roots? Let me check using Rational Root Theorem on the cubic equation 5x³ - 6x² + 24x + 1 = 0.

Possible rational roots are ±1, ±1/5. Let me test x = -1/5:

5*(-1/5)^3 -6*(-1/5)^2 +24*(-1/5) +1 = 5*(-1/125) -6*(1/25) -24/5 +1 = (-1/25) - 6/25 - 24/5 +1 = (-7/25) - 24/5 +1. Converting to 25 denominator: -7/25 - 120/25 + 25/25 = (-7 -120 +25)/25 = (-102)/25 ≠ 0.

x = 1/5:

5*(1/5)^3 -6*(1/5)^2 +24*(1/5) +1 = 5*(1/125) -6*(1/25) +24/5 +1 = 1/25 - 6/25 + 24/5 +1 = (-5/25) + 24/5 +1 = -1/5 + 24/5 +1 = (23/5) +1 = 28/5 ≠ 0.

x = 1: 5 -6 +24 +1 = 24 ≠ 0.

x = -1: -5 -6 -24 +1 = -34 ≠ 0.

So no rational roots. Therefore, maybe the cubic doesn't factor nicely, but that's okay. Vieta's formula still applies regardless.

Therefore, unless there's some mistake in my reasoning, the answer should be 6/5.

But let me check once more. Maybe I made a mistake in the sign when moving terms. Let me go back:

Original equation after multiplying by x: 1 + 5x³ = 6x² -24x.

So bringing all terms to the left: 1 + 5x³ -6x² +24x = 0. Which is 5x³ -6x² +24x +1 = 0. Yes, that's correct.

Alternatively, if I write it as 5x³ -6x² +24x +1 = 0.

So Vieta's formula says sum of roots is -(-6)/5 = 6/5.

Therefore, I think that is correct.

But maybe the problem is expecting integer or fractional answer? 6/5 is 1.2, which is a fraction.

Alternatively, maybe I made a mistake in the original equation transformation? Let me check again:

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Yes. Then moving all terms to left: 5x³ -6x² +24x +1 = 0.

Yes.

Alternatively, maybe the problem is written differently? Let me check the original problem again:

"Compute a+b+c, given that a, b, and c are the roots of 1/x + 5x² = 6x - 24."

Yes, that's what I used.

Alternatively, maybe there's a typo in my calculation? Let me check once again:

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Yes. Then subtract 6x² and add 24x to both sides: 1 + 5x³ -6x² +24x = 0.

Yes. So 5x³ -6x² +24x +1 = 0.

Therefore, the coefficients are correct.

Therefore, unless there's some other trick here, I think the answer is 6/5.

But let me think again. Maybe the original equation is supposed to be 1/x + 5x² = 6x - 24, but perhaps there's a different interpretation? For example, maybe the equation is supposed to be 1/(x + 5x²) = 6x -24? But that would be a different equation. However, the original problem is written as 1/x + 5x² = 6x -24, so I think the correct interpretation is as written.

Alternatively, maybe the equation is supposed to be 1/(x + 5x²) = 6x -24? But that would be a different equation. However, given the original problem is written as \frac{1}{x} + 5x^2 = 6x - 24, so it's definitely 1/x + 5x² equals that.

Therefore, I think my approach is correct.

Alternatively, maybe the problem is expecting me to consider that multiplying by x could introduce extraneous roots? But as discussed earlier, since x ≠ 0, and we checked that x=0 is not a solution, so all roots of the cubic equation are valid roots of the original equation.

Therefore, I think the answer is 6/5.

But just to be thorough, let me check if there's another way to approach this problem. For example, maybe using substitution or something else.

Alternatively, suppose I let y = x, then the equation is 1/y + 5y² = 6y -24. But that's the same as before.

Alternatively, maybe rearrange the original equation:

1/x = 6x -24 -5x².

Then, 1/x = -5x² +6x -24.

Then, cross multiplying: 1 = (-5x² +6x -24)x = -5x³ +6x² -24x.

Then, bringing all terms to left: 1 +5x³ -6x² +24x = 0, which is same as before: 5x³ -6x² +24x +1 = 0.

Same equation. Therefore, same result.

Therefore, I think my answer is correct.

Therefore, the sum a + b + c is 6/5, so the answer is \boxed{\dfrac{6}{5}}.

**Final Answer**
\boxed{\dfrac{6}{5}}

To solve the problem, we are given the equation:

$$
\frac{1}{x} + 5x^2 = 6x - 24
$$

and asked to compute $ a + b + c $, where $ a, b, c $ are the roots of this equation.

---

### Step 1: Eliminate the Fraction

The equation contains a term $ \frac{1}{x} $, which suggests that the equation is not a polynomial. To make it a polynomial, we multiply both sides of the equation by $ x $, assuming $ x \ne 0 $ (since $ x = 0 $ would make the original equation undefined):

$$
x \cdot \left( \frac{1}{x} + 5x^2 \right) = x \cdot (6x - 24)
$$

Simplifying both sides:

$$
1 + 5x^3 = 6x^2 - 24x
$$

---

### Step 2: Rearrange into a Polynomial Equation

Bring all terms to one side to form a standard polynomial equation:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

This is a cubic polynomial in $ x $, and we are told that $ a, b, c $ are the roots of the original equation. Since we have transformed the original equation into a cubic polynomial, and verified that $ x = 0 $ is not a solution, the roots of the cubic equation are exactly the roots of the original equation.

---

### Step 3: Apply Vieta's Formula

For a cubic polynomial of the form:

$$
Ax^3 + Bx^2 + Cx + D = 0
$$

the sum of the roots $ a + b + c $ is given by:

$$
a + b + c = -\frac{B}{A}
$$

In our case, the polynomial is:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

So, $ A = 5 $, $ B = -6 $, and thus:

$$
a + b + c = -\frac{-6}{5} = \frac{6}{5}
$$

---

### Final Answer

$$
\boxed{\dfrac{6}{5}}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of a + b + c, where a, b, and c are the roots of the equation given: 1/x + 5x² = 6x - 24. Hmm, let me think about how to approach this.

First, I remember that when dealing with equations involving roots, Vieta's formulas are usually helpful. Vieta's relates the coefficients of a polynomial to sums and products of its roots. But wait, the equation here isn't a polynomial equation yet. It has a term with 1/x, which complicates things. So maybe I need to manipulate the equation to convert it into a polynomial equation first?

Let me write down the original equation again:

1/x + 5x² = 6x - 24.

Okay, so to eliminate the fraction, I can multiply both sides of the equation by x. That should get rid of the denominator. Let me try that:

Multiplying both sides by x:

x*(1/x) + x*(5x²) = x*(6x) - x*(24)

Simplifying each term:

1 + 5x³ = 6x² - 24x

Now, let me bring all terms to one side to set the equation to zero:

5x³ - 6x² + 24x + 1 = 0

Wait, let me check that again. Starting from:

Left side after multiplying by x: 1 + 5x³

Right side after multiplying by x: 6x² - 24x

So bringing everything to the left side:

1 + 5x³ - 6x² + 24x = 0

Which can be written as:

5x³ - 6x² + 24x + 1 = 0

Wait, hold on. Let me check the signs again. If I subtract the right side from both sides:

Left side - Right side = 0:

(1 + 5x³) - (6x² - 24x) = 0

Which is 1 + 5x³ - 6x² + 24x = 0. Yes, that's correct. So rearranged:

5x³ - 6x² + 24x + 1 = 0

So now we have a cubic equation: 5x³ - 6x² + 24x + 1 = 0.

But wait, the original equation had 1/x, so when we multiplied by x, we assumed that x ≠ 0. Therefore, we need to check if x = 0 is a solution. Plugging x = 0 into the original equation: 1/0 is undefined, so x = 0 is not a solution. Therefore, our transformed equation 5x³ - 6x² + 24x + 1 = 0 should have all the roots of the original equation.

But the problem states that a, b, c are the roots of the original equation. However, the transformed equation is a cubic, so it has three roots. Therefore, a, b, c are the roots of this cubic equation. Therefore, we can apply Vieta's formula here.

Vieta's formula for a cubic equation ax³ + bx² + cx + d = 0 states that the sum of the roots is -b/a.

So in our case, the cubic equation is 5x³ - 6x² + 24x + 1 = 0. Therefore, the coefficients are:

a_cubic = 5 (coefficient of x³)

b_cubic = -6 (coefficient of x²)

c_cubic = 24 (coefficient of x)

d_cubic = 1 (constant term)

Therefore, the sum of the roots a + b + c = -b_cubic / a_cubic = -(-6)/5 = 6/5.

Wait, so is that the answer? 6/5? But let me make sure I didn't make any mistakes in transforming the original equation.

Original equation: 1/x + 5x² = 6x - 24.

Multiply both sides by x: 1 + 5x³ = 6x² - 24x.

Bring all terms to left: 5x³ - 6x² + 24x + 1 = 0. Yes, that seems correct.

Therefore, the cubic equation is correct, and Vieta's formula gives the sum of roots as 6/5. Therefore, a + b + c = 6/5.

But wait, hold on. Let me check again. Maybe I made a mistake in the sign when applying Vieta's formula. Let me recall Vieta's formula for cubic equations.

For a general cubic equation: ax³ + bx² + cx + d = 0, the sum of roots is -b/a.

In our case, the equation is 5x³ - 6x² + 24x + 1 = 0. Therefore, the coefficient of x² is -6, so the sum of roots is -(-6)/5 = 6/5. Yes, that's correct.

But let me think again. Suppose I have a cubic equation with roots a, b, c. Then:

(x - a)(x - b)(x - c) = x³ - (a + b + c)x² + (ab + ac + bc)x - abc.

But our equation is 5x³ - 6x² + 24x + 1 = 0. To match the standard form, we can divide both sides by 5:

x³ - (6/5)x² + (24/5)x + 1/5 = 0.

Comparing with the standard cubic equation x³ - (a + b + c)x² + (ab + ac + bc)x - abc = 0, we can see that:

-(a + b + c) = -6/5 => a + b + c = 6/5.

Yes, that's correct. Therefore, the sum of the roots is indeed 6/5.

But wait, the problem says "Compute a + b + c, given that a, b, and c are the roots of [equation]." So according to this, the answer is 6/5.

But let me check if there's any possibility that the original equation might have more roots or something? For example, when we multiplied both sides by x, we might have introduced extraneous roots? Wait, but we already checked that x = 0 is not a solution, so the transformed equation should have exactly the same roots as the original equation.

But let me check with an example. Suppose there is a root x = k of the original equation. Then, multiplying by x gives 1 + 5k³ = 6k² - 24k, which is equivalent to 5k³ - 6k² + 24k + 1 = 0. Therefore, k is a root of the cubic equation. Conversely, if x is a root of the cubic equation, then 5x³ - 6x² + 24x + 1 = 0, which implies 1 + 5x³ = 6x² - 24x, so dividing both sides by x (since x ≠ 0) gives 1/x + 5x² = 6x - 24, which is the original equation. Therefore, the roots of the cubic equation are exactly the roots of the original equation. Therefore, there are no extraneous roots introduced.

Therefore, the sum of the roots is indeed 6/5.

But wait, the problem says "a, b, and c are the roots". However, a cubic equation can have up to three roots (real or complex). But the original equation is 1/x + 5x² = 6x - 24. Depending on the context, maybe the problem assumes that a, b, c are real roots? But even so, Vieta's formula applies regardless of whether the roots are real or complex.

But maybe I need to check if the cubic equation actually has three real roots? Let me think.

But the problem doesn't specify, so I think we can assume that a, b, c are all roots (real or complex) of the equation, and Vieta's formula still applies. Therefore, the answer is 6/5.

But let me check if I did the algebra correctly again. Let me re-derive the equation step by step.

Original equation: 1/x + 5x² = 6x - 24.

Multiply both sides by x: 1 + 5x³ = 6x² - 24x.

Bring all terms to left: 5x³ - 6x² + 24x + 1 = 0.

Yes, that's correct.

Alternatively, maybe I can check with substituting a value. Let me try to see if there are any obvious roots. For example, let me try x = -1:

Left side of original equation: 1/(-1) + 5*(-1)^2 = -1 + 5(1) = -1 + 5 = 4.

Right side: 6*(-1) -24 = -6 -24 = -30. Not equal.

x = 1: Left side: 1/1 + 5(1)^2 = 1 + 5 = 6. Right side: 6(1) -24 = 6 -24 = -18. Not equal.

x = 2: Left side: 1/2 + 5*(4) = 0.5 + 20 = 20.5. Right side: 6*2 -24 = 12 -24 = -12. Not equal.

x = -2: Left side: 1/(-2) + 5*(4) = -0.5 + 20 = 19.5. Right side: 6*(-2) -24 = -12 -24 = -36. Not equal.

x = 1/2: Left side: 1/(1/2) + 5*(1/2)^2 = 2 + 5*(1/4) = 2 + 1.25 = 3.25. Right side: 6*(1/2) -24 = 3 -24 = -21. Not equal.

Hmm, maybe there are no rational roots? Let me check using Rational Root Theorem on the cubic equation 5x³ - 6x² + 24x + 1 = 0.

Possible rational roots are ±1, ±1/5. Let me test x = -1/5:

5*(-1/5)^3 -6*(-1/5)^2 +24*(-1/5) +1 = 5*(-1/125) -6*(1/25) -24/5 +1 = (-1/25) - 6/25 - 24/5 +1 = (-7/25) - 24/5 +1. Converting to 25 denominator: -7/25 - 120/25 + 25/25 = (-7 -120 +25)/25 = (-102)/25 ≠ 0.

x = 1/5:

5*(1/5)^3 -6*(1/5)^2 +24*(1/5) +1 = 5*(1/125) -6*(1/25) +24/5 +1 = 1/25 - 6/25 + 24/5 +1 = (-5/25) + 24/5 +1 = -1/5 + 24/5 +1 = (23/5) +1 = 28/5 ≠ 0.

x = 1: 5 -6 +24 +1 = 24 ≠ 0.

x = -1: -5 -6 -24 +1 = -34 ≠ 0.

So no rational roots. Therefore, maybe the cubic doesn't factor nicely, but that's okay. Vieta's formula still applies regardless.

Therefore, unless there's some mistake in my reasoning, the answer should be 6/5.

But let me check once more. Maybe I made a mistake in the sign when moving terms. Let me go back:

Original equation after multiplying by x: 1 + 5x³ = 6x² -24x.

So bringing all terms to the left: 1 + 5x³ -6x² +24x = 0. Which is 5x³ -6x² +24x +1 = 0. Yes, that's correct.

Alternatively, if I write it as 5x³ -6x² +24x +1 = 0.

So Vieta's formula says sum of roots is -(-6)/5 = 6/5.

Therefore, I think that is correct.

But maybe the problem is expecting integer or fractional answer? 6/5 is 1.2, which is a fraction.

Alternatively, maybe I made a mistake in the original equation transformation? Let me check again:

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Yes. Then moving all terms to left: 5x³ -6x² +24x +1 = 0.

Yes.

Alternatively, maybe the problem is written differently? Let me check the original problem again:

"Compute a+b+c, given that a, b, and c are the roots of 1/x + 5x² = 6x - 24."

Yes, that's what I used.

Alternatively, maybe there's a typo in my calculation? Let me check once again:

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Yes. Then subtract 6x² and add 24x to both sides: 1 + 5x³ -6x² +24x = 0.

Yes. So 5x³ -6x² +24x +1 = 0.

Therefore, the coefficients are correct.

Therefore, unless there's some other trick here, I think the answer is 6/5.

But let me think again. Maybe the original equation is supposed to be 1/x + 5x² = 6x - 24, but perhaps there's a different interpretation? For example, maybe the equation is supposed to be 1/(x + 5x²) = 6x -24? But that would be a different equation. However, the original problem is written as 1/x + 5x² = 6x -24, so I think the correct interpretation is as written.

Alternatively, maybe the equation is supposed to be 1/(x + 5x²) = 6x -24? But that would be a different equation. However, given the original problem is written as \frac{1}{x} + 5x^2 = 6x - 24, so it's definitely 1/x + 5x² equals that.

Therefore, I think my approach is correct.

Alternatively, maybe the problem is expecting me to consider that multiplying by x could introduce extraneous roots? But as discussed earlier, since x ≠ 0, and we checked that x=0 is not a solution, so all roots of the cubic equation are valid roots of the original equation.

Therefore, I think the answer is 6/5.

But just to be thorough, let me check if there's another way to approach this problem. For example, maybe using substitution or something else.

Alternatively, suppose I let y = x, then the equation is 1/y + 5y² = 6y -24. But that's the same as before.

Alternatively, maybe rearrange the original equation:

1/x = 6x -24 -5x².

Then, 1/x = -5x² +6x -24.

Then, cross multiplying: 1 = (-5x² +6x -24)x = -5x³ +6x² -24x.

Then, bringing all terms to left: 1 +5x³ -6x² +24x = 0, which is same as before: 5x³ -6x² +24x +1 = 0.

Same equation. Therefore, same result.

Therefore, I think my answer is correct.

Therefore, the sum a + b + c is 6/5, so the answer is \boxed{\dfrac{6}{5}}.

**Final Answer**
\boxed{\dfrac{6}{5}}

To solve the problem, we are given the equation:

$$
\frac{1}{x} + 5x^2 = 6x - 24
$$

and asked to compute $ a + b + c $, where $ a, b, c $ are the roots of this equation.

---

### Step 1: Eliminate the Fraction

The equation contains a term $ \frac{1}{x} $, which suggests that the equation is not a polynomial. To make it a polynomial, we multiply both sides of the equation by $ x $, assuming $ x \ne 0 $ (since $ x = 0 $ would make the original equation undefined):

$$
x \cdot \left( \frac{1}{x} + 5x^2 \right) = x \cdot (6x - 24)
$$

Simplifying both sides:

$$
1 + 5x^3 = 6x^2 - 24x
$$

---

### Step 2: Rearrange into a Polynomial Equation

Bring all terms to one side to form a standard polynomial equation:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

This is a cubic polynomial in $ x $, and we are told that $ a, b, c $ are the roots of the original equation. Since we have transformed the original equation into a cubic polynomial, and verified that $ x = 0 $ is not a solution, the roots of the cubic equation are exactly the roots of the original equation.

---

### Step 3: Apply Vieta's Formula

For a cubic polynomial of the form:

$$
Ax^3 + Bx^2 + Cx + D = 0
$$

the sum of the roots $ a + b + c $ is given by:

$$
a + b + c = -\frac{B}{A}
$$

In our case, the polynomial is:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

So, $ A = 5 $, $ B = -6 $, and thus:

$$
a + b + c = -\frac{-6}{5} = \frac{6}{5}
$$

---

### Final Answer

$$
\boxed{\dfrac{6}{5}}
$$

Solution: We want to apply Vieta's formulas, but the given equation is not a polynomial equation, because of the $\frac1x$ term. To turn this equation into an equivalent polynomial equation, we multiply both sides by $x$ and rearrange: \[\begin{aligned} 1+5x^3 &= 6x^2 - 24x \\ 5x^3 - 6x^2 + 24x + 1 &= 0 .\end{aligned}\]Now we can use Vieta: the sum of the roots is $a+b+c=\boxed{\frac65}.$


Explanation:
Okay, so I need to find two consecutive positive even integers whose product is 288, and then determine the greater one. Let me think about how to approach this.

First, let me recall that consecutive even integers are numbers like 2 and 4, 6 and 8, etc. They differ by 2. So if I let the first integer be some number, the next one would be that number plus 2.

Let me denote the smaller integer as x. Then the next consecutive even integer would be x + 2. Their product is given as 288. So, the equation would be:

x * (x + 2) = 288

Alright, so expanding that equation:

x² + 2x = 288

To solve this quadratic equation, I need to bring all terms to one side to set the equation equal to zero. Let me subtract 288 from both sides:

x² + 2x - 288 = 0

Now, I need to solve this quadratic equation. Let me check if I can factor it. To factor, I need two numbers that multiply to -288 and add up to 2. Hmm, let me think.

Since the product is negative, one of the numbers has to be positive and the other negative. Also, their sum is positive 2. So the positive number must be larger in magnitude than the negative one.

Let me list the factors of 288. Let me think... 288 divided by 2 is 144, 288 divided by 3 is 96, 288 divided by 4 is 72, 6 is 48, 8 is 36, 12 is 24, 16 is 18. So factors are 1, 2, 3, 4, 6, 8, 12, 16, 18, 24, 36, 48, 72, 96, 144, 288.

Looking for two numbers that multiply to -288 and add to 2. Let me try some combinations. Let me think of positive and negative pairs.

Suppose I take 18 and -16. Their product is -288? 18 * (-16) = -288. Yes! And 18 + (-16) = 2. Perfect! So that works.

Therefore, the quadratic factors as:

(x + 18)(x - 16) = 0

Wait, let me check that. If the factors are (x + 18)(x - 16), then expanding gives x² -16x + 18x - 288 = x² + 2x - 288. Yes, that's correct.

So setting each factor equal to zero:

x + 18 = 0 => x = -18

or

x - 16 = 0 => x = 16

But the problem states that the integers are positive, so x must be positive. Therefore, x = 16.

Therefore, the two integers are 16 and 16 + 2 = 18. The greater of the two is 18.

Wait, let me check if that's correct. 16 * 18 = 288? Let me compute 16*18. 16*10=160, 16*8=128, so 160+128=288. Yes, that's correct.

But let me just make sure I didn't make a mistake in factoring. Let me try another way. Maybe using the quadratic formula?

The quadratic equation is x² + 2x - 288 = 0. The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / 2a. Here, a = 1, b = 2, c = -288.

So discriminant D = b² - 4ac = (2)^2 - 4(1)(-288) = 4 + 1152 = 1156.

sqrt(1156) is 34, since 34*34 = 1156.

Therefore, x = [-2 ± 34]/2.

So two solutions:

x = (-2 + 34)/2 = 32/2 = 16

x = (-2 - 34)/2 = -36/2 = -18

Same results as before. So x is 16 or -18. Since we need positive integers, x = 16. Therefore, the numbers are 16 and 18. The greater one is 18. So the answer is 18.

But let me think again if there's another way to approach this problem. Maybe by letting the numbers be n and n+2, but maybe starting with a different variable? Probably not necessary, but just to confirm.

Alternatively, maybe approximate the square root of 288? Since the product of two numbers close to each other is 288. The square root of 288 is approx sqrt(289) is 17, so sqrt(288) is approx 16.97. So the numbers are around 17. But since they are even integers, the closest even numbers would be 16 and 18. Which is exactly what we found. So that's another way to check.

Alternatively, if I didn't remember the factors, maybe I could have tried dividing 288 by some even numbers to see if the result is the next even number.

For example, let's try dividing 288 by 16: 288 / 16 = 18. Which is the next even number. So that works. If I tried 14: 288 /14 is about 20.57, not integer. 12: 288 /12 = 24, which is 12 and 24. But 24 - 12 = 12, which is not consecutive even integers. So that's not. Similarly, 18: 288 /18 = 16, which is the same as before. So yeah, 16 and 18.

So all methods lead to the same answer. Therefore, I can be confident that the greater integer is 18.

**Final Answer**
The greater of the two integers is \boxed{18}.

To solve the problem, we are given that the product of two **consecutive positive even integers** is 288, and we are to find the **greater** of the two integers.

---

### Step 1: Define the Variables

Let the smaller even integer be $ x $. Then the next consecutive even integer is $ x + 2 $.

According to the problem:

$$
x(x + 2) = 288
$$

---

### Step 2: Expand and Rearrange the Equation

$$
x^2 + 2x = 288
$$

Bring all terms to one side to form a standard quadratic equation:

$$
x^2 + 2x - 288 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We can solve this quadratic equation using factoring or the quadratic formula. Let's try factoring first.

We need two numbers that multiply to $-288$ and add to $2$. These numbers are $18$ and $-16$:

$$
(x + 18)(x - 16) = 0
$$

Setting each factor equal to zero:

$$
x + 18 = 0 \Rightarrow x = -18 \\
x - 16 = 0 \Rightarrow x = 16
$$

Since the problem specifies **positive** integers, we discard $x = -18$ and take $x = 16$.

---

### Step 4: Find the Greater Integer

If the smaller integer is $16$, then the next consecutive even integer is:

$$
16 + 2 = 18
$$

---

### Step 5: Verify the Solution

Check the product:

$$
16 \times 18 = 288
$$

This confirms the solution is correct.

---

### Final Answer

$$
\boxed{18}
$$

[Assistant]
Explanation:
Okay, so I need to find two consecutive positive even integers whose product is 288, and then determine the greater one. Let me think about how to approach this.

First, let me recall that consecutive even integers are numbers like 2 and 4, 6 and 8, etc. They differ by 2. So if I let the first integer be some number, the next one would be that number plus 2.

Let me denote the smaller integer as x. Then the next consecutive even integer would be x + 2. Their product is given as 288. So, the equation would be:

x * (x + 2) = 288

Alright, so expanding that equation:

x² + 2x = 288

To solve this quadratic equation, I need to bring all terms to one side to set the equation equal to zero. Let me subtract 288 from both sides:

x² + 2x - 288 = 0

Now, I need to solve this quadratic equation. Let me check if I can factor it. To factor, I need two numbers that multiply to -288 and add up to 2. Hmm, let me think.

Since the product is negative, one of the numbers has to be positive and the other negative. Also, their sum is positive 2. So the positive number must be larger in magnitude than the negative one.

Let me list the factors of 288. Let me think... 288 divided by 2 is 144, 288 divided by 3 is 96, 288 divided by 4 is 72, 6 is 48, 8 is 36, 12 is 24, 16 is 18. So factors are 1, 2, 3, 4, 6, 8, 12, 16, 18, 24, 36, 48, 72, 96, 144, 288.

Looking for two numbers that multiply to -288 and add to 2. Let me try some combinations. Let me think of positive and negative pairs.

Suppose I take 18 and -16. Their product is -288? 18 * (-16) = -288. Yes! And 18 + (-16) = 2. Perfect! So that works.

Therefore, the quadratic factors as:

(x + 18)(x - 16) = 0

Wait, let me check that. If the factors are (x + 18)(x - 16), then expanding gives x² -16x + 18x - 288 = x² + 2x - 288. Yes, that's correct.

So setting each factor equal to zero:

x + 18 = 0 => x = -18

or

x - 16 = 0 => x = 16

But the problem states that the integers are positive, so x must be positive. Therefore, x = 16.

Therefore, the two integers are 16 and 16 + 2 = 18. The greater of the two is 18.

Wait, let me check if that's correct. 16 * 18 = 288? Let me compute 16*18. 16*10=160, 16*8=128, so 160+128=288. Yes, that's correct.

But let me just make sure I didn't make a mistake in factoring. Let me try another way. Maybe using the quadratic formula?

The quadratic equation is x² + 2x - 288 = 0. The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / 2a. Here, a = 1, b = 2, c = -288.

So discriminant D = b² - 4ac = (2)^2 - 4(1)(-288) = 4 + 1152 = 1156.

sqrt(1156) is 34, since 34*34 = 1156.

Therefore, x = [-2 ± 34]/2.

So two solutions:

x = (-2 + 34)/2 = 32/2 = 16

x = (-2 - 34)/2 = -36/2 = -18

Same results as before. So x is 16 or -18. Since we need positive integers, x = 16. Therefore, the numbers are 16 and 18. The greater one is 18. So the answer is 18.

But let me think again if there's another way to approach this problem. Maybe by letting the numbers be n and n+2, but maybe starting with a different variable? Probably not necessary, but just to confirm.

Alternatively, maybe approximate the square root of 288? Since the product of two numbers close to each other is 288. The square root of 288 is approx sqrt(289) is 17, so sqrt(288) is approx 16.97. So the numbers are around 17. But since they are even integers, the closest even numbers would be 16 and 18. Which is exactly what we found. So that's another way to check.

Alternatively, if I didn't remember the factors, maybe I could have tried dividing 288 by some even numbers to see if the result is the next even number.

For example, let's try dividing 288 by 16: 288 / 16 = 18. Which is the next even number. So that works. If I tried 14: 288 /14 is about 20.57, not integer. 12: 288 /12 = 24, which is 12 and 24. But 24 - 12 = 12, which is not consecutive even integers. So that's not. Similarly, 18: 288 /18 = 16, which is the same as before. So yeah, 16 and 18.

So all methods lead to the same answer. Therefore, I can be confident that the greater integer is 18.

**Final Answer**
The greater of the two integers is \boxed{18}.

To solve the problem, we are given that the product of two **consecutive positive even integers** is 288, and we are to find the **greater** of the two integers.

---

### Step 1: Define the Variables

Let the smaller even integer be $ x $. Then the next consecutive even integer is $ x + 2 $.

According to the problem:

$$
x(x + 2) = 288
$$

---

### Step 2: Expand and Rearrange the Equation

$$
x^2 + 2x = 288
$$

Bring all terms to one side to form a standard quadratic equation:

$$
x^2 + 2x - 288 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We can solve this quadratic equation using factoring or the quadratic formula. Let's try factoring first.

We need two numbers that multiply to $-288$ and add to $2$. These numbers are $18$ and $-16$:

$$
(x + 18)(x - 16) = 0
$$

Setting each factor equal to zero:

$$
x + 18 = 0 \Rightarrow x = -18 \\
x - 16 = 0 \Rightarrow x = 16
$$

Since the problem specifies **positive** integers, we discard $x = -18$ and take $x = 16$.

---

### Step 4: Find the Greater Integer

If the smaller integer is $16$, then the next consecutive even integer is:

$$
16 + 2 = 18
$$

---

### Step 5: Verify the Solution

Check the product:

$$
16 \times 18 = 288
$$

This confirms the solution is correct.

---

### Final Answer

$$
\boxed{18}
$$

Solution: First we find the prime factorization of 288 to be $2^5\cdot 3^2$, and we have to split these factors among two consecutive even integers. The 3 must be with at least one 2 for the integer to be even, meaning one of the factors must be a multiple of $6.$ After some playing around, we find that when one factor is 18, that leaves us with $2^4=16$. So, our two integers are 16 and 18, with the greater integer being $\boxed{18}$.


Explanation:
Okay, so I need to evaluate log base 2 of 64. Hmm, let me think. Logarithms can sometimes be tricky, but maybe I can remember what log base 2 means. It's asking the question: 2 raised to what power gives me 64? So, in other words, I need to find an exponent x such that 2^x = 64.

Alright, let me recall the powers of 2. Let me start from the beginning. 2^1 is 2, 2^2 is 4, 2^3 is 8, 2^4 is 16, 2^5 is 32, and then 2^6 should be 64. Wait, is that right? Let me check. 2^5 is 32, so multiplying by 2 again would be 32*2=64. Yes, so 2^6 equals 64. Therefore, if 2^6 = 64, then log base 2 of 64 is 6.

But let me make sure I didn't skip any steps or make a mistake. Maybe I can write out the powers step by step. Let's see:

- 2^1 = 2
- 2^2 = 2*2 = 4
- 2^3 = 4*2 = 8
- 2^4 = 8*2 = 16
- 2^5 = 16*2 = 32
- 2^6 = 32*2 = 64

Yep, that's correct. So each time the exponent increases by 1, the result doubles. So after 6 doublings, starting from 1 (which is 2^0), we get to 64. Wait, actually, 2^0 is 1, right? So maybe I should check if 2^6 is indeed 64. Let me calculate 2^6:

2*2=4, 4*2=8, 8*2=16, 16*2=32, 32*2=64. Yep, that's five multiplications, which would be 2^5? Wait, no. Wait, starting from 2^1, each multiplication is adding 1 to the exponent. So 2^1 is 2, then 2^2 is 4, so 2^6 would be 2 multiplied by itself 6 times. Let me count:

First multiplication: 2^1 = 2

Second: 2^2 = 4

Third: 2^3 = 8

Fourth: 2^4 = 16

Fifth: 2^5 = 32

Sixth: 2^6 = 64

Yes, so that's six multiplications, which gives 2^6 = 64. Therefore, log base 2 of 64 is 6.

Alternatively, maybe I can use logarithm properties to verify. For example, I know that log_b(a) = ln(a)/ln(b), so maybe I can compute it using natural logarithms? Let me try that.

So log_2(64) = ln(64)/ln(2). Let me calculate ln(64). But I need to remember that 64 is 2^6, so ln(64) is ln(2^6) which is 6*ln(2). Therefore, ln(64)/ln(2) = 6*ln(2)/ln(2) = 6. So that confirms it.

Alternatively, if I didn't remember that 64 is 2^6, maybe I can factor 64 into powers of 2. Let me try that. 64 divided by 2 is 32, divided by 2 is 16, divided by 2 is 8, divided by 2 is 4, divided by 2 is 2, divided by 2 is 1. So that's 6 divisions by 2, which means 2^6 = 64. Therefore, log base 2 of 64 is 6.

Another way: maybe using exponents. Let me think. If I have log_2(64) = x, then 2^x = 64. So, solving for x. Since 64 is a power of 2, this should be straightforward. But if I didn't know that, maybe I can express 64 as a power of 2. Let me think. 2^6 is 64, as we saw.

Alternatively, maybe I can use prime factorization. 64 is 2*2*2*2*2*2, which is 2^6. Therefore, log base 2 of 2^6 is 6.

Wait, so all these methods lead me to the same answer, 6. So I think that's solid.

But just to make sure, let me check with another logarithm base. For example, if I convert log base 2 of 64 to log base 10. So log_2(64) = log10(64)/log10(2). Let me approximate these values.

I know that log10(2) is approximately 0.3010, and log10(64)... Well, 10^1.8 is approximately 63.1, since 10^1.8 ≈ 10^(1 + 0.8) = 10*10^0.8. And 10^0.8 is approximately 6.3 (since 10^0.778 ≈ 6, because log10(6) ≈ 0.778). Therefore, 10^1.8 ≈ 10*6.3 = 63. So log10(64) is approximately 1.806, since 10^1.806 ≈ 64.

Therefore, log10(64)/log10(2) ≈ 1.806 / 0.3010 ≈ 6. So that gives me approximately 6, which matches the exact value.

Alternatively, if I use natural logs, ln(64) is ln(2^6) = 6 ln 2, so ln(64)/ln(2) = 6.

So all methods confirm that log base 2 of 64 is 6.

Is there any chance I could have made a mistake here? Let me think. Maybe if I confused the base or the argument? For example, if I thought log base 64 of 2, that would be 1/6, but the question is log base 2 of 64, which is different.

Alternatively, if I miscalculated the exponents? Let me check again:

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

Yes, that's correct. So 2^6 is 64. Therefore, log_2(64) = 6.

Another way: think about binary numbers. 64 in binary is 1000000. The number of digits minus one gives the exponent. Wait, in binary, 64 is 1 followed by six zeros, so that's 7 digits. But the exponent is 6. So that's consistent. Because 2^6 is 64, which is represented as 1000000 in binary, which has 7 digits. So the exponent is one less than the number of digits in binary. So that's another way to think about it.

Alternatively, if I didn't remember all these properties, maybe I can use logarithm change of base formula. For example, log_2(64) = log_10(64)/log_10(2). If I don't remember the exact values, but I know that 2^6 = 64, then log_10(64) = log_10(2^6) = 6 log_10(2). Therefore, dividing by log_10(2) gives 6.

So regardless of the method, it seems that the answer is 6.

I think I've verified it multiple ways: direct computation of powers of 2, using logarithm properties, prime factorization, change of base formula, and even thinking about binary representations. All of them lead to the conclusion that log base 2 of 64 is 6.

Therefore, I can confidently say that the answer is 6.

**Final Answer**
The value of $\log_2 64$ is \boxed{6}.

To evaluate the expression $\log_2 64$, we interpret it as asking: **"To what power must 2 be raised to obtain 64?"**

---

### Step 1: Understand the Definition of a Logarithm

The logarithm $\log_b a$ is the exponent $x$ such that:

$$
b^x = a
$$

In this case, we are looking for:

$$
\log_2 64 = x \quad \text{such that} \quad 2^x = 64
$$

---

### Step 2: Compute Powers of 2

Let’s compute successive powers of 2 to find the exponent that gives 64:

$$
\begin{align*}
2^1 &= 2 \\
2^2 &= 4 \\
2^3 &= 8 \\
2^4 &= 16 \\
2^5 &= 32 \\
2^6 &= 64
\end{align*}
$$

We see that $2^6 = 64$, so the exponent $x$ is 6.

---

### Step 3: Confirm Using Logarithmic Properties

We can also use the **change of base formula**:

$$
\log_2 64 = \frac{\log 64}{\log 2}
$$

Since $64 = 2^6$, we can write:

$$
\log 64 = \log(2^6) = 6 \log 2
$$

So:

$$
\log_2 64 = \frac{6 \log 2}{\log 2} = 6
$$

This confirms the result.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to evaluate log base 2 of 64. Hmm, let me think. Logarithms can sometimes be tricky, but maybe I can remember what log base 2 means. It's asking the question: 2 raised to what power gives me 64? So, in other words, I need to find an exponent x such that 2^x = 64.

Alright, let me recall the powers of 2. Let me start from the beginning. 2^1 is 2, 2^2 is 4, 2^3 is 8, 2^4 is 16, 2^5 is 32, and then 2^6 should be 64. Wait, is that right? Let me check. 2^5 is 32, so multiplying by 2 again would be 32*2=64. Yes, so 2^6 equals 64. Therefore, if 2^6 = 64, then log base 2 of 64 is 6.

But let me make sure I didn't skip any steps or make a mistake. Maybe I can write out the powers step by step. Let's see:

- 2^1 = 2
- 2^2 = 2*2 = 4
- 2^3 = 4*2 = 8
- 2^4 = 8*2 = 16
- 2^5 = 16*2 = 32
- 2^6 = 32*2 = 64

Yep, that's correct. So each time the exponent increases by 1, the result doubles. So after 6 doublings, starting from 1 (which is 2^0), we get to 64. Wait, actually, 2^0 is 1, right? So maybe I should check if 2^6 is indeed 64. Let me calculate 2^6:

2*2=4, 4*2=8, 8*2=16, 16*2=32, 32*2=64. Yep, that's five multiplications, which would be 2^5? Wait, no. Wait, starting from 2^1, each multiplication is adding 1 to the exponent. So 2^1 is 2, then 2^2 is 4, so 2^6 would be 2 multiplied by itself 6 times. Let me count:

First multiplication: 2^1 = 2

Second: 2^2 = 4

Third: 2^3 = 8

Fourth: 2^4 = 16

Fifth: 2^5 = 32

Sixth: 2^6 = 64

Yes, so that's six multiplications, which gives 2^6 = 64. Therefore, log base 2 of 64 is 6.

Alternatively, maybe I can use logarithm properties to verify. For example, I know that log_b(a) = ln(a)/ln(b), so maybe I can compute it using natural logarithms? Let me try that.

So log_2(64) = ln(64)/ln(2). Let me calculate ln(64). But I need to remember that 64 is 2^6, so ln(64) is ln(2^6) which is 6*ln(2). Therefore, ln(64)/ln(2) = 6*ln(2)/ln(2) = 6. So that confirms it.

Alternatively, if I didn't remember that 64 is 2^6, maybe I can factor 64 into powers of 2. Let me try that. 64 divided by 2 is 32, divided by 2 is 16, divided by 2 is 8, divided by 2 is 4, divided by 2 is 2, divided by 2 is 1. So that's 6 divisions by 2, which means 2^6 = 64. Therefore, log base 2 of 64 is 6.

Another way: maybe using exponents. Let me think. If I have log_2(64) = x, then 2^x = 64. So, solving for x. Since 64 is a power of 2, this should be straightforward. But if I didn't know that, maybe I can express 64 as a power of 2. Let me think. 2^6 is 64, as we saw.

Alternatively, maybe I can use prime factorization. 64 is 2*2*2*2*2*2, which is 2^6. Therefore, log base 2 of 2^6 is 6.

Wait, so all these methods lead me to the same answer, 6. So I think that's solid.

But just to make sure, let me check with another logarithm base. For example, if I convert log base 2 of 64 to log base 10. So log_2(64) = log10(64)/log10(2). Let me approximate these values.

I know that log10(2) is approximately 0.3010, and log10(64)... Well, 10^1.8 is approximately 63.1, since 10^1.8 ≈ 10^(1 + 0.8) = 10*10^0.8. And 10^0.8 is approximately 6.3 (since 10^0.778 ≈ 6, because log10(6) ≈ 0.778). Therefore, 10^1.8 ≈ 10*6.3 = 63. So log10(64) is approximately 1.806, since 10^1.806 ≈ 64.

Therefore, log10(64)/log10(2) ≈ 1.806 / 0.3010 ≈ 6. So that gives me approximately 6, which matches the exact value.

Alternatively, if I use natural logs, ln(64) is ln(2^6) = 6 ln 2, so ln(64)/ln(2) = 6.

So all methods confirm that log base 2 of 64 is 6.

Is there any chance I could have made a mistake here? Let me think. Maybe if I confused the base or the argument? For example, if I thought log base 64 of 2, that would be 1/6, but the question is log base 2 of 64, which is different.

Alternatively, if I miscalculated the exponents? Let me check again:

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

Yes, that's correct. So 2^6 is 64. Therefore, log_2(64) = 6.

Another way: think about binary numbers. 64 in binary is 1000000. The number of digits minus one gives the exponent. Wait, in binary, 64 is 1 followed by six zeros, so that's 7 digits. But the exponent is 6. So that's consistent. Because 2^6 is 64, which is represented as 1000000 in binary, which has 7 digits. So the exponent is one less than the number of digits in binary. So that's another way to think about it.

Alternatively, if I didn't remember all these properties, maybe I can use logarithm change of base formula. For example, log_2(64) = log_10(64)/log_10(2). If I don't remember the exact values, but I know that 2^6 = 64, then log_10(64) = log_10(2^6) = 6 log_10(2). Therefore, dividing by log_10(2) gives 6.

So regardless of the method, it seems that the answer is 6.

I think I've verified it multiple ways: direct computation of powers of 2, using logarithm properties, prime factorization, change of base formula, and even thinking about binary representations. All of them lead to the conclusion that log base 2 of 64 is 6.

Therefore, I can confidently say that the answer is 6.

**Final Answer**
The value of $\log_2 64$ is \boxed{6}.

To evaluate the expression $\log_2 64$, we interpret it as asking: **"To what power must 2 be raised to obtain 64?"**

---

### Step 1: Understand the Definition of a Logarithm

The logarithm $\log_b a$ is the exponent $x$ such that:

$$
b^x = a
$$

In this case, we are looking for:

$$
\log_2 64 = x \quad \text{such that} \quad 2^x = 64
$$

---

### Step 2: Compute Powers of 2

Let’s compute successive powers of 2 to find the exponent that gives 64:

$$
\begin{align*}
2^1 &= 2 \\
2^2 &= 4 \\
2^3 &= 8 \\
2^4 &= 16 \\
2^5 &= 32 \\
2^6 &= 64
\end{align*}
$$

We see that $2^6 = 64$, so the exponent $x$ is 6.

---

### Step 3: Confirm Using Logarithmic Properties

We can also use the **change of base formula**:

$$
\log_2 64 = \frac{\log 64}{\log 2}
$$

Since $64 = 2^6$, we can write:

$$
\log 64 = \log(2^6) = 6 \log 2
$$

So:

$$
\log_2 64 = \frac{6 \log 2}{\log 2} = 6
$$

This confirms the result.

---

### Final Answer

$$
\boxed{6}
$$

Solution: We have $2^6=64$, so $\log_2 64 = \boxed{6}$.


Explanation:
Okay, so I need to simplify the expression tan 100° + 4 sin 100°. Hmm, let me think about how to approach this. First, I remember that tan θ is sin θ over cos θ, so maybe I can rewrite the expression in terms of sine and cosine. Let me try that.

So, tan 100° is sin 100° divided by cos 100°, right? So the expression becomes:

sin 100° / cos 100° + 4 sin 100°

Hmm, maybe I can factor out sin 100°? Let me see:

sin 100° [1 / cos 100° + 4]

But I don't know if that helps. Maybe I need to find a common denominator or combine the terms somehow. Let me write it as:

[sin 100° + 4 sin 100° cos 100°] / cos 100°

Wait, that's combining the two terms over a common denominator. Let me check:

Original expression: sin 100° / cos 100° + 4 sin 100°

To combine them, the second term needs to have denominator cos 100°, so multiply numerator and denominator by cos 100°:

[sin 100° + 4 sin 100° cos 100°] / cos 100°

Yes, that's correct. Now, maybe I can factor sin 100° from the numerator:

sin 100° [1 + 4 cos 100°] / cos 100°

Hmm, not sure if that helps. Maybe I need to use some trigonometric identities here. Let me recall some identities. For example, maybe double-angle formulas or something else?

Alternatively, maybe I can convert degrees to radians? Wait, but I don't know if that's necessary. Let me think about the angles. 100° is in the second quadrant, so sin 100° is positive, cos 100° is negative. Also, tan 100° is negative because tan is sin/cos, and cos is negative there. So tan 100° is negative, and 4 sin 100° is positive. So the whole expression is negative plus positive. Not sure if that helps.

Alternatively, maybe I can express tan 100° as tan(180° - 80°) which is -tan 80°, since tan(180° - θ) = -tan θ. Similarly, sin 100° is sin(80°), because sin(180° - θ) = sin θ. So maybe rewriting the expression in terms of 80°?

Let me try that. So tan 100° = -tan 80°, and sin 100° = sin 80°. Therefore, the original expression becomes:

- tan 80° + 4 sin 80°

Hmm, maybe that's easier? Let me write that:

- tan 80° + 4 sin 80°

Since tan 80° is sin 80° / cos 80°, so substituting that in:

- [sin 80° / cos 80°] + 4 sin 80°

Again, combining terms:

sin 80° [ -1 / cos 80° + 4 ]

Which is similar to before. Maybe I can combine the terms:

sin 80° [ ( -1 + 4 cos 80° ) / cos 80° ]

So that's [ sin 80° ( -1 + 4 cos 80° ) ] / cos 80°

Hmm, not sure. Maybe I need to compute numerical values? Wait, but the problem says to simplify, not necessarily compute numerically. Maybe there's an exact value?

Alternatively, perhaps using some trigonometric identities. Let me think. Let me recall that 4 sin θ can be related to some multiple-angle formulas?

Alternatively, maybe express everything in terms of sin and cos and then combine. Let me try again:

Original expression: tan 100° + 4 sin 100° = -tan 80° + 4 sin 80°

Let me write tan 80° as sin 80° / cos 80°, so:

- sin 80° / cos 80° + 4 sin 80° = sin 80° [ -1 / cos 80° + 4 ]

Let me compute the term in the brackets:

-1 / cos 80° + 4 = 4 - sec 80°

Not sure. Maybe I can write 4 as 4 cos 80° / cos 80°, so:

[4 cos 80° - 1] / cos 80°

Therefore, the entire expression becomes:

sin 80° [4 cos 80° - 1] / cos 80°

Which is [ sin 80° (4 cos 80° - 1) ] / cos 80°

Hmm, maybe I can compute 4 cos 80° - 1? Let me think. Let me compute 4 cos 80° - 1 numerically? Wait, but maybe there's an exact value?

Alternatively, maybe I can use some identity. Let me think. Let me recall that 80° is related to 60° and 20°, but not sure. Alternatively, maybe using product-to-sum formulas?

Alternatively, think about writing 4 cos 80° as something. Wait, 4 cos 80° is 4 cos(60° + 20°). Maybe expand that?

Let me try expanding cos(60° + 20°):

cos 60° cos 20° - sin 60° sin 20°

Which is (0.5) cos 20° - (√3/2) sin 20°

So 4 cos 80° = 4 [ 0.5 cos 20° - (√3/2) sin 20° ] = 2 cos 20° - 2√3 sin 20°

Therefore, 4 cos 80° - 1 = 2 cos 20° - 2√3 sin 20° - 1

Hmm, not sure if that helps. Maybe not. Let me think of another approach.

Alternatively, maybe consider writing the entire expression as a single trigonometric function. Let me think. Suppose I have:

Let me denote θ = 80°, so the expression is -tan θ + 4 sin θ. Let me write that:

- tan θ + 4 sin θ = - (sin θ / cos θ) + 4 sin θ = sin θ ( -1 / cos θ + 4 )

Which is the same as before. Maybe I can write this as:

sin θ (4 - sec θ )

Alternatively, maybe think of 4 as 4 cos θ / cos θ, so:

sin θ (4 cos θ - 1 ) / cos θ

So, [ sin θ (4 cos θ - 1 ) ] / cos θ

Hmm. Maybe I can compute 4 cos θ - 1? Let me think. If θ is 80°, then 4 cos 80° - 1. Let me compute cos 80° numerically? Wait, but maybe there's an exact value?

Alternatively, maybe think of 4 cos θ - 1 as something. Let me think. Maybe relate to some known angle?

Alternatively, think of the expression as a combination of sine and cosine terms. Let me think. Suppose I have:

Let me consider the expression [ sin θ (4 cos θ - 1 ) ] / cos θ. Let me write that as:

sin θ * (4 cos θ - 1 ) / cos θ = 4 sin θ - (sin θ / cos θ ) = 4 sin θ - tan θ, which is the original expression. So that's going in circles.

Wait, maybe I need to think of another identity. Let me think. Maybe express tan θ in terms of sin and cos and then combine with the other term. Wait, that's what I did before.

Alternatively, maybe think of tan θ + something. Wait, but the original expression is tan 100° + 4 sin 100°, which we converted to -tan 80° + 4 sin 80°. Maybe I can think of this as a combination that can be expressed as a single trigonometric function?

Alternatively, think of 4 sin θ - tan θ. Let me write that as 4 sin θ - sin θ / cos θ = sin θ (4 - 1 / cos θ ). Hmm, same as before.

Alternatively, maybe write 4 as 4 cos θ / cos θ, so:

sin θ (4 cos θ - 1 ) / cos θ. So, if I can write 4 cos θ - 1 as something... Maybe relate to another angle?

Alternatively, think of 4 cos θ - 1 as a multiple of some sine or cosine. For example, maybe using the formula for a sin θ + b cos θ? Wait, but here it's 4 cos θ - 1. Maybe not directly.

Alternatively, think of 4 cos θ - 1 as 4 cos θ - 1 = 0? Wait, no. Maybe not helpful.

Alternatively, think of specific values. Let me compute approximate values to see if I can recognize the result?

Let me try that. Let me compute tan 100° + 4 sin 100° numerically.

First, convert degrees to radians if needed, but maybe I can use calculator approximations.

First, tan 100°. 100° is in the second quadrant. Let me compute tan 100°:

tan(100°) ≈ tan(90° + 10°) = -cot 10° ≈ -5.671

Wait, cot 10° is 1/tan 10° ≈ 1 / 0.1763 ≈ 5.671, so tan 100° ≈ -5.671

Then, sin 100° is sin(80°) ≈ 0.9848

Therefore, 4 sin 100° ≈ 4 * 0.9848 ≈ 3.9392

Therefore, tan 100° + 4 sin 100° ≈ -5.671 + 3.9392 ≈ -1.7318

Hmm, approximately -1.732. Wait, that's approximately -√3, since √3 ≈ 1.732. So maybe the exact value is -√3?

Wait, that seems promising. Let me check with more accurate calculations.

First, let me compute tan(100°):

Using calculator, tan(100°):

First, 100° in radians is 100 * π / 180 ≈ 1.74533 radians.

tan(1.74533) ≈ tan(π - 2π/9) since π ≈ 3.1416, π - 1.74533 ≈ 1.3963 radians, which is 80°, so tan(π - 80°) = -tan(80°). Let me compute tan(80°):

tan(80°) ≈ 5.671, so tan(100°) ≈ -5.671.

Now, sin(100°) = sin(80°) ≈ 0.984807753012208.

Therefore, 4 * sin(100°) ≈ 4 * 0.984807753 ≈ 3.939231012.

Adding tan(100°) + 4 sin(100°) ≈ -5.671 + 3.93923 ≈ -1.73177.

Which is approximately -√3 ≈ -1.73205. So that's very close. Therefore, the exact value is likely -√3. Therefore, the simplified form is -√3.

But how to show that algebraically?

So, if tan 100° + 4 sin 100° = -√3, then we need to prove that.

Let me try to verify this.

Let me consider θ = 80°, so tan(100°) = -tan(80°), and sin(100°) = sin(80°). Therefore, the expression is -tan(80°) + 4 sin(80°). Let me compute this:

Let me write tan(80°) as sin(80°)/cos(80°), so:

- sin(80°)/cos(80°) + 4 sin(80°) = sin(80°) [ -1/cos(80°) + 4 ].

Let me compute the term in the brackets:

4 - 1/cos(80°). Let me compute 1/cos(80°). cos(80°) ≈ 0.1736, so 1 / 0.1736 ≈ 5.7588. Therefore, 4 - 5.7588 ≈ -1.7588. Then multiply by sin(80°) ≈ 0.9848, so -1.7588 * 0.9848 ≈ -1.732, which is -√3. So that seems to check out.

But how to show this algebraically?

Let me think. Let me suppose that -tan(80°) + 4 sin(80°) = -√3. Let me verify this equation.

Let me write tan(80°) = sin(80°)/cos(80°), so:

- sin(80°)/cos(80°) + 4 sin(80°) = -√3

Multiply both sides by cos(80°):

- sin(80°) + 4 sin(80°) cos(80°) = -√3 cos(80°)

Let me compute the left-hand side:

sin(80°) [ -1 + 4 cos(80°) ]

So:

sin(80°) [4 cos(80°) - 1] = -√3 cos(80°)

Let me compute 4 cos(80°) - 1. Let me recall that cos(80°) is sin(10°), since cos(90° - θ) = sin θ. So cos(80°) = sin(10°). Therefore:

4 sin(10°) - 1. Hmm, not sure. Alternatively, maybe use exact values?

Alternatively, let me consider using known angles. Let me think. Let me recall that 80° is 60° + 20°, or 30° + 50°, but not sure. Alternatively, think of 80° as 3*20°, but again not sure.

Alternatively, maybe use the sine of some angle. Let me think. Suppose I consider the equation:

sin(80°)(4 cos(80°) - 1) = -√3 cos(80°)

Let me divide both sides by cos(80°):

sin(80°)(4 cos(80°) - 1)/cos(80°) = -√3

Which is:

sin(80°)(4 - sec(80°)) = -√3

But this seems not helpful. Alternatively, maybe square both sides? Although that might complicate things.

Alternatively, let me compute sin(80°)(4 cos(80°) - 1) + √3 cos(80°) = 0

Let me compute this expression:

sin(80°)(4 cos(80°) - 1) + √3 cos(80°) = 0

Let me expand:

4 sin(80°) cos(80°) - sin(80°) + √3 cos(80°) = 0

Note that 4 sin(80°) cos(80°) is 2 * 2 sin(80°) cos(80°) = 2 sin(160°), since sin(2θ) = 2 sin θ cos θ. Therefore:

2 sin(160°) - sin(80°) + √3 cos(80°) = 0

But sin(160°) = sin(180° - 20°) = sin(20°). Therefore:

2 sin(20°) - sin(80°) + √3 cos(80°) = 0

Now, let me compute 2 sin(20°) - sin(80°). Let me recall that sin(80°) = sin(60° + 20°) = sin(60°)cos(20°) + cos(60°)sin(20°) = (√3/2) cos(20°) + (1/2) sin(20°)

Therefore, 2 sin(20°) - sin(80°) = 2 sin(20°) - [ (√3/2) cos(20°) + (1/2) sin(20°) ] = 2 sin(20°) - (1/2) sin(20°) - (√3/2) cos(20°) = (3/2) sin(20°) - (√3/2) cos(20°)

Therefore, the entire expression becomes:

(3/2) sin(20°) - (√3/2) cos(20°) + √3 cos(80°) = 0

Now, let me note that cos(80°) = sin(10°), and maybe express everything in terms of sin(20°) and cos(20°). Alternatively, let me compute √3 cos(80°):

√3 cos(80°) = √3 sin(10°). Hmm, not sure. Alternatively, maybe express cos(80°) as cos(60° + 20°):

cos(60° + 20°) = cos(60°)cos(20°) - sin(60°)sin(20°) = (1/2) cos(20°) - (√3/2) sin(20°)

Therefore, √3 cos(80°) = √3 [ (1/2) cos(20°) - (√3/2) sin(20°) ] = (√3/2) cos(20°) - (3/2) sin(20°)

Therefore, substituting back into the expression:

(3/2 sin(20°) - √3/2 cos(20°)) + (√3/2 cos(20°) - 3/2 sin(20°)) = 0

Let me compute term by term:

First term: 3/2 sin(20°) - √3/2 cos(20°)

Second term: + √3/2 cos(20°) - 3/2 sin(20°)

Adding them together:

(3/2 sin(20°) - 3/2 sin(20°)) + (-√3/2 cos(20°) + √3/2 cos(20°)) = 0 + 0 = 0

Wow! So the entire expression equals zero, which means our original assumption that -tan(80°) + 4 sin(80°) = -√3 is correct. Therefore, tan(100°) + 4 sin(100°) = -√3.

Therefore, the simplified form is -√3.

But let me check once again to make sure I didn't make any mistakes in the algebra.

Starting from:

Original expression: tan(100°) + 4 sin(100°) = -tan(80°) + 4 sin(80°)

Then, through manipulations, we arrived at an identity that equals zero, which confirms that the original expression equals -√3. Therefore, the answer is -√3.

Alternatively, maybe there's a more straightforward way to see this? Let me think.

Suppose we consider the expression tan θ + 4 sin θ where θ = 100°, but we saw that θ = 100° is equivalent to -tan(80°) + 4 sin(80°). Maybe there's a way to relate this to a known triangle or identity?

Alternatively, think of 80° as 60° + 20°, and use angle addition formulas. But I think the previous method was sufficient.

Alternatively, think of the expression as follows:

Let me consider that -tan(80°) + 4 sin(80°) = - [ tan(80°) - 4 sin(80°) ]

But not sure. Alternatively, think of tan(80°) = sin(80°)/cos(80°), so:

- sin(80°)/cos(80°) + 4 sin(80°) = sin(80°) [ -1/cos(80°) + 4 ] = sin(80°) [ 4 - sec(80°) ]

But if we can show that 4 - sec(80°) equals -√3 / sin(80°), then the whole expression would be -√3. Let me check:

If 4 - sec(80°) = -√3 / sin(80°), then multiplying both sides by sin(80°):

4 sin(80°) - sec(80°) sin(80°) = -√3

Which is 4 sin(80°) - 1 / cos(80°) * sin(80°) = 4 sin(80°) - tan(80°) = original expression, which we already know equals -√3. So that's consistent.

But again, this is going in circles. However, since the numerical calculation shows that it's approximately -√3 and the algebraic manipulation confirms that the expression equals zero when assuming it's -√3, I think we can be confident that the exact value is indeed -√3.

Therefore, the simplified form of tan 100° + 4 sin 100° is -√3.

**Final Answer**
\boxed{-\sqrt{3}}

To simplify the expression:

$$
\tan 100^\circ + 4 \sin 100^\circ
$$

we begin by recognizing that $ 100^\circ $ lies in the second quadrant, where:

- $ \tan 100^\circ = -\tan 80^\circ $ (since $ \tan(180^\circ - \theta) = -\tan \theta $)
- $ \sin 100^\circ = \sin 80^\circ $ (since $ \sin(180^\circ - \theta) = \sin \theta $)

So, the expression becomes:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\tan 80^\circ + 4 \sin 80^\circ
$$

Now, recall that:

$$
\tan \theta = \frac{\sin \theta}{\cos \theta}
$$

So:

$$
-\tan 80^\circ + 4 \sin 80^\circ = -\frac{\sin 80^\circ}{\cos 80^\circ} + 4 \sin 80^\circ
$$

We can factor out $ \sin 80^\circ $:

$$
\sin 80^\circ \left( -\frac{1}{\cos 80^\circ} + 4 \right)
$$

Now, combine the terms in the parentheses:

$$
\sin 80^\circ \left( 4 - \frac{1}{\cos 80^\circ} \right)
$$

This is equivalent to:

$$
\sin 80^\circ \left( 4 - \sec 80^\circ \right)
$$

Now, we evaluate this expression numerically to check for a possible exact value. Using known approximations:

- $ \sin 80^\circ \approx 0.9848 $
- $ \cos 80^\circ \approx 0.1736 \Rightarrow \sec 80^\circ \approx 5.7588 $

So:

$$
4 - \sec 80^\circ \approx 4 - 5.7588 = -1.7588
$$

Then:

$$
\sin 80^\circ \cdot (-1.7588) \approx 0.9848 \cdot (-1.7588) \approx -1.732
$$

This is very close to $ -\sqrt{3} \approx -1.73205 $, suggesting that the exact value is:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\sqrt{3}
$$

---

### Final Answer:

$$
\boxed{-\sqrt{3}}
$$

[Assistant]
Explanation:
Okay, so I need to simplify the expression tan 100° + 4 sin 100°. Hmm, let me think about how to approach this. First, I remember that tan θ is sin θ over cos θ, so maybe I can rewrite the expression in terms of sine and cosine. Let me try that.

So, tan 100° is sin 100° divided by cos 100°, right? So the expression becomes:

sin 100° / cos 100° + 4 sin 100°

Hmm, maybe I can factor out sin 100°? Let me see:

sin 100° [1 / cos 100° + 4]

But I don't know if that helps. Maybe I need to find a common denominator or combine the terms somehow. Let me write it as:

[sin 100° + 4 sin 100° cos 100°] / cos 100°

Wait, that's combining the two terms over a common denominator. Let me check:

Original expression: sin 100° / cos 100° + 4 sin 100°

To combine them, the second term needs to have denominator cos 100°, so multiply numerator and denominator by cos 100°:

[sin 100° + 4 sin 100° cos 100°] / cos 100°

Yes, that's correct. Now, maybe I can factor sin 100° from the numerator:

sin 100° [1 + 4 cos 100°] / cos 100°

Hmm, not sure if that helps. Maybe I need to use some trigonometric identities here. Let me recall some identities. For example, maybe double-angle formulas or something else?

Alternatively, maybe I can convert degrees to radians? Wait, but I don't know if that's necessary. Let me think about the angles. 100° is in the second quadrant, so sin 100° is positive, cos 100° is negative. Also, tan 100° is negative because tan is sin/cos, and cos is negative there. So tan 100° is negative, and 4 sin 100° is positive. So the whole expression is negative plus positive. Not sure if that helps.

Alternatively, maybe I can express tan 100° as tan(180° - 80°) which is -tan 80°, since tan(180° - θ) = -tan θ. Similarly, sin 100° is sin(80°), because sin(180° - θ) = sin θ. So maybe rewriting the expression in terms of 80°?

Let me try that. So tan 100° = -tan 80°, and sin 100° = sin 80°. Therefore, the original expression becomes:

- tan 80° + 4 sin 80°

Hmm, maybe that's easier? Let me write that:

- tan 80° + 4 sin 80°

Since tan 80° is sin 80° / cos 80°, so substituting that in:

- [sin 80° / cos 80°] + 4 sin 80°

Again, combining terms:

sin 80° [ -1 / cos 80° + 4 ]

Which is similar to before. Maybe I can combine the terms:

sin 80° [ ( -1 + 4 cos 80° ) / cos 80° ]

So that's [ sin 80° ( -1 + 4 cos 80° ) ] / cos 80°

Hmm, not sure. Maybe I need to compute numerical values? Wait, but the problem says to simplify, not necessarily compute numerically. Maybe there's an exact value?

Alternatively, perhaps using some trigonometric identities. Let me think. Let me recall that 4 sin θ can be related to some multiple-angle formulas?

Alternatively, maybe express everything in terms of sin and cos and then combine. Let me try again:

Original expression: tan 100° + 4 sin 100° = -tan 80° + 4 sin 80°

Let me write tan 80° as sin 80° / cos 80°, so:

- sin 80° / cos 80° + 4 sin 80° = sin 80° [ -1 / cos 80° + 4 ]

Let me compute the term in the brackets:

-1 / cos 80° + 4 = 4 - sec 80°

Not sure. Maybe I can write 4 as 4 cos 80° / cos 80°, so:

[4 cos 80° - 1] / cos 80°

Therefore, the entire expression becomes:

sin 80° [4 cos 80° - 1] / cos 80°

Which is [ sin 80° (4 cos 80° - 1) ] / cos 80°

Hmm, maybe I can compute 4 cos 80° - 1? Let me think. Let me compute 4 cos 80° - 1 numerically? Wait, but maybe there's an exact value?

Alternatively, maybe I can use some identity. Let me think. Let me recall that 80° is related to 60° and 20°, but not sure. Alternatively, maybe using product-to-sum formulas?

Alternatively, think about writing 4 cos 80° as something. Wait, 4 cos 80° is 4 cos(60° + 20°). Maybe expand that?

Let me try expanding cos(60° + 20°):

cos 60° cos 20° - sin 60° sin 20°

Which is (0.5) cos 20° - (√3/2) sin 20°

So 4 cos 80° = 4 [ 0.5 cos 20° - (√3/2) sin 20° ] = 2 cos 20° - 2√3 sin 20°

Therefore, 4 cos 80° - 1 = 2 cos 20° - 2√3 sin 20° - 1

Hmm, not sure if that helps. Maybe not. Let me think of another approach.

Alternatively, maybe consider writing the entire expression as a single trigonometric function. Let me think. Suppose I have:

Let me denote θ = 80°, so the expression is -tan θ + 4 sin θ. Let me write that:

- tan θ + 4 sin θ = - (sin θ / cos θ) + 4 sin θ = sin θ ( -1 / cos θ + 4 )

Which is the same as before. Maybe I can write this as:

sin θ (4 - sec θ )

Alternatively, maybe think of 4 as 4 cos θ / cos θ, so:

sin θ (4 cos θ - 1 ) / cos θ

So, [ sin θ (4 cos θ - 1 ) ] / cos θ

Hmm. Maybe I can compute 4 cos θ - 1? Let me think. If θ is 80°, then 4 cos 80° - 1. Let me compute cos 80° numerically? Wait, but maybe there's an exact value?

Alternatively, maybe think of 4 cos θ - 1 as something. Let me think. Maybe relate to some known angle?

Alternatively, think of the expression as a combination of sine and cosine terms. Let me think. Suppose I have:

Let me consider the expression [ sin θ (4 cos θ - 1 ) ] / cos θ. Let me write that as:

sin θ * (4 cos θ - 1 ) / cos θ = 4 sin θ - (sin θ / cos θ ) = 4 sin θ - tan θ, which is the original expression. So that's going in circles.

Wait, maybe I need to think of another identity. Let me think. Maybe express tan θ in terms of sin and cos and then combine with the other term. Wait, that's what I did before.

Alternatively, maybe think of tan θ + something. Wait, but the original expression is tan 100° + 4 sin 100°, which we converted to -tan 80° + 4 sin 80°. Maybe I can think of this as a combination that can be expressed as a single trigonometric function?

Alternatively, think of 4 sin θ - tan θ. Let me write that as 4 sin θ - sin θ / cos θ = sin θ (4 - 1 / cos θ ). Hmm, same as before.

Alternatively, maybe write 4 as 4 cos θ / cos θ, so:

sin θ (4 cos θ - 1 ) / cos θ. So, if I can write 4 cos θ - 1 as something... Maybe relate to another angle?

Alternatively, think of 4 cos θ - 1 as a multiple of some sine or cosine. For example, maybe using the formula for a sin θ + b cos θ? Wait, but here it's 4 cos θ - 1. Maybe not directly.

Alternatively, think of 4 cos θ - 1 as 4 cos θ - 1 = 0? Wait, no. Maybe not helpful.

Alternatively, think of specific values. Let me compute approximate values to see if I can recognize the result?

Let me try that. Let me compute tan 100° + 4 sin 100° numerically.

First, convert degrees to radians if needed, but maybe I can use calculator approximations.

First, tan 100°. 100° is in the second quadrant. Let me compute tan 100°:

tan(100°) ≈ tan(90° + 10°) = -cot 10° ≈ -5.671

Wait, cot 10° is 1/tan 10° ≈ 1 / 0.1763 ≈ 5.671, so tan 100° ≈ -5.671

Then, sin 100° is sin(80°) ≈ 0.9848

Therefore, 4 sin 100° ≈ 4 * 0.9848 ≈ 3.9392

Therefore, tan 100° + 4 sin 100° ≈ -5.671 + 3.9392 ≈ -1.7318

Hmm, approximately -1.732. Wait, that's approximately -√3, since √3 ≈ 1.732. So maybe the exact value is -√3?

Wait, that seems promising. Let me check with more accurate calculations.

First, let me compute tan(100°):

Using calculator, tan(100°):

First, 100° in radians is 100 * π / 180 ≈ 1.74533 radians.

tan(1.74533) ≈ tan(π - 2π/9) since π ≈ 3.1416, π - 1.74533 ≈ 1.3963 radians, which is 80°, so tan(π - 80°) = -tan(80°). Let me compute tan(80°):

tan(80°) ≈ 5.671, so tan(100°) ≈ -5.671.

Now, sin(100°) = sin(80°) ≈ 0.984807753012208.

Therefore, 4 * sin(100°) ≈ 4 * 0.984807753 ≈ 3.939231012.

Adding tan(100°) + 4 sin(100°) ≈ -5.671 + 3.93923 ≈ -1.73177.

Which is approximately -√3 ≈ -1.73205. So that's very close. Therefore, the exact value is likely -√3. Therefore, the simplified form is -√3.

But how to show that algebraically?

So, if tan 100° + 4 sin 100° = -√3, then we need to prove that.

Let me try to verify this.

Let me consider θ = 80°, so tan(100°) = -tan(80°), and sin(100°) = sin(80°). Therefore, the expression is -tan(80°) + 4 sin(80°). Let me compute this:

Let me write tan(80°) as sin(80°)/cos(80°), so:

- sin(80°)/cos(80°) + 4 sin(80°) = sin(80°) [ -1/cos(80°) + 4 ].

Let me compute the term in the brackets:

4 - 1/cos(80°). Let me compute 1/cos(80°). cos(80°) ≈ 0.1736, so 1 / 0.1736 ≈ 5.7588. Therefore, 4 - 5.7588 ≈ -1.7588. Then multiply by sin(80°) ≈ 0.9848, so -1.7588 * 0.9848 ≈ -1.732, which is -√3. So that seems to check out.

But how to show this algebraically?

Let me think. Let me suppose that -tan(80°) + 4 sin(80°) = -√3. Let me verify this equation.

Let me write tan(80°) = sin(80°)/cos(80°), so:

- sin(80°)/cos(80°) + 4 sin(80°) = -√3

Multiply both sides by cos(80°):

- sin(80°) + 4 sin(80°) cos(80°) = -√3 cos(80°)

Let me compute the left-hand side:

sin(80°) [ -1 + 4 cos(80°) ]

So:

sin(80°) [4 cos(80°) - 1] = -√3 cos(80°)

Let me compute 4 cos(80°) - 1. Let me recall that cos(80°) is sin(10°), since cos(90° - θ) = sin θ. So cos(80°) = sin(10°). Therefore:

4 sin(10°) - 1. Hmm, not sure. Alternatively, maybe use exact values?

Alternatively, let me consider using known angles. Let me think. Let me recall that 80° is 60° + 20°, or 30° + 50°, but not sure. Alternatively, think of 80° as 3*20°, but again not sure.

Alternatively, maybe use the sine of some angle. Let me think. Suppose I consider the equation:

sin(80°)(4 cos(80°) - 1) = -√3 cos(80°)

Let me divide both sides by cos(80°):

sin(80°)(4 cos(80°) - 1)/cos(80°) = -√3

Which is:

sin(80°)(4 - sec(80°)) = -√3

But this seems not helpful. Alternatively, maybe square both sides? Although that might complicate things.

Alternatively, let me compute sin(80°)(4 cos(80°) - 1) + √3 cos(80°) = 0

Let me compute this expression:

sin(80°)(4 cos(80°) - 1) + √3 cos(80°) = 0

Let me expand:

4 sin(80°) cos(80°) - sin(80°) + √3 cos(80°) = 0

Note that 4 sin(80°) cos(80°) is 2 * 2 sin(80°) cos(80°) = 2 sin(160°), since sin(2θ) = 2 sin θ cos θ. Therefore:

2 sin(160°) - sin(80°) + √3 cos(80°) = 0

But sin(160°) = sin(180° - 20°) = sin(20°). Therefore:

2 sin(20°) - sin(80°) + √3 cos(80°) = 0

Now, let me compute 2 sin(20°) - sin(80°). Let me recall that sin(80°) = sin(60° + 20°) = sin(60°)cos(20°) + cos(60°)sin(20°) = (√3/2) cos(20°) + (1/2) sin(20°)

Therefore, 2 sin(20°) - sin(80°) = 2 sin(20°) - [ (√3/2) cos(20°) + (1/2) sin(20°) ] = 2 sin(20°) - (1/2) sin(20°) - (√3/2) cos(20°) = (3/2) sin(20°) - (√3/2) cos(20°)

Therefore, the entire expression becomes:

(3/2) sin(20°) - (√3/2) cos(20°) + √3 cos(80°) = 0

Now, let me note that cos(80°) = sin(10°), and maybe express everything in terms of sin(20°) and cos(20°). Alternatively, let me compute √3 cos(80°):

√3 cos(80°) = √3 sin(10°). Hmm, not sure. Alternatively, maybe express cos(80°) as cos(60° + 20°):

cos(60° + 20°) = cos(60°)cos(20°) - sin(60°)sin(20°) = (1/2) cos(20°) - (√3/2) sin(20°)

Therefore, √3 cos(80°) = √3 [ (1/2) cos(20°) - (√3/2) sin(20°) ] = (√3/2) cos(20°) - (3/2) sin(20°)

Therefore, substituting back into the expression:

(3/2 sin(20°) - √3/2 cos(20°)) + (√3/2 cos(20°) - 3/2 sin(20°)) = 0

Let me compute term by term:

First term: 3/2 sin(20°) - √3/2 cos(20°)

Second term: + √3/2 cos(20°) - 3/2 sin(20°)

Adding them together:

(3/2 sin(20°) - 3/2 sin(20°)) + (-√3/2 cos(20°) + √3/2 cos(20°)) = 0 + 0 = 0

Wow! So the entire expression equals zero, which means our original assumption that -tan(80°) + 4 sin(80°) = -√3 is correct. Therefore, tan(100°) + 4 sin(100°) = -√3.

Therefore, the simplified form is -√3.

But let me check once again to make sure I didn't make any mistakes in the algebra.

Starting from:

Original expression: tan(100°) + 4 sin(100°) = -tan(80°) + 4 sin(80°)

Then, through manipulations, we arrived at an identity that equals zero, which confirms that the original expression equals -√3. Therefore, the answer is -√3.

Alternatively, maybe there's a more straightforward way to see this? Let me think.

Suppose we consider the expression tan θ + 4 sin θ where θ = 100°, but we saw that θ = 100° is equivalent to -tan(80°) + 4 sin(80°). Maybe there's a way to relate this to a known triangle or identity?

Alternatively, think of 80° as 60° + 20°, and use angle addition formulas. But I think the previous method was sufficient.

Alternatively, think of the expression as follows:

Let me consider that -tan(80°) + 4 sin(80°) = - [ tan(80°) - 4 sin(80°) ]

But not sure. Alternatively, think of tan(80°) = sin(80°)/cos(80°), so:

- sin(80°)/cos(80°) + 4 sin(80°) = sin(80°) [ -1/cos(80°) + 4 ] = sin(80°) [ 4 - sec(80°) ]

But if we can show that 4 - sec(80°) equals -√3 / sin(80°), then the whole expression would be -√3. Let me check:

If 4 - sec(80°) = -√3 / sin(80°), then multiplying both sides by sin(80°):

4 sin(80°) - sec(80°) sin(80°) = -√3

Which is 4 sin(80°) - 1 / cos(80°) * sin(80°) = 4 sin(80°) - tan(80°) = original expression, which we already know equals -√3. So that's consistent.

But again, this is going in circles. However, since the numerical calculation shows that it's approximately -√3 and the algebraic manipulation confirms that the expression equals zero when assuming it's -√3, I think we can be confident that the exact value is indeed -√3.

Therefore, the simplified form of tan 100° + 4 sin 100° is -√3.

**Final Answer**
\boxed{-\sqrt{3}}

To simplify the expression:

$$
\tan 100^\circ + 4 \sin 100^\circ
$$

we begin by recognizing that $ 100^\circ $ lies in the second quadrant, where:

- $ \tan 100^\circ = -\tan 80^\circ $ (since $ \tan(180^\circ - \theta) = -\tan \theta $)
- $ \sin 100^\circ = \sin 80^\circ $ (since $ \sin(180^\circ - \theta) = \sin \theta $)

So, the expression becomes:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\tan 80^\circ + 4 \sin 80^\circ
$$

Now, recall that:

$$
\tan \theta = \frac{\sin \theta}{\cos \theta}
$$

So:

$$
-\tan 80^\circ + 4 \sin 80^\circ = -\frac{\sin 80^\circ}{\cos 80^\circ} + 4 \sin 80^\circ
$$

We can factor out $ \sin 80^\circ $:

$$
\sin 80^\circ \left( -\frac{1}{\cos 80^\circ} + 4 \right)
$$

Now, combine the terms in the parentheses:

$$
\sin 80^\circ \left( 4 - \frac{1}{\cos 80^\circ} \right)
$$

This is equivalent to:

$$
\sin 80^\circ \left( 4 - \sec 80^\circ \right)
$$

Now, we evaluate this expression numerically to check for a possible exact value. Using known approximations:

- $ \sin 80^\circ \approx 0.9848 $
- $ \cos 80^\circ \approx 0.1736 \Rightarrow \sec 80^\circ \approx 5.7588 $

So:

$$
4 - \sec 80^\circ \approx 4 - 5.7588 = -1.7588
$$

Then:

$$
\sin 80^\circ \cdot (-1.7588) \approx 0.9848 \cdot (-1.7588) \approx -1.732
$$

This is very close to $ -\sqrt{3} \approx -1.73205 $, suggesting that the exact value is:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\sqrt{3}
$$

---

### Final Answer:

$$
\boxed{-\sqrt{3}}
$$

Solution:
We have that
\begin{align*}
\tan 100^\circ + 4 \sin 100^\circ &= \frac{\sin 100^\circ}{\cos 100^\circ} + 4 \sin 100^\circ \\
&= \frac{\sin 80^\circ}{-\cos 80^\circ} + 4 \sin 80^\circ \\
&= -\frac{\cos 10^\circ}{\sin 10^\circ} + 4 \cos 10^\circ \\
&= \frac{4 \cos 10^\circ \sin 10^\circ - \cos 10^\circ}{\sin 10^\circ}.
\end{align*}By double angle formula,
\begin{align*}
\frac{4 \cos 10^\circ \sin 10^\circ - \cos 10^\circ}{\sin 10^\circ} &= \frac{2 \sin 20^\circ - \cos 10^\circ}{\sin 10^\circ} \\
&= \frac{\sin 20^\circ + \sin 20^\circ - \sin 80^\circ}{\sin 10^\circ}.
\end{align*}By sum-to-product,
\[\sin 20^\circ - \sin 80^\circ = 2 \cos 50^\circ \sin (-30^\circ) = -\cos 50^\circ,\]so
\begin{align*}
\frac{\sin 20^\circ + \sin 20^\circ - \sin 80^\circ}{\sin 10^\circ} &= \frac{\sin 20^\circ - \cos 50^\circ}{\sin 10^\circ} \\
&= \frac{\sin 20^\circ - \sin 40^\circ}{\sin 10^\circ}.
\end{align*}By sum-to-product,
\[\sin 20^\circ - \sin 40^\circ = 2 \cos 30^\circ \sin (-10^\circ) = -\sqrt{3} \sin 10^\circ,\]so $\frac{\sin 20^\circ - \sin 40^\circ}{\sin 10^\circ} = \boxed{-\sqrt{3}}.$