Explanation:
Okay, so I need to find the greatest common factor (GCF) of 6432 and 132, and then add 11 to that result. Let me think about how to approach this.

First, I remember that the GCF of two numbers is the largest number that divides both of them without leaving a remainder. To find the GCF, one method is to use the prime factorization of both numbers and then multiply the common prime factors with the lowest exponents. Alternatively, I can use the Euclidean algorithm, which might be more efficient for larger numbers. Let me try both methods and see which one works better here.

Starting with prime factorization. Let me factorize 132 first because it seems smaller.

132 can be divided by 2 first. 132 ÷ 2 = 66. Then 66 ÷ 2 = 33. Now 33 can be divided by 3: 33 ÷ 3 = 11. And 11 is a prime number. So the prime factors of 132 are 2² × 3 × 11.

Now let me factorize 6432. Hmm, this might take a bit longer. Let me start dividing by small primes.

First, check if it's even. 6432 ÷ 2 = 3216. Still even, so divide by 2 again: 3216 ÷ 2 = 1608. Again, divide by 2: 1608 ÷ 2 = 804. Again: 804 ÷ 2 = 402. Again: 402 ÷ 2 = 201. Okay, so that's five divisions by 2, so 2⁵. Now 201. Let me check divisibility by 3. 2 + 0 + 1 = 3, which is divisible by 3. So 201 ÷ 3 = 67. Is 67 a prime number? Yes, because it's not divisible by 2, 3, 5, 7, or 11. Let me check 7: 7*9=63, 7*10=70, so no. 11*6=66, 11*7=77, so no. So 67 is prime.

Therefore, the prime factors of 6432 are 2⁵ × 3 × 67.

Now, comparing the prime factors of 132 (which is 2² × 3 × 11) and 6432 (which is 2⁵ × 3 × 67), the common prime factors are 2² and 3. So the GCF would be 2² × 3. Let me calculate that: 2² is 4, times 3 is 12. So the GCF is 12? Wait, but let me check if that's correct.

Wait, hold on. Let me verify this with another method to make sure I didn't make a mistake. Maybe I made an error in prime factorization. Let me check 6432 again.

Starting over with 6432:

Divide by 2: 6432 / 2 = 3216

Divide by 2 again: 3216 / 2 = 1608

Divide by 2 again: 1608 / 2 = 804

Divide by 2 again: 804 / 2 = 402

Divide by 2 again: 402 / 2 = 201. So that's five 2s. Then 201 divided by 3 is 67. So yes, 2⁵ × 3 × 67.

And 132 is 2² × 3 × 11. So common factors are 2² and 3. So 2² × 3 = 4 × 3 = 12. So GCF is 12.

But let me check with the Euclidean algorithm to confirm. The Euclidean algorithm says that GCF(a, b) = GCF(b, a mod b). So let me apply that.

First, take 6432 and 132.

Compute 6432 divided by 132. Let me find how many times 132 goes into 6432.

Well, 132 × 50 = 6600, which is more than 6432. Let me try 132 × 48. Let me compute 132 × 40 = 5280, 132 × 8 = 1056. So 5280 + 1056 = 6336. Then 6432 - 6336 = 96. So 6432 = 132 × 48 + 96.

So GCF(6432, 132) = GCF(132, 96).

Now compute GCF(132, 96).

Divide 132 by 96. 96 × 1 = 96, remainder 132 - 96 = 36. So GCF(96, 36).

Now GCF(96, 36). Divide 96 by 36. 36 × 2 = 72, remainder 96 - 72 = 24. So GCF(36, 24).

GCF(36, 24). Divide 36 by 24. 24 × 1 = 24, remainder 12. So GCF(24, 12).

GCF(24, 12). Divide 24 by 12. Remainder 0. So GCF is 12.

So that confirms it. The GCF is indeed 12.

Therefore, the problem says to increase this GCF by 11. So 12 + 11 = 23.

Wait, but let me just make sure I didn't make any calculation errors in the Euclidean algorithm steps. Let me go through them again.

First step: 6432 divided by 132. Let me compute 132 × 48.

Well, 132 × 50 is 6600, so subtract 132 × 2 = 264, so 6600 - 264 = 6336. Then 6432 - 6336 = 96. Correct.

Then GCF(132, 96).

132 divided by 96 is 1 with remainder 36. Correct.

Then GCF(96, 36).

96 divided by 36 is 2 with remainder 24. Correct.

GCF(36, 24).

36 divided by 24 is 1 with remainder 12. Correct.

GCF(24, 12).

24 divided by 12 is 2 with remainder 0. So GCF is 12.

Yes, that's correct.

Alternatively, maybe I can check with another method. Let me try dividing both numbers by 12 and see if they are integers.

6432 ÷ 12. Let me compute that. 12 × 500 = 6000. 6432 - 6000 = 432. 432 ÷ 12 = 36. So total is 500 + 36 = 536. So 12 × 536 = 6432.

132 ÷ 12 = 11. So yes, both numbers are divisible by 12.

Is there a larger number that divides both? Let me check 24.

6432 ÷ 24. Well, 6432 ÷ 12 = 536, so ÷2 is 268. So 24 × 268 = 6432.

132 ÷ 24 = 5.5. Which is not an integer. So 24 doesn't divide 132. Therefore, 12 is indeed the GCF.

Therefore, adding 11 to the GCF: 12 + 11 = 23.

Wait, but just to be thorough, let me check if there's a higher common factor. For example, maybe 6? Well, 6 is a factor of 12, so if 12 is the GCF, 6 would be a common factor, but not the greatest.

Alternatively, maybe I made a mistake in prime factors? Let me check 6432 again.

Earlier, I had 2⁵ × 3 × 67. Let me check 2⁵ is 32. 32 × 3 is 96. 96 × 67. Let me compute that.

Well, 96 × 60 = 5760, 96 × 7 = 672. So 5760 + 672 = 6432. Correct.

So 6432 is indeed 2⁵ × 3 × 67.

And 132 is 2² × 3 × 11.

So common factors are 2² and 3. So 4 × 3 = 12.

Therefore, the GCF is 12.

So adding 11 gives 23.

Therefore, the answer should be 23.

But just to make sure, let me check with another approach. Let me list the factors of 132 and see what's the highest common with 6432.

Factors of 132: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.

Now check which of these divide 6432.

Start with 132: 6432 ÷ 132 = 48.9... Wait, earlier we saw that 132 × 48 = 6336, and 6432 - 6336 = 96, so remainder 96. Therefore, 132 does not divide 6432.

Next, 66: 6432 ÷ 66. Let me compute 66 × 97 = 66 × 100 - 66 × 3 = 6600 - 198 = 6402. Then 6432 - 6402 = 30. So remainder 30. Not divisible.

Next, 44: 6432 ÷ 44. Let me divide 6432 by 4. 6432 ÷ 4 = 1608. Then divide by 11: 1608 ÷ 11. 11 × 146 = 1606. Remainder 2. Not divisible.

Next, 33: 6432 ÷ 33. Let me divide 6432 by 3: 2144. Then divide by 11: 2144 ÷ 11. 11 × 194 = 2134. Remainder 10. Not divisible.

Next, 22: 6432 ÷ 22. Let me divide by 2: 3216. Then divide by 11: 3216 ÷ 11. 11 × 292 = 3212. Remainder 4. Not divisible.

Next, 12: 6432 ÷ 12 = 536, which is integer.

So the highest common factor is 12.

Therefore, adding 11 gives 23.

So I think that's solid.

But just to make sure, let me check if 23 is the correct answer.

Alternatively, maybe I miscalculated something? Let me check once again.

Original numbers: 6432 and 132.

GCF is 12. 12 + 11 = 23.

Yes.

Alternatively, maybe the question is a trick question? Like, maybe the GCF is something else? But no, both methods gave me 12.

Alternatively, maybe I made a mistake in the Euclidean algorithm? Let me check again.

First step: GCF(6432, 132).

Compute 6432 divided by 132.

As above, 132 × 48 = 6336. 6432 - 6336 = 96. So GCF(132, 96).

Then GCF(132, 96).

132 divided by 96 is 1 with remainder 36. So GCF(96, 36).

Then GCF(96, 36).

96 divided by 36 is 2 with remainder 24. GCF(36, 24).

GCF(36, 24).

36 divided by 24 is 1 with remainder 12. GCF(24, 12).

24 divided by 12 is 2 with remainder 0. So GCF is 12.

Yes, that's correct.

Therefore, the answer is 12 + 11 = 23.

I think that's solid.

**Final Answer**
The result is \boxed{23}.

To solve the problem, we are asked to find the **greatest common factor (GCF)** of the numbers **6432** and **132**, and then **add 11** to that result.

---

### Step 1: Find the GCF of 6432 and 132

We can use the **Euclidean Algorithm**, which is efficient for large numbers.

#### Apply the Euclidean Algorithm:

1. **Divide 6432 by 132**:
   $$
   6432 \div 132 = 48 \text{ remainder } 96
   $$
   So, $ \text{GCF}(6432, 132) = \text{GCF}(132, 96) $

2. **Divide 132 by 96**:
   $$
   132 \div 96 = 1 \text{ remainder } 36
   $$
   So, $ \text{GCF}(132, 96) = \text{GCF}(96, 36) $

3. **Divide 96 by 36**:
   $$
   96 \div 36 = 2 \text{ remainder } 24
   $$
   So, $ \text{GCF}(96, 36) = \text{GCF}(36, 24) $

4. **Divide 36 by 24**:
   $$
   36 \div 24 = 1 \text{ remainder } 12
   $$
   So, $ \text{GCF}(36, 24) = \text{GCF}(24, 12) $

5. **Divide 24 by 12**:
   $$
   24 \div 12 = 2 \text{ remainder } 0
   $$
   So, $ \text{GCF}(24, 12) = 12 $

Thus, the **GCF of 6432 and 132 is 12**.

---

### Step 2: Add 11 to the GCF

Now, we are asked to **increase the GCF by 11**:

$$
12 + 11 = 23
$$

---

### Final Answer

$$
\boxed{23}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest common factor (GCF) of 6432 and 132, and then add 11 to that result. Let me think about how to approach this.

First, I remember that the GCF of two numbers is the largest number that divides both of them without leaving a remainder. To find the GCF, one method is to use the prime factorization of both numbers and then multiply the common prime factors with the lowest exponents. Alternatively, I can use the Euclidean algorithm, which might be more efficient for larger numbers. Let me try both methods and see which one works better here.

Starting with prime factorization. Let me factorize 132 first because it seems smaller.

132 can be divided by 2 first. 132 ÷ 2 = 66. Then 66 ÷ 2 = 33. Now 33 can be divided by 3: 33 ÷ 3 = 11. And 11 is a prime number. So the prime factors of 132 are 2² × 3 × 11.

Now let me factorize 6432. Hmm, this might take a bit longer. Let me start dividing by small primes.

First, check if it's even. 6432 ÷ 2 = 3216. Still even, so divide by 2 again: 3216 ÷ 2 = 1608. Again, divide by 2: 1608 ÷ 2 = 804. Again: 804 ÷ 2 = 402. Again: 402 ÷ 2 = 201. Okay, so that's five divisions by 2, so 2⁵. Now 201. Let me check divisibility by 3. 2 + 0 + 1 = 3, which is divisible by 3. So 201 ÷ 3 = 67. Is 67 a prime number? Yes, because it's not divisible by 2, 3, 5, 7, or 11. Let me check 7: 7*9=63, 7*10=70, so no. 11*6=66, 11*7=77, so no. So 67 is prime.

Therefore, the prime factors of 6432 are 2⁵ × 3 × 67.

Now, comparing the prime factors of 132 (which is 2² × 3 × 11) and 6432 (which is 2⁵ × 3 × 67), the common prime factors are 2² and 3. So the GCF would be 2² × 3. Let me calculate that: 2² is 4, times 3 is 12. So the GCF is 12? Wait, but let me check if that's correct.

Wait, hold on. Let me verify this with another method to make sure I didn't make a mistake. Maybe I made an error in prime factorization. Let me check 6432 again.

Starting over with 6432:

Divide by 2: 6432 / 2 = 3216

Divide by 2 again: 3216 / 2 = 1608

Divide by 2 again: 1608 / 2 = 804

Divide by 2 again: 804 / 2 = 402

Divide by 2 again: 402 / 2 = 201. So that's five 2s. Then 201 divided by 3 is 67. So yes, 2⁵ × 3 × 67.

And 132 is 2² × 3 × 11. So common factors are 2² and 3. So 2² × 3 = 4 × 3 = 12. So GCF is 12.

But let me check with the Euclidean algorithm to confirm. The Euclidean algorithm says that GCF(a, b) = GCF(b, a mod b). So let me apply that.

First, take 6432 and 132.

Compute 6432 divided by 132. Let me find how many times 132 goes into 6432.

Well, 132 × 50 = 6600, which is more than 6432. Let me try 132 × 48. Let me compute 132 × 40 = 5280, 132 × 8 = 1056. So 5280 + 1056 = 6336. Then 6432 - 6336 = 96. So 6432 = 132 × 48 + 96.

So GCF(6432, 132) = GCF(132, 96).

Now compute GCF(132, 96).

Divide 132 by 96. 96 × 1 = 96, remainder 132 - 96 = 36. So GCF(96, 36).

Now GCF(96, 36). Divide 96 by 36. 36 × 2 = 72, remainder 96 - 72 = 24. So GCF(36, 24).

GCF(36, 24). Divide 36 by 24. 24 × 1 = 24, remainder 12. So GCF(24, 12).

GCF(24, 12). Divide 24 by 12. Remainder 0. So GCF is 12.

So that confirms it. The GCF is indeed 12.

Therefore, the problem says to increase this GCF by 11. So 12 + 11 = 23.

Wait, but let me just make sure I didn't make any calculation errors in the Euclidean algorithm steps. Let me go through them again.

First step: 6432 divided by 132. Let me compute 132 × 48.

Well, 132 × 50 is 6600, so subtract 132 × 2 = 264, so 6600 - 264 = 6336. Then 6432 - 6336 = 96. Correct.

Then GCF(132, 96).

132 divided by 96 is 1 with remainder 36. Correct.

Then GCF(96, 36).

96 divided by 36 is 2 with remainder 24. Correct.

GCF(36, 24).

36 divided by 24 is 1 with remainder 12. Correct.

GCF(24, 12).

24 divided by 12 is 2 with remainder 0. So GCF is 12.

Yes, that's correct.

Alternatively, maybe I can check with another method. Let me try dividing both numbers by 12 and see if they are integers.

6432 ÷ 12. Let me compute that. 12 × 500 = 6000. 6432 - 6000 = 432. 432 ÷ 12 = 36. So total is 500 + 36 = 536. So 12 × 536 = 6432.

132 ÷ 12 = 11. So yes, both numbers are divisible by 12.

Is there a larger number that divides both? Let me check 24.

6432 ÷ 24. Well, 6432 ÷ 12 = 536, so ÷2 is 268. So 24 × 268 = 6432.

132 ÷ 24 = 5.5. Which is not an integer. So 24 doesn't divide 132. Therefore, 12 is indeed the GCF.

Therefore, adding 11 to the GCF: 12 + 11 = 23.

Wait, but just to be thorough, let me check if there's a higher common factor. For example, maybe 6? Well, 6 is a factor of 12, so if 12 is the GCF, 6 would be a common factor, but not the greatest.

Alternatively, maybe I made a mistake in prime factors? Let me check 6432 again.

Earlier, I had 2⁵ × 3 × 67. Let me check 2⁵ is 32. 32 × 3 is 96. 96 × 67. Let me compute that.

Well, 96 × 60 = 5760, 96 × 7 = 672. So 5760 + 672 = 6432. Correct.

So 6432 is indeed 2⁵ × 3 × 67.

And 132 is 2² × 3 × 11.

So common factors are 2² and 3. So 4 × 3 = 12.

Therefore, the GCF is 12.

So adding 11 gives 23.

Therefore, the answer should be 23.

But just to make sure, let me check with another approach. Let me list the factors of 132 and see what's the highest common with 6432.

Factors of 132: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.

Now check which of these divide 6432.

Start with 132: 6432 ÷ 132 = 48.9... Wait, earlier we saw that 132 × 48 = 6336, and 6432 - 6336 = 96, so remainder 96. Therefore, 132 does not divide 6432.

Next, 66: 6432 ÷ 66. Let me compute 66 × 97 = 66 × 100 - 66 × 3 = 6600 - 198 = 6402. Then 6432 - 6402 = 30. So remainder 30. Not divisible.

Next, 44: 6432 ÷ 44. Let me divide 6432 by 4. 6432 ÷ 4 = 1608. Then divide by 11: 1608 ÷ 11. 11 × 146 = 1606. Remainder 2. Not divisible.

Next, 33: 6432 ÷ 33. Let me divide 6432 by 3: 2144. Then divide by 11: 2144 ÷ 11. 11 × 194 = 2134. Remainder 10. Not divisible.

Next, 22: 6432 ÷ 22. Let me divide by 2: 3216. Then divide by 11: 3216 ÷ 11. 11 × 292 = 3212. Remainder 4. Not divisible.

Next, 12: 6432 ÷ 12 = 536, which is integer.

So the highest common factor is 12.

Therefore, adding 11 gives 23.

So I think that's solid.

But just to make sure, let me check if 23 is the correct answer.

Alternatively, maybe I miscalculated something? Let me check once again.

Original numbers: 6432 and 132.

GCF is 12. 12 + 11 = 23.

Yes.

Alternatively, maybe the question is a trick question? Like, maybe the GCF is something else? But no, both methods gave me 12.

Alternatively, maybe I made a mistake in the Euclidean algorithm? Let me check again.

First step: GCF(6432, 132).

Compute 6432 divided by 132.

As above, 132 × 48 = 6336. 6432 - 6336 = 96. So GCF(132, 96).

Then GCF(132, 96).

132 divided by 96 is 1 with remainder 36. So GCF(96, 36).

Then GCF(96, 36).

96 divided by 36 is 2 with remainder 24. GCF(36, 24).

GCF(36, 24).

36 divided by 24 is 1 with remainder 12. GCF(24, 12).

24 divided by 12 is 2 with remainder 0. So GCF is 12.

Yes, that's correct.

Therefore, the answer is 12 + 11 = 23.

I think that's solid.

**Final Answer**
The result is \boxed{23}.

To solve the problem, we are asked to find the **greatest common factor (GCF)** of the numbers **6432** and **132**, and then **add 11** to that result.

---

### Step 1: Find the GCF of 6432 and 132

We can use the **Euclidean Algorithm**, which is efficient for large numbers.

#### Apply the Euclidean Algorithm:

1. **Divide 6432 by 132**:
   $$
   6432 \div 132 = 48 \text{ remainder } 96
   $$
   So, $ \text{GCF}(6432, 132) = \text{GCF}(132, 96) $

2. **Divide 132 by 96**:
   $$
   132 \div 96 = 1 \text{ remainder } 36
   $$
   So, $ \text{GCF}(132, 96) = \text{GCF}(96, 36) $

3. **Divide 96 by 36**:
   $$
   96 \div 36 = 2 \text{ remainder } 24
   $$
   So, $ \text{GCF}(96, 36) = \text{GCF}(36, 24) $

4. **Divide 36 by 24**:
   $$
   36 \div 24 = 1 \text{ remainder } 12
   $$
   So, $ \text{GCF}(36, 24) = \text{GCF}(24, 12) $

5. **Divide 24 by 12**:
   $$
   24 \div 12 = 2 \text{ remainder } 0
   $$
   So, $ \text{GCF}(24, 12) = 12 $

Thus, the **GCF of 6432 and 132 is 12**.

---

### Step 2: Add 11 to the GCF

Now, we are asked to **increase the GCF by 11**:

$$
12 + 11 = 23
$$

---

### Final Answer

$$
\boxed{23}
$$

Solution: We first recognize that $132=11\times 12$, so its prime factorization is $132 = 2^2 \cdot 3 \cdot 11$. We only need to see if these three prime factors will divide into $6432$. Indeed, $6432$ will satisfy the divisibility properties for both $3$ and $4$, and we can long divide to see that $11$ does not divide into $6432$. Thus, the greatest common factor is $3 \times 4 = 12$. The greatest common factor increased by 11 is $12+11 = \boxed{23}$.


Explanation:
Okay, so I need to find tan A in this right triangle ABC where angle B is 90 degrees, and they told me that sin A is equal to 2 times cos A. Hmm, let me think.

First, let me recall some basic trigonometric identities and properties of right triangles. In a right triangle, the sine of an angle is the opposite side over the hypotenuse, and the cosine is the adjacent side over the hypotenuse. Also, tangent is sine over cosine, so tan A would be opposite over adjacent.

But here, they gave me a relationship between sin A and cos A. Specifically, sin A = 2 cos A. Maybe I can use this to find the ratio of the sides, and then figure out tan A?

Let me denote the sides. Let me call the sides of the triangle. Since angle B is the right angle, then sides AB and BC are the legs, and AC is the hypotenuse. Let me label them properly. Let me think: angle A is at vertex A, so the sides opposite and adjacent to angle A would be BC and AB respectively. So, sin A is opposite over hypotenuse, which is BC / AC, and cos A is adjacent over hypotenuse, which is AB / AC.

So according to the problem, sin A = 2 cos A. Therefore, BC / AC = 2 * (AB / AC). If I multiply both sides by AC, that cancels out the denominators, so BC = 2 AB. So that tells me that the side opposite angle A is twice the length of the side adjacent to angle A.

So if I let AB be some length, say, let me assign variables. Let me let AB = x. Then BC = 2x. Then, since it's a right triangle, I can find AC using the Pythagorean theorem.

So AC² = AB² + BC² = x² + (2x)² = x² + 4x² = 5x². Therefore, AC = sqrt(5x²) = x*sqrt(5).

So now, if I want tan A, which is opposite over adjacent, that would be BC / AB = (2x)/x = 2. Wait, so tan A is 2? But let me check if that's consistent with the given condition sin A = 2 cos A.

Let me verify. If tan A is 2, then that means sin A / cos A = 2, so sin A = 2 cos A, which is exactly the given condition. So that seems to check out.

But let me go through it again step by step to make sure I didn't skip anything.

Starting with sin A = 2 cos A. Since in a right triangle, sin A = opposite / hypotenuse, cos A = adjacent / hypotenuse. So if I let the sides be AB, BC, AC with angle A, then:

sin A = BC / AC

cos A = AB / AC

Therefore, BC / AC = 2 * (AB / AC) => BC = 2 AB.

So BC is twice AB. Then, using Pythagoras, AC² = AB² + BC² = AB² + (2 AB)² = AB² + 4 AB² = 5 AB². Therefore, AC = AB * sqrt(5).

Therefore, tan A = BC / AB = 2 AB / AB = 2.

So tan A is 2.

But let me think if there's another way to approach this, maybe using trigonometric identities.

Since sin A = 2 cos A, and we know that sin² A + cos² A = 1. Maybe I can substitute sin A with 2 cos A in that identity.

So substituting, (2 cos A)^2 + cos² A = 1 => 4 cos² A + cos² A = 1 => 5 cos² A = 1 => cos² A = 1/5 => cos A = 1/√5 or -1/√5. But since angle A is in a right triangle, it must be between 0 and 90 degrees, so cosine is positive. Therefore, cos A = 1/√5.

Then sin A = 2 cos A = 2/√5.

Therefore, tan A = sin A / cos A = (2/√5) / (1/√5) = 2.

Same result. So that confirms it.

Alternatively, maybe using the definition of tan A directly. Since tan A = sin A / cos A, and they told us sin A = 2 cos A, so tan A = 2 cos A / cos A = 2.

Wait, that's even simpler. If sin A = 2 cos A, then dividing both sides by cos A (assuming cos A ≠ 0, which it isn't in a right triangle because angle A is acute), we get tan A = 2.

So that's straightforward.

But maybe I need to check if angle A is acute. Since it's a right triangle with angle B = 90°, angles A and C must be acute, so yes, angle A is between 0 and 90°, so tan A is positive, which matches our result.

So all methods lead to tan A = 2.

But let me think again if there's any possible mistake here. For example, could there be a case where sin A = 2 cos A is not possible? Let me check with actual values.

Suppose angle A is such that tan A = 2. Then angle A is arctangent of 2, which is approximately 63.43 degrees. Then sin(arctan 2) is 2 / sqrt(1 + 2²) = 2 / sqrt(5), and cos(arctan 2) is 1 / sqrt(5). Therefore, sin A is indeed twice cos A.

So that works.

Alternatively, if I think of a 1-2-sqrt(5) triangle. If AB = 1, BC = 2, then AC = sqrt(1 + 4) = sqrt(5). Then sin A = 2 / sqrt(5), cos A = 1 / sqrt(5), so sin A = 2 cos A.

Therefore, tan A = 2.

So all these confirmations lead me to believe that tan A is indeed 2.

I think that's solid. So the answer is 2.

**Final Answer**
\boxed{2}

To solve the problem, we are given a right triangle $ \triangle ABC $ with $ \angle B = 90^\circ $, and the condition:

$$
\sin A = 2 \cos A
$$

We are asked to find $ \tan A $.

---

### Step 1: Use the definition of tangent

Recall the fundamental trigonometric identity:

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

Given that $ \sin A = 2 \cos A $, we can substitute this into the expression for $ \tan A $:

$$
\tan A = \frac{2 \cos A}{\cos A} = 2
$$

This is a direct and valid approach, assuming $ \cos A \neq 0 $, which is true in a right triangle where angle $ A $ is acute (i.e., between $ 0^\circ $ and $ 90^\circ $).

---

### Step 2: Verify using the Pythagorean identity

To double-check, we can use the identity:

$$
\sin^2 A + \cos^2 A = 1
$$

Substitute $ \sin A = 2 \cos A $ into the identity:

$$
(2 \cos A)^2 + \cos^2 A = 1 \Rightarrow 4 \cos^2 A + \cos^2 A = 1 \Rightarrow 5 \cos^2 A = 1
$$

$$
\cos^2 A = \frac{1}{5} \Rightarrow \cos A = \frac{1}{\sqrt{5}} \quad (\text{since } \cos A > 0 \text{ for acute angles})
$$

Then:

$$
\sin A = 2 \cos A = \frac{2}{\sqrt{5}}
$$

Now compute $ \tan A $:

$$
\tan A = \frac{\sin A}{\cos A} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2
$$

This confirms our earlier result.

---

### Step 3: Geometric interpretation

Let’s consider the triangle with angle $ A $, and sides:

- Opposite side to $ A $: $ BC $
- Adjacent side to $ A $: $ AB $
- Hypotenuse: $ AC $

From the condition $ \sin A = 2 \cos A $, we have:

$$
\frac{BC}{AC} = 2 \cdot \frac{AB}{AC} \Rightarrow BC = 2 AB
$$

Using the Pythagorean theorem:

$$
AC^2 = AB^2 + BC^2 = AB^2 + (2 AB)^2 = AB^2 + 4 AB^2 = 5 AB^2 \Rightarrow AC = AB \sqrt{5}
$$

Then:

$$
\tan A = \frac{BC}{AB} = \frac{2 AB}{AB} = 2
$$

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to find tan A in this right triangle ABC where angle B is 90 degrees, and they told me that sin A is equal to 2 times cos A. Hmm, let me think.

First, let me recall some basic trigonometric identities and properties of right triangles. In a right triangle, the sine of an angle is the opposite side over the hypotenuse, and the cosine is the adjacent side over the hypotenuse. Also, tangent is sine over cosine, so tan A would be opposite over adjacent.

But here, they gave me a relationship between sin A and cos A. Specifically, sin A = 2 cos A. Maybe I can use this to find the ratio of the sides, and then figure out tan A?

Let me denote the sides. Let me call the sides of the triangle. Since angle B is the right angle, then sides AB and BC are the legs, and AC is the hypotenuse. Let me label them properly. Let me think: angle A is at vertex A, so the sides opposite and adjacent to angle A would be BC and AB respectively. So, sin A is opposite over hypotenuse, which is BC / AC, and cos A is adjacent over hypotenuse, which is AB / AC.

So according to the problem, sin A = 2 cos A. Therefore, BC / AC = 2 * (AB / AC). If I multiply both sides by AC, that cancels out the denominators, so BC = 2 AB. So that tells me that the side opposite angle A is twice the length of the side adjacent to angle A.

So if I let AB be some length, say, let me assign variables. Let me let AB = x. Then BC = 2x. Then, since it's a right triangle, I can find AC using the Pythagorean theorem.

So AC² = AB² + BC² = x² + (2x)² = x² + 4x² = 5x². Therefore, AC = sqrt(5x²) = x*sqrt(5).

So now, if I want tan A, which is opposite over adjacent, that would be BC / AB = (2x)/x = 2. Wait, so tan A is 2? But let me check if that's consistent with the given condition sin A = 2 cos A.

Let me verify. If tan A is 2, then that means sin A / cos A = 2, so sin A = 2 cos A, which is exactly the given condition. So that seems to check out.

But let me go through it again step by step to make sure I didn't skip anything.

Starting with sin A = 2 cos A. Since in a right triangle, sin A = opposite / hypotenuse, cos A = adjacent / hypotenuse. So if I let the sides be AB, BC, AC with angle A, then:

sin A = BC / AC

cos A = AB / AC

Therefore, BC / AC = 2 * (AB / AC) => BC = 2 AB.

So BC is twice AB. Then, using Pythagoras, AC² = AB² + BC² = AB² + (2 AB)² = AB² + 4 AB² = 5 AB². Therefore, AC = AB * sqrt(5).

Therefore, tan A = BC / AB = 2 AB / AB = 2.

So tan A is 2.

But let me think if there's another way to approach this, maybe using trigonometric identities.

Since sin A = 2 cos A, and we know that sin² A + cos² A = 1. Maybe I can substitute sin A with 2 cos A in that identity.

So substituting, (2 cos A)^2 + cos² A = 1 => 4 cos² A + cos² A = 1 => 5 cos² A = 1 => cos² A = 1/5 => cos A = 1/√5 or -1/√5. But since angle A is in a right triangle, it must be between 0 and 90 degrees, so cosine is positive. Therefore, cos A = 1/√5.

Then sin A = 2 cos A = 2/√5.

Therefore, tan A = sin A / cos A = (2/√5) / (1/√5) = 2.

Same result. So that confirms it.

Alternatively, maybe using the definition of tan A directly. Since tan A = sin A / cos A, and they told us sin A = 2 cos A, so tan A = 2 cos A / cos A = 2.

Wait, that's even simpler. If sin A = 2 cos A, then dividing both sides by cos A (assuming cos A ≠ 0, which it isn't in a right triangle because angle A is acute), we get tan A = 2.

So that's straightforward.

But maybe I need to check if angle A is acute. Since it's a right triangle with angle B = 90°, angles A and C must be acute, so yes, angle A is between 0 and 90°, so tan A is positive, which matches our result.

So all methods lead to tan A = 2.

But let me think again if there's any possible mistake here. For example, could there be a case where sin A = 2 cos A is not possible? Let me check with actual values.

Suppose angle A is such that tan A = 2. Then angle A is arctangent of 2, which is approximately 63.43 degrees. Then sin(arctan 2) is 2 / sqrt(1 + 2²) = 2 / sqrt(5), and cos(arctan 2) is 1 / sqrt(5). Therefore, sin A is indeed twice cos A.

So that works.

Alternatively, if I think of a 1-2-sqrt(5) triangle. If AB = 1, BC = 2, then AC = sqrt(1 + 4) = sqrt(5). Then sin A = 2 / sqrt(5), cos A = 1 / sqrt(5), so sin A = 2 cos A.

Therefore, tan A = 2.

So all these confirmations lead me to believe that tan A is indeed 2.

I think that's solid. So the answer is 2.

**Final Answer**
\boxed{2}

To solve the problem, we are given a right triangle $ \triangle ABC $ with $ \angle B = 90^\circ $, and the condition:

$$
\sin A = 2 \cos A
$$

We are asked to find $ \tan A $.

---

### Step 1: Use the definition of tangent

Recall the fundamental trigonometric identity:

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

Given that $ \sin A = 2 \cos A $, we can substitute this into the expression for $ \tan A $:

$$
\tan A = \frac{2 \cos A}{\cos A} = 2
$$

This is a direct and valid approach, assuming $ \cos A \neq 0 $, which is true in a right triangle where angle $ A $ is acute (i.e., between $ 0^\circ $ and $ 90^\circ $).

---

### Step 2: Verify using the Pythagorean identity

To double-check, we can use the identity:

$$
\sin^2 A + \cos^2 A = 1
$$

Substitute $ \sin A = 2 \cos A $ into the identity:

$$
(2 \cos A)^2 + \cos^2 A = 1 \Rightarrow 4 \cos^2 A + \cos^2 A = 1 \Rightarrow 5 \cos^2 A = 1
$$

$$
\cos^2 A = \frac{1}{5} \Rightarrow \cos A = \frac{1}{\sqrt{5}} \quad (\text{since } \cos A > 0 \text{ for acute angles})
$$

Then:

$$
\sin A = 2 \cos A = \frac{2}{\sqrt{5}}
$$

Now compute $ \tan A $:

$$
\tan A = \frac{\sin A}{\cos A} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2
$$

This confirms our earlier result.

---

### Step 3: Geometric interpretation

Let’s consider the triangle with angle $ A $, and sides:

- Opposite side to $ A $: $ BC $
- Adjacent side to $ A $: $ AB $
- Hypotenuse: $ AC $

From the condition $ \sin A = 2 \cos A $, we have:

$$
\frac{BC}{AC} = 2 \cdot \frac{AB}{AC} \Rightarrow BC = 2 AB
$$

Using the Pythagorean theorem:

$$
AC^2 = AB^2 + BC^2 = AB^2 + (2 AB)^2 = AB^2 + 4 AB^2 = 5 AB^2 \Rightarrow AC = AB \sqrt{5}
$$

Then:

$$
\tan A = \frac{BC}{AB} = \frac{2 AB}{AB} = 2
$$

---

### Final Answer

$$
\boxed{2}
$$

Solution:
The triangle is shown below:

[asy]
pair A,B,C;
A = (0,0);
B = (5,0);
C = (5,10);
draw(A--B--C--A);
draw(rightanglemark(C,B,A,16));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
[/asy]

We have $\sin A = \frac{BC}{AC}$ and $\cos A = \frac{AB}{AC}$, so $\sin A = 2\cos A$ gives us $\frac{BC}{AC} = 2\cdot\frac{AB}{AC}$.  Multiplying both sides by $AC$ gives $BC = 2AB$, so $\frac{BC}{AB} = 2$.  Finally, we have $\tan A = \frac{BC}{AB} = \boxed{2}$.

We also could have noted that $\tan A = \frac{\sin A}{\cos A} = \frac{2\cos A}{\cos A } =\boxed{2}$.


Explanation:
Okay, so I need to find the value of N where the sum of the first N positive odd numbers is 121. Hmm, let me think. I remember that there's a formula for the sum of the first N odd numbers. Wait, isn't it something like N squared? Let me check that.

Let me start by recalling some examples. The first odd number is 1. If N is 1, the sum is 1. According to the formula, 1 squared is 1. That works. The first two odd numbers are 1 and 3. Their sum is 4, which is 2 squared. The first three odd numbers: 1 + 3 + 5 = 9, which is 3 squared. Oh, right! So it seems like the sum of the first N odd numbers is N². So if the sum is 121, then N² = 121. Therefore, N is the square root of 121. The square root of 121 is 11. So N should be 11. Let me verify that.

Let me add the first 11 odd numbers to make sure. The sequence is 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21. Let me add them step by step.

Starting with 1 + 3 = 4. Then 4 + 5 = 9. 9 + 7 = 16. 16 + 9 = 25. 25 + 11 = 36. 36 + 13 = 49. 49 + 15 = 64. 64 + 17 = 81. 81 + 19 = 100. 100 + 21 = 121. Yes! That adds up to 121. So N is indeed 11.

But wait, let me make sure I didn't skip any numbers. The first odd number is 1, then each subsequent odd number increases by 2. So for N=11, the last term should be 1 + (11-1)*2. Let me calculate that. 1 + 10*2 = 1 + 20 = 21. Which matches the last term I added. So the sequence is correct.

Alternatively, I can use the formula for the sum of an arithmetic series. The sum of an arithmetic series is (number of terms)/2 times (first term + last term). For the first N odd numbers, the first term is 1, the last term is 2N - 1. So the sum is N/2 * (1 + (2N - 1)) = N/2 * (2N) = N². Yep, that confirms the formula. So if the sum is 121, then N² = 121, so N = sqrt(121) = 11. Since N has to be a positive integer, 11 is the answer.

I think that's solid. Let me check if there's another way someone might approach this problem, maybe by trial and error? Let's see. If I didn't remember the formula, I could try adding odd numbers until I reach 121. Let's try that.

Start with N=1: sum=1. N=2: 1+3=4. N=3: 4+5=9. N=4: 9+7=16. N=5: 16+9=25. N=6: 25+11=36. N=7: 36+13=49. N=8: 49+15=64. N=9: 64+17=81. N=10: 81+19=100. N=11: 100+21=121. Yep, same result. So even if I didn't remember the formula, I could have figured it out by adding them up.

Alternatively, maybe someone might think that the sum of odd numbers is something else? Let me check for N=4. The sum is 1+3+5+7=16, which is 4². So that's correct. So the formula holds.

Another way to think about it: each time you add an odd number, you're essentially completing a square. Like, 1 is 1x1, adding 3 gives 2x2, adding 5 gives 3x3, etc. So each new odd number is the difference between consecutive squares. For example, 2² - 1² = 4 - 1 = 3, which is the second odd number. 3² - 2² = 9 - 4 = 5, the third odd number. So the nth odd number is n² - (n-1)² = 2n - 1. Which makes sense. Therefore, the sum of the first N odd numbers is N².

Therefore, if the sum is 121, N is 11.

I think that's thorough. I can't think of any other way this could be approached. Maybe someone might make a mistake in calculating the square root? For example, if they thought 121 is 10 squared or 12 squared? But 10 squared is 100, 11 squared is 121, 12 squared is 144. So 121 is definitely 11 squared.

Alternatively, if someone thought the formula was different, maybe they might miscalculate. But given that multiple methods confirm that the sum is N squared, and verifying by adding up the numbers, I think the answer is definitely 11.

So, yeah, N is 11.

**Final Answer**
The value of $ N $ is \boxed{11}.

To solve the problem, we are given that the sum of the first $ N $ positive odd numbers is 121, and we are to find the value of $ N $.

---

### Step 1: Understanding the Pattern

The first few positive odd numbers are:

$$
1, 3, 5, 7, 9, \dots
$$

Let’s compute the sum of the first few terms:

- $ N = 1 $: $ 1 = 1^2 $
- $ N = 2 $: $ 1 + 3 = 4 = 2^2 $
- $ N = 3 $: $ 1 + 3 + 5 = 9 = 3^2 $
- $ N = 4 $: $ 1 + 3 + 5 + 7 = 16 = 4^2 $

This pattern suggests that the sum of the first $ N $ positive odd numbers is:

$$
\text{Sum} = N^2
$$

---

### Step 2: Applying the Formula

We are told that the sum is 121. Using the formula:

$$
N^2 = 121
$$

Taking the square root of both sides:

$$
N = \sqrt{121} = 11
$$

---

### Step 3: Verification

To confirm, we can manually add the first 11 odd numbers:

$$
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121
$$

This confirms that the sum of the first 11 odd numbers is indeed 121.

---

### Final Answer

$$
\boxed{11}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of N where the sum of the first N positive odd numbers is 121. Hmm, let me think. I remember that there's a formula for the sum of the first N odd numbers. Wait, isn't it something like N squared? Let me check that.

Let me start by recalling some examples. The first odd number is 1. If N is 1, the sum is 1. According to the formula, 1 squared is 1. That works. The first two odd numbers are 1 and 3. Their sum is 4, which is 2 squared. The first three odd numbers: 1 + 3 + 5 = 9, which is 3 squared. Oh, right! So it seems like the sum of the first N odd numbers is N². So if the sum is 121, then N² = 121. Therefore, N is the square root of 121. The square root of 121 is 11. So N should be 11. Let me verify that.

Let me add the first 11 odd numbers to make sure. The sequence is 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21. Let me add them step by step.

Starting with 1 + 3 = 4. Then 4 + 5 = 9. 9 + 7 = 16. 16 + 9 = 25. 25 + 11 = 36. 36 + 13 = 49. 49 + 15 = 64. 64 + 17 = 81. 81 + 19 = 100. 100 + 21 = 121. Yes! That adds up to 121. So N is indeed 11.

But wait, let me make sure I didn't skip any numbers. The first odd number is 1, then each subsequent odd number increases by 2. So for N=11, the last term should be 1 + (11-1)*2. Let me calculate that. 1 + 10*2 = 1 + 20 = 21. Which matches the last term I added. So the sequence is correct.

Alternatively, I can use the formula for the sum of an arithmetic series. The sum of an arithmetic series is (number of terms)/2 times (first term + last term). For the first N odd numbers, the first term is 1, the last term is 2N - 1. So the sum is N/2 * (1 + (2N - 1)) = N/2 * (2N) = N². Yep, that confirms the formula. So if the sum is 121, then N² = 121, so N = sqrt(121) = 11. Since N has to be a positive integer, 11 is the answer.

I think that's solid. Let me check if there's another way someone might approach this problem, maybe by trial and error? Let's see. If I didn't remember the formula, I could try adding odd numbers until I reach 121. Let's try that.

Start with N=1: sum=1. N=2: 1+3=4. N=3: 4+5=9. N=4: 9+7=16. N=5: 16+9=25. N=6: 25+11=36. N=7: 36+13=49. N=8: 49+15=64. N=9: 64+17=81. N=10: 81+19=100. N=11: 100+21=121. Yep, same result. So even if I didn't remember the formula, I could have figured it out by adding them up.

Alternatively, maybe someone might think that the sum of odd numbers is something else? Let me check for N=4. The sum is 1+3+5+7=16, which is 4². So that's correct. So the formula holds.

Another way to think about it: each time you add an odd number, you're essentially completing a square. Like, 1 is 1x1, adding 3 gives 2x2, adding 5 gives 3x3, etc. So each new odd number is the difference between consecutive squares. For example, 2² - 1² = 4 - 1 = 3, which is the second odd number. 3² - 2² = 9 - 4 = 5, the third odd number. So the nth odd number is n² - (n-1)² = 2n - 1. Which makes sense. Therefore, the sum of the first N odd numbers is N².

Therefore, if the sum is 121, N is 11.

I think that's thorough. I can't think of any other way this could be approached. Maybe someone might make a mistake in calculating the square root? For example, if they thought 121 is 10 squared or 12 squared? But 10 squared is 100, 11 squared is 121, 12 squared is 144. So 121 is definitely 11 squared.

Alternatively, if someone thought the formula was different, maybe they might miscalculate. But given that multiple methods confirm that the sum is N squared, and verifying by adding up the numbers, I think the answer is definitely 11.

So, yeah, N is 11.

**Final Answer**
The value of $ N $ is \boxed{11}.

To solve the problem, we are given that the sum of the first $ N $ positive odd numbers is 121, and we are to find the value of $ N $.

---

### Step 1: Understanding the Pattern

The first few positive odd numbers are:

$$
1, 3, 5, 7, 9, \dots
$$

Let’s compute the sum of the first few terms:

- $ N = 1 $: $ 1 = 1^2 $
- $ N = 2 $: $ 1 + 3 = 4 = 2^2 $
- $ N = 3 $: $ 1 + 3 + 5 = 9 = 3^2 $
- $ N = 4 $: $ 1 + 3 + 5 + 7 = 16 = 4^2 $

This pattern suggests that the sum of the first $ N $ positive odd numbers is:

$$
\text{Sum} = N^2
$$

---

### Step 2: Applying the Formula

We are told that the sum is 121. Using the formula:

$$
N^2 = 121
$$

Taking the square root of both sides:

$$
N = \sqrt{121} = 11
$$

---

### Step 3: Verification

To confirm, we can manually add the first 11 odd numbers:

$$
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121
$$

This confirms that the sum of the first 11 odd numbers is indeed 121.

---

### Final Answer

$$
\boxed{11}
$$

Solution: The first $N$ positive odd integers are 1, 3, $\dots$, $2N - 1$.  The sum of an arithmetic series is equal to the average of the first and last term, multiplied by the number of terms, so the sum of the first $N$ positive odd integers is \[\frac{1 + (2N - 1)}{2} \cdot N = N^2.\]If $N^2 = 121$, then $N = \boxed{11}$.


Explanation:
Okay, so I need to find the value of $ a $ given that the cubic equation $ 8x^3 + 4ax^2 + 2bx + a = 0 $ has three distinct positive roots, and the sum of the base-2 logarithms of the roots is 5. Hmm, let me start by recalling some algebra concepts.

First, since the equation is a cubic with three distinct positive roots, I can use Vieta's formulas to relate the coefficients of the polynomial to the sums and products of its roots. Let me denote the roots as $ r $, $ s $, and $ t $. All of them are positive and distinct.

Vieta's formulas for a cubic equation $ px^3 + qx^2 + rx + s = 0 $ tell us that:

1. $ r + s + t = -q/p $
2. $ rs + rt + st = r/p $
3. $ rst = -s/p $

Wait, let me check that again. For a general cubic equation $ ax^3 + bx^2 + cx + d = 0 $, the sum of roots is $ -b/a $, the sum of products two at a time is $ c/a $, and the product of roots is $ -d/a $. So applying that to our equation:

Given the equation $ 8x^3 + 4a x^2 + 2b x + a = 0 $, so here $ p = 8 $, $ q = 4a $, $ r = 2b $, and $ s = a $. Therefore:

1. Sum of roots: $ r + s + t = -q/p = -4a/8 = -a/2 $
2. Sum of products two at a time: $ rs + rt + st = r/p = 2b/8 = b/4 $
3. Product of roots: $ rst = -s/p = -a/8 $

But wait, the problem says the roots are positive. So the sum of the roots is $ -a/2 $. But since the roots are positive, their sum must be positive. Therefore, $ -a/2 > 0 $ implies that $ a < 0 $. So $ a $ is negative. Got that.

Also, the product of the roots is $ -a/8 $. Since the roots are positive, their product is positive, so $ -a/8 > 0 $, which again implies $ a < 0 $. So that's consistent.

Now, the problem says that the sum of the base-2 logarithms of the roots is 5. Let me write that down:

$ \log_2 r + \log_2 s + \log_2 t = 5 $

Using logarithm properties, this is equivalent to:

$ \log_2 (rst) = 5 $

Which means:

$ rst = 2^5 = 32 $

But from Vieta's formula, we have $ rst = -a/8 $. Therefore:

$ -a/8 = 32 $

Solving for $ a $:

Multiply both sides by 8:

$ -a = 256 $

Multiply both sides by -1:

$ a = -256 $

Wait, is that all? Let me check if there are any other conditions or if I missed something.

Wait, the problem mentions that the roots are distinct and positive. So, we need to make sure that with $ a = -256 $, the equation actually has three distinct positive roots. But maybe there's more to it? Let me think.

But the problem says "for some real numbers $ a $ and $ b $", so maybe there's a relation between $ a $ and $ b $ that we need to check? Wait, but the question only asks for $ a $, so maybe once we find $ a $, we can check if such $ b $ exists? But perhaps the given condition about the sum of logs already gives us $ a $, and the rest is just ensuring that the equation has three distinct positive roots. However, since the problem says "has three distinct positive roots", maybe there's more constraints?

Wait, but maybe I need to check if with $ a = -256 $, the equation can have three distinct positive roots. Let me think. Let me recall that for a cubic equation to have three distinct real roots, its discriminant must be positive. But calculating the discriminant of a cubic might be complicated. Alternatively, maybe we can use Vieta's formulas and other relations?

Alternatively, maybe there's another condition from the sum of the roots and the sum of products?

Wait, let me check again. The problem gives me that the sum of the logs is 5, which gives me $ rst = 32 $, and from Vieta's formula, $ rst = -a/8 $, so $ a = -256 $. But then, what about the other Vieta's formulas? For example, the sum of roots is $ -a/2 = -(-256)/2 = 128 $. So the sum of the roots is 128, and the product is 32. Also, the sum of products two at a time is $ b/4 $, so $ b = 4(rs + rt + st) $. But since the problem doesn't give us any information about $ b $, maybe we don't need to worry about it? The problem only asks for $ a $, so maybe even if there are multiple possible $ b $, as long as there exists some $ b $ such that the equation has three distinct positive roots, then $ a = -256 $ is acceptable?

But maybe there's something else? Let me think. For example, if the roots are positive, then the coefficients have to satisfy certain conditions. For example, the coefficients of the polynomial must be such that the polynomial can be factored with positive roots. But since we already used Vieta's formulas, and the sum and product, maybe that's sufficient?

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

Given $ \log_2 r + \log_2 s + \log_2 t = 5 $, so $ \log_2 (rst) = 5 $, so $ rst = 2^5 = 32 $. Then Vieta's formula says $ rst = -a/8 $, so $ -a/8 = 32 $, so $ a = -256 $. That seems straightforward.

But let me check if there's another condition. For example, the sum of the roots is $ r + s + t = -a/2 = 128 $, and the product is 32. So, if we have three positive numbers with sum 128 and product 32, is that possible? Well, for example, if two roots are very small and one is almost 128, but then the product would be small. Wait, but 32 is quite small compared to 128. Wait, but maybe they are all fractions?

Wait, but if all roots are positive, then their product is 32, and their sum is 128. Let me think of an example. Suppose all three roots are equal? Then each would be 128/3 ≈ 42.666..., and the product would be (128/3)^3 ≈ 237, which is way larger than 32. So if they are equal, the product is too big. Therefore, to get a smaller product, the roots must be smaller. But how can they have a sum of 128 and product 32? Let me think.

Suppose two roots are very small, approaching zero, then the third root approaches 128, and the product approaches zero. So to get a product of 32, the roots can't be too small. Maybe there's a balance. Let me try to think of specific numbers. Let me suppose that two roots are equal? Wait, but the problem says three distinct roots. So they can't be equal. But even so, maybe there's a way to have three distinct positive numbers with sum 128 and product 32. For example, let me take one root as 1, another as 2, then the third root would be 128 - 1 - 2 = 125. Then the product is 1*2*125 = 250, which is still bigger than 32. If I take roots as 1, 1, 126, but they need to be distinct. Let me take 1, 2, 125 as before. If I take smaller numbers, say 0.5, 0.5, 127, but again, they need to be distinct. Let me take 0.5, 1, 126.5. The product is 0.5*1*126.5 = 63.25, still bigger than 32. If I take even smaller numbers, say 0.25, 0.25, 127.5, but again, distinctness. Let me take 0.25, 0.5, 127.25. Product is 0.25*0.5*127.25 ≈ 15.9, which is less than 32. So somewhere between 0.25 and 1, maybe?

But maybe there's a way to have three distinct positive numbers with sum 128 and product 32. Since the product is fixed, and the sum is fixed, maybe such numbers exist? For example, using the AM ≥ GM inequality. The arithmetic mean of the roots is 128/3 ≈ 42.666..., and the geometric mean is (32)^(1/3) ≈ 3.1748. Since AM is much larger than GM, which is possible because the numbers are not all equal. So, yes, such numbers can exist. Therefore, there is no contradiction here. Therefore, even though the numbers seem counterintuitive, they can exist. Therefore, with $ a = -256 $, there exists such roots. Therefore, maybe the answer is $ a = -256 $.

But let me check again if I made a mistake in Vieta's formula. Let me re-derive Vieta's formula for the given polynomial.

Given the cubic equation $ 8x^3 + 4a x^2 + 2b x + a = 0 $. Let me write it as $ 8x^3 + 4a x^2 + 2b x + a = 0 $. Let me divide both sides by 8 to make it monic:

$ x^3 + \frac{4a}{8}x^2 + \frac{2b}{8}x + \frac{a}{8} = 0 $

Simplifying:

$ x^3 + \frac{a}{2}x^2 + \frac{b}{4}x + \frac{a}{8} = 0 $

So, in standard form $ x^3 + px^2 + qx + r = 0 $, we have:

- $ p = \frac{a}{2} $
- $ q = \frac{b}{4} $
- $ r = \frac{a}{8} $

Then Vieta's formula says:

Sum of roots: $ r + s + t = -p = -\frac{a}{2} $

Sum of products two at a time: $ rs + rt + st = q = \frac{b}{4} $

Product of roots: $ rst = -r = -\frac{a}{8} $

Wait, hold on! Wait, in the standard cubic equation $ x^3 + px^2 + qx + r = 0 $, the product of roots is $ -r $. Wait, no, let me check again. For cubic equation $ x^3 + ax^2 + bx + c = 0 $, the product of roots is $ -c $. So in our case, the equation is $ x^3 + \frac{a}{2}x^2 + \frac{b}{4}x + \frac{a}{8} = 0 $, so the product of roots is $ -\frac{a}{8} $. Therefore, $ rst = -\frac{a}{8} $. Which is what I had before. Therefore, since $ rst = 32 $, then $ -\frac{a}{8} = 32 $, so $ a = -256 $. So that's correct.

Therefore, unless there is some other condition that I haven't considered, this should be the answer. However, the problem mentions that the roots are distinct. So, even though we have the product and sum, we need to ensure that there are three distinct positive roots. But how can we be sure? Maybe there's a way that with $ a = -256 $, the equation has three distinct positive roots?

Alternatively, maybe there's a relation between $ a $ and $ b $ that we need to check? For example, if we have the sum of roots and the sum of products, maybe we can find $ b $, but since the problem doesn't give us more information, maybe it's not necessary. However, perhaps there's a condition on the discriminant?

Alternatively, maybe the problem is designed so that once we have $ a = -256 $, the equation will have three distinct positive roots regardless of $ b $? But that seems unlikely. Wait, but $ b $ is related to the sum of products of roots. So, if we have $ rs + rt + st = \frac{b}{4} $, so $ b = 4(rs + rt + st) $. Since $ r, s, t $ are positive, $ rs + rt + st $ is positive, so $ b $ is positive. But since the problem says "for some real numbers $ a $ and $ b $", so as long as there exists some $ b $ such that the equation has three distinct positive roots, then $ a = -256 $ is acceptable. Therefore, even if for some $ b $, the equation might not have three distinct roots, but since the problem says "for some real numbers $ a $ and $ b $", meaning that there exists at least one such $ b $, then $ a = -256 $ is acceptable.

But maybe there's a way to check if with $ a = -256 $, there exists such $ b $ that the equation has three distinct positive roots. For example, if we can find $ b $ such that the cubic has three distinct positive roots. But how?

Alternatively, maybe we can use the fact that if the cubic has three distinct real roots, then its derivative has two distinct real roots, and the function changes sign between them. But this might be complicated.

Alternatively, since we already have the sum and product of roots, maybe we can use the fact that for three positive numbers with given sum and product, there are multiple possibilities for the sum of products. Therefore, as long as the sum of products is positive (which it is, since all roots are positive), then there exists such $ b $. Therefore, since $ rs + rt + st $ is positive, $ b $ is positive, so there exists such $ b $. Therefore, the answer is $ a = -256 $.

But let me check again if there's a mistake in the problem statement. Wait, the original equation is $ 8x^3 + 4a x^2 + 2b x + a = 0 $. So when I divided by 8, I got $ x^3 + (a/2)x^2 + (b/4)x + a/8 = 0 $. So Vieta's formula for product of roots is $ -a/8 $, which is correct. So if $ rst = 32 $, then $ -a/8 = 32 $, so $ a = -256 $. Therefore, I think that is correct.

Alternatively, maybe the problem is expecting a positive value for $ a $? But according to the logic, since the sum of roots is $ -a/2 $, which is positive, so $ a $ must be negative. Therefore, $ a = -256 $ is negative, which is consistent.

Alternatively, maybe I made a mistake in interpreting the sum of the logarithms? Let me check again. The problem says "the sum of the base-2 logarithms of the roots is 5". So $ \log_2 r + \log_2 s + \log_2 t = 5 $. Which is $ \log_2 (rst) = 5 $, so $ rst = 2^5 = 32 $. That's correct.

Therefore, I think my answer is correct. Therefore, the value of $ a $ is $ -256 $.

But let me check with an example. Suppose that the roots are such that their product is 32 and their sum is 128. Let me try to find such roots. Let me assume that one of the roots is 1, then the other two roots would have product 32 and sum 127. Let me call them $ m $ and $ n $, so $ m + n = 127 $, $ mn = 32 $. Then solving quadratic equation $ x^2 - 127x + 32 = 0 $. The discriminant is $ 127^2 - 4*1*32 = 16129 - 128 = 16001 $. Which is positive, so two real roots. Therefore, there are two real roots, so with the third root being 1, we have three real roots. However, are they distinct? Well, since the quadratic equation has two distinct roots (since discriminant is positive), and 1 is different from them? Let me check. Suppose the roots of the quadratic are $ m $ and $ n $. If 1 is not equal to either $ m $ or $ n $, then all three roots are distinct. Let me check if 1 is a root of the quadratic equation $ x^2 - 127x + 32 = 0 $. Plugging in x=1: 1 - 127 + 32 = -94 ≠ 0. Therefore, 1 is not a root, so all three roots are distinct. Therefore, such roots exist. Therefore, with $ a = -256 $, there exists such roots. Therefore, the answer is correct.

Therefore, I think the answer is $ \boxed{-256} $.

**Final Answer**
\boxed{-256}

To solve the problem, we are given the cubic equation:

$$
8x^3 + 4ax^2 + 2bx + a = 0
$$

and told that it has **three distinct positive roots**. Additionally, we are told that the **sum of the base-2 logarithms of the roots is 5**. Our goal is to find the value of $ a $.

---

### Step 1: Use the Logarithmic Condition

Let the roots of the cubic be $ r, s, t $, all positive and distinct. The condition:

$$
\log_2 r + \log_2 s + \log_2 t = 5
$$

can be rewritten using logarithmic properties:

$$
\log_2 (rst) = 5 \Rightarrow rst = 2^5 = 32
$$

So, the **product of the roots** is 32.

---

### Step 2: Apply Vieta's Formulas

We now apply **Vieta's formulas** to the cubic equation:

$$
8x^3 + 4ax^2 + 2bx + a = 0
$$

Let’s divide the entire equation by 8 to make it monic:

$$
x^3 + \frac{a}{2}x^2 + \frac{b}{4}x + \frac{a}{8} = 0
$$

Now, using Vieta’s formulas for a cubic equation $ x^3 + px^2 + qx + r = 0 $, we have:

- Sum of roots: $ r + s + t = -p = -\frac{a}{2} $
- Sum of products of roots two at a time: $ rs + rt + st = q = \frac{b}{4} $
- Product of roots: $ rst = -r = -\frac{a}{8} $

From the product of roots:

$$
rst = -\frac{a}{8}
$$

But we already found that $ rst = 32 $, so:

$$
-\frac{a}{8} = 32 \Rightarrow a = -256
$$

---

### Step 3: Verify Consistency with the Given Conditions

We now check whether this value of $ a $ is consistent with the rest of the problem.

- The sum of the roots is $ r + s + t = -\frac{a}{2} = \frac{256}{2} = 128 $
- The product of the roots is $ rst = 32 $
- The sum of the logs is $ \log_2(rst) = \log_2(32) = 5 $, which matches the given condition

So, the values are consistent.

---

### Step 4: Ensure the Roots Are Distinct and Positive

We are told the roots are **distinct and positive**. Since the product $ rst = 32 $ and the sum $ r + s + t = 128 $, we can check whether such a set of positive numbers exists.

Let’s suppose two of the roots are small, say $ r = 1 $, $ s = 2 $, then the third root $ t = 128 - 1 - 2 = 125 $, and the product is $ 1 \cdot 2 \cdot 125 = 250 $, which is **greater than 32**.

To get a smaller product, we can try smaller values. For example, if $ r = 0.5 $, $ s = 0.5 $, then $ t = 128 - 0.5 - 0.5 = 127 $, and the product is $ 0.5 \cdot 0.5 \cdot 127 = 31.75 $, which is **close to 32**.

Thus, it is **possible** to find three **distinct positive** numbers with sum 128 and product 32. Therefore, the cubic equation with $ a = -256 $ can indeed have three distinct positive roots.

---

### Final Answer

$$
\boxed{-256}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ a $ given that the cubic equation $ 8x^3 + 4ax^2 + 2bx + a = 0 $ has three distinct positive roots, and the sum of the base-2 logarithms of the roots is 5. Hmm, let me start by recalling some algebra concepts.

First, since the equation is a cubic with three distinct positive roots, I can use Vieta's formulas to relate the coefficients of the polynomial to the sums and products of its roots. Let me denote the roots as $ r $, $ s $, and $ t $. All of them are positive and distinct.

Vieta's formulas for a cubic equation $ px^3 + qx^2 + rx + s = 0 $ tell us that:

1. $ r + s + t = -q/p $
2. $ rs + rt + st = r/p $
3. $ rst = -s/p $

Wait, let me check that again. For a general cubic equation $ ax^3 + bx^2 + cx + d = 0 $, the sum of roots is $ -b/a $, the sum of products two at a time is $ c/a $, and the product of roots is $ -d/a $. So applying that to our equation:

Given the equation $ 8x^3 + 4a x^2 + 2b x + a = 0 $, so here $ p = 8 $, $ q = 4a $, $ r = 2b $, and $ s = a $. Therefore:

1. Sum of roots: $ r + s + t = -q/p = -4a/8 = -a/2 $
2. Sum of products two at a time: $ rs + rt + st = r/p = 2b/8 = b/4 $
3. Product of roots: $ rst = -s/p = -a/8 $

But wait, the problem says the roots are positive. So the sum of the roots is $ -a/2 $. But since the roots are positive, their sum must be positive. Therefore, $ -a/2 > 0 $ implies that $ a < 0 $. So $ a $ is negative. Got that.

Also, the product of the roots is $ -a/8 $. Since the roots are positive, their product is positive, so $ -a/8 > 0 $, which again implies $ a < 0 $. So that's consistent.

Now, the problem says that the sum of the base-2 logarithms of the roots is 5. Let me write that down:

$ \log_2 r + \log_2 s + \log_2 t = 5 $

Using logarithm properties, this is equivalent to:

$ \log_2 (rst) = 5 $

Which means:

$ rst = 2^5 = 32 $

But from Vieta's formula, we have $ rst = -a/8 $. Therefore:

$ -a/8 = 32 $

Solving for $ a $:

Multiply both sides by 8:

$ -a = 256 $

Multiply both sides by -1:

$ a = -256 $

Wait, is that all? Let me check if there are any other conditions or if I missed something.

Wait, the problem mentions that the roots are distinct and positive. So, we need to make sure that with $ a = -256 $, the equation actually has three distinct positive roots. But maybe there's more to it? Let me think.

But the problem says "for some real numbers $ a $ and $ b $", so maybe there's a relation between $ a $ and $ b $ that we need to check? Wait, but the question only asks for $ a $, so maybe once we find $ a $, we can check if such $ b $ exists? But perhaps the given condition about the sum of logs already gives us $ a $, and the rest is just ensuring that the equation has three distinct positive roots. However, since the problem says "has three distinct positive roots", maybe there's more constraints?

Wait, but maybe I need to check if with $ a = -256 $, the equation can have three distinct positive roots. Let me think. Let me recall that for a cubic equation to have three distinct real roots, its discriminant must be positive. But calculating the discriminant of a cubic might be complicated. Alternatively, maybe we can use Vieta's formulas and other relations?

Alternatively, maybe there's another condition from the sum of the roots and the sum of products?

Wait, let me check again. The problem gives me that the sum of the logs is 5, which gives me $ rst = 32 $, and from Vieta's formula, $ rst = -a/8 $, so $ a = -256 $. But then, what about the other Vieta's formulas? For example, the sum of roots is $ -a/2 = -(-256)/2 = 128 $. So the sum of the roots is 128, and the product is 32. Also, the sum of products two at a time is $ b/4 $, so $ b = 4(rs + rt + st) $. But since the problem doesn't give us any information about $ b $, maybe we don't need to worry about it? The problem only asks for $ a $, so maybe even if there are multiple possible $ b $, as long as there exists some $ b $ such that the equation has three distinct positive roots, then $ a = -256 $ is acceptable?

But maybe there's something else? Let me think. For example, if the roots are positive, then the coefficients have to satisfy certain conditions. For example, the coefficients of the polynomial must be such that the polynomial can be factored with positive roots. But since we already used Vieta's formulas, and the sum and product, maybe that's sufficient?

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

Given $ \log_2 r + \log_2 s + \log_2 t = 5 $, so $ \log_2 (rst) = 5 $, so $ rst = 2^5 = 32 $. Then Vieta's formula says $ rst = -a/8 $, so $ -a/8 = 32 $, so $ a = -256 $. That seems straightforward.

But let me check if there's another condition. For example, the sum of the roots is $ r + s + t = -a/2 = 128 $, and the product is 32. So, if we have three positive numbers with sum 128 and product 32, is that possible? Well, for example, if two roots are very small and one is almost 128, but then the product would be small. Wait, but 32 is quite small compared to 128. Wait, but maybe they are all fractions?

Wait, but if all roots are positive, then their product is 32, and their sum is 128. Let me think of an example. Suppose all three roots are equal? Then each would be 128/3 ≈ 42.666..., and the product would be (128/3)^3 ≈ 237, which is way larger than 32. So if they are equal, the product is too big. Therefore, to get a smaller product, the roots must be smaller. But how can they have a sum of 128 and product 32? Let me think.

Suppose two roots are very small, approaching zero, then the third root approaches 128, and the product approaches zero. So to get a product of 32, the roots can't be too small. Maybe there's a balance. Let me try to think of specific numbers. Let me suppose that two roots are equal? Wait, but the problem says three distinct roots. So they can't be equal. But even so, maybe there's a way to have three distinct positive numbers with sum 128 and product 32. For example, let me take one root as 1, another as 2, then the third root would be 128 - 1 - 2 = 125. Then the product is 1*2*125 = 250, which is still bigger than 32. If I take roots as 1, 1, 126, but they need to be distinct. Let me take 1, 2, 125 as before. If I take smaller numbers, say 0.5, 0.5, 127, but again, they need to be distinct. Let me take 0.5, 1, 126.5. The product is 0.5*1*126.5 = 63.25, still bigger than 32. If I take even smaller numbers, say 0.25, 0.25, 127.5, but again, distinctness. Let me take 0.25, 0.5, 127.25. Product is 0.25*0.5*127.25 ≈ 15.9, which is less than 32. So somewhere between 0.25 and 1, maybe?

But maybe there's a way to have three distinct positive numbers with sum 128 and product 32. Since the product is fixed, and the sum is fixed, maybe such numbers exist? For example, using the AM ≥ GM inequality. The arithmetic mean of the roots is 128/3 ≈ 42.666..., and the geometric mean is (32)^(1/3) ≈ 3.1748. Since AM is much larger than GM, which is possible because the numbers are not all equal. So, yes, such numbers can exist. Therefore, there is no contradiction here. Therefore, even though the numbers seem counterintuitive, they can exist. Therefore, with $ a = -256 $, there exists such roots. Therefore, maybe the answer is $ a = -256 $.

But let me check again if I made a mistake in Vieta's formula. Let me re-derive Vieta's formula for the given polynomial.

Given the cubic equation $ 8x^3 + 4a x^2 + 2b x + a = 0 $. Let me write it as $ 8x^3 + 4a x^2 + 2b x + a = 0 $. Let me divide both sides by 8 to make it monic:

$ x^3 + \frac{4a}{8}x^2 + \frac{2b}{8}x + \frac{a}{8} = 0 $

Simplifying:

$ x^3 + \frac{a}{2}x^2 + \frac{b}{4}x + \frac{a}{8} = 0 $

So, in standard form $ x^3 + px^2 + qx + r = 0 $, we have:

- $ p = \frac{a}{2} $
- $ q = \frac{b}{4} $
- $ r = \frac{a}{8} $

Then Vieta's formula says:

Sum of roots: $ r + s + t = -p = -\frac{a}{2} $

Sum of products two at a time: $ rs + rt + st = q = \frac{b}{4} $

Product of roots: $ rst = -r = -\frac{a}{8} $

Wait, hold on! Wait, in the standard cubic equation $ x^3 + px^2 + qx + r = 0 $, the product of roots is $ -r $. Wait, no, let me check again. For cubic equation $ x^3 + ax^2 + bx + c = 0 $, the product of roots is $ -c $. So in our case, the equation is $ x^3 + \frac{a}{2}x^2 + \frac{b}{4}x + \frac{a}{8} = 0 $, so the product of roots is $ -\frac{a}{8} $. Therefore, $ rst = -\frac{a}{8} $. Which is what I had before. Therefore, since $ rst = 32 $, then $ -\frac{a}{8} = 32 $, so $ a = -256 $. So that's correct.

Therefore, unless there is some other condition that I haven't considered, this should be the answer. However, the problem mentions that the roots are distinct. So, even though we have the product and sum, we need to ensure that there are three distinct positive roots. But how can we be sure? Maybe there's a way that with $ a = -256 $, the equation has three distinct positive roots?

Alternatively, maybe there's a relation between $ a $ and $ b $ that we need to check? For example, if we have the sum of roots and the sum of products, maybe we can find $ b $, but since the problem doesn't give us more information, maybe it's not necessary. However, perhaps there's a condition on the discriminant?

Alternatively, maybe the problem is designed so that once we have $ a = -256 $, the equation will have three distinct positive roots regardless of $ b $? But that seems unlikely. Wait, but $ b $ is related to the sum of products of roots. So, if we have $ rs + rt + st = \frac{b}{4} $, so $ b = 4(rs + rt + st) $. Since $ r, s, t $ are positive, $ rs + rt + st $ is positive, so $ b $ is positive. But since the problem says "for some real numbers $ a $ and $ b $", so as long as there exists some $ b $ such that the equation has three distinct positive roots, then $ a = -256 $ is acceptable. Therefore, even if for some $ b $, the equation might not have three distinct roots, but since the problem says "for some real numbers $ a $ and $ b $", meaning that there exists at least one such $ b $, then $ a = -256 $ is acceptable.

But maybe there's a way to check if with $ a = -256 $, there exists such $ b $ that the equation has three distinct positive roots. For example, if we can find $ b $ such that the cubic has three distinct positive roots. But how?

Alternatively, maybe we can use the fact that if the cubic has three distinct real roots, then its derivative has two distinct real roots, and the function changes sign between them. But this might be complicated.

Alternatively, since we already have the sum and product of roots, maybe we can use the fact that for three positive numbers with given sum and product, there are multiple possibilities for the sum of products. Therefore, as long as the sum of products is positive (which it is, since all roots are positive), then there exists such $ b $. Therefore, since $ rs + rt + st $ is positive, $ b $ is positive, so there exists such $ b $. Therefore, the answer is $ a = -256 $.

But let me check again if there's a mistake in the problem statement. Wait, the original equation is $ 8x^3 + 4a x^2 + 2b x + a = 0 $. So when I divided by 8, I got $ x^3 + (a/2)x^2 + (b/4)x + a/8 = 0 $. So Vieta's formula for product of roots is $ -a/8 $, which is correct. So if $ rst = 32 $, then $ -a/8 = 32 $, so $ a = -256 $. Therefore, I think that is correct.

Alternatively, maybe the problem is expecting a positive value for $ a $? But according to the logic, since the sum of roots is $ -a/2 $, which is positive, so $ a $ must be negative. Therefore, $ a = -256 $ is negative, which is consistent.

Alternatively, maybe I made a mistake in interpreting the sum of the logarithms? Let me check again. The problem says "the sum of the base-2 logarithms of the roots is 5". So $ \log_2 r + \log_2 s + \log_2 t = 5 $. Which is $ \log_2 (rst) = 5 $, so $ rst = 2^5 = 32 $. That's correct.

Therefore, I think my answer is correct. Therefore, the value of $ a $ is $ -256 $.

But let me check with an example. Suppose that the roots are such that their product is 32 and their sum is 128. Let me try to find such roots. Let me assume that one of the roots is 1, then the other two roots would have product 32 and sum 127. Let me call them $ m $ and $ n $, so $ m + n = 127 $, $ mn = 32 $. Then solving quadratic equation $ x^2 - 127x + 32 = 0 $. The discriminant is $ 127^2 - 4*1*32 = 16129 - 128 = 16001 $. Which is positive, so two real roots. Therefore, there are two real roots, so with the third root being 1, we have three real roots. However, are they distinct? Well, since the quadratic equation has two distinct roots (since discriminant is positive), and 1 is different from them? Let me check. Suppose the roots of the quadratic are $ m $ and $ n $. If 1 is not equal to either $ m $ or $ n $, then all three roots are distinct. Let me check if 1 is a root of the quadratic equation $ x^2 - 127x + 32 = 0 $. Plugging in x=1: 1 - 127 + 32 = -94 ≠ 0. Therefore, 1 is not a root, so all three roots are distinct. Therefore, such roots exist. Therefore, with $ a = -256 $, there exists such roots. Therefore, the answer is correct.

Therefore, I think the answer is $ \boxed{-256} $.

**Final Answer**
\boxed{-256}

To solve the problem, we are given the cubic equation:

$$
8x^3 + 4ax^2 + 2bx + a = 0
$$

and told that it has **three distinct positive roots**. Additionally, we are told that the **sum of the base-2 logarithms of the roots is 5**. Our goal is to find the value of $ a $.

---

### Step 1: Use the Logarithmic Condition

Let the roots of the cubic be $ r, s, t $, all positive and distinct. The condition:

$$
\log_2 r + \log_2 s + \log_2 t = 5
$$

can be rewritten using logarithmic properties:

$$
\log_2 (rst) = 5 \Rightarrow rst = 2^5 = 32
$$

So, the **product of the roots** is 32.

---

### Step 2: Apply Vieta's Formulas

We now apply **Vieta's formulas** to the cubic equation:

$$
8x^3 + 4ax^2 + 2bx + a = 0
$$

Let’s divide the entire equation by 8 to make it monic:

$$
x^3 + \frac{a}{2}x^2 + \frac{b}{4}x + \frac{a}{8} = 0
$$

Now, using Vieta’s formulas for a cubic equation $ x^3 + px^2 + qx + r = 0 $, we have:

- Sum of roots: $ r + s + t = -p = -\frac{a}{2} $
- Sum of products of roots two at a time: $ rs + rt + st = q = \frac{b}{4} $
- Product of roots: $ rst = -r = -\frac{a}{8} $

From the product of roots:

$$
rst = -\frac{a}{8}
$$

But we already found that $ rst = 32 $, so:

$$
-\frac{a}{8} = 32 \Rightarrow a = -256
$$

---

### Step 3: Verify Consistency with the Given Conditions

We now check whether this value of $ a $ is consistent with the rest of the problem.

- The sum of the roots is $ r + s + t = -\frac{a}{2} = \frac{256}{2} = 128 $
- The product of the roots is $ rst = 32 $
- The sum of the logs is $ \log_2(rst) = \log_2(32) = 5 $, which matches the given condition

So, the values are consistent.

---

### Step 4: Ensure the Roots Are Distinct and Positive

We are told the roots are **distinct and positive**. Since the product $ rst = 32 $ and the sum $ r + s + t = 128 $, we can check whether such a set of positive numbers exists.

Let’s suppose two of the roots are small, say $ r = 1 $, $ s = 2 $, then the third root $ t = 128 - 1 - 2 = 125 $, and the product is $ 1 \cdot 2 \cdot 125 = 250 $, which is **greater than 32**.

To get a smaller product, we can try smaller values. For example, if $ r = 0.5 $, $ s = 0.5 $, then $ t = 128 - 0.5 - 0.5 = 127 $, and the product is $ 0.5 \cdot 0.5 \cdot 127 = 31.75 $, which is **close to 32**.

Thus, it is **possible** to find three **distinct positive** numbers with sum 128 and product 32. Therefore, the cubic equation with $ a = -256 $ can indeed have three distinct positive roots.

---

### Final Answer

$$
\boxed{-256}
$$

Solution:
Let $r_1, r_2$, and $r_3$ be the roots. Then \[
5= \log_2r_1 + \log_2 r_2 + \log_2 r_3 = \log_2r_1r_2r_3,
\]so $r_1r_2r_3 = 2^5 = 32$. Since \[
8x^{3}+4ax^{2}+2bx+a=8(x-r_1)(x-r_2)(x-r_3),
\]it follows that $a = -8r_1r_2r_3= \boxed{-256}$.


Explanation:
Okay, so I need to expand and simplify this algebraic expression: x times (x times (1 + x) plus 2x) minus 3 times (x squared minus x plus 2). Let me take it step by step. First, I should probably start by looking at the innermost parentheses and work my way out.

Starting with the first part: x(1 + x). Wait, no, actually, the expression is x multiplied by [x(1 + x) + 2x]. So maybe I should first handle the part inside the brackets. Let me write that down.

So inside the brackets, there's x(1 + x) + 2x. Let me expand x(1 + x) first. Multiplying x by 1 gives x, and multiplying x by x gives x squared. So that part becomes x + x². Then we add 2x to that. So combining like terms, x + 2x is 3x. Therefore, the entire expression inside the brackets simplifies to x² + 3x.

Wait, let me check that again. Original inside the brackets: x(1 + x) + 2x. Expanding x(1 + x) is x*1 + x*x = x + x². Then adding 2x gives x + x² + 2x. Combining x and 2x gives 3x, so yes, it's x² + 3x. Okay, that seems right.

So now the original expression becomes x multiplied by (x² + 3x) minus 3 times (x² - x + 2). Let me write that down:

x(x² + 3x) - 3(x² - x + 2)

Now I need to expand both of these terms. Let me start with the first term: x multiplied by (x² + 3x). Distribute the x to each term inside the parentheses.

Multiplying x by x² gives x³, and multiplying x by 3x gives 3x². So the first part becomes x³ + 3x².

Now the second part is -3 multiplied by (x² - x + 2). Let me distribute the -3 to each term inside the parentheses.

First term: -3 * x² = -3x²

Second term: -3 * (-x) = +3x

Third term: -3 * 2 = -6

So putting that together, the second part is -3x² + 3x - 6.

Now, combining both parts: the first part was x³ + 3x² and the second part is -3x² + 3x - 6. So adding them together:

x³ + 3x² - 3x² + 3x - 6

Now let me combine like terms. The 3x² and -3x² cancel each other out because 3x² - 3x² is 0. So that leaves us with x³ + 3x - 6.

Wait, is that all? Let me check again.

Original expression after expansion:

First part: x³ + 3x²

Second part: -3x² + 3x - 6

Adding them together:

x³ + 3x² - 3x² + 3x - 6

Yes, the 3x² and -3x² cancel. So we have x³ + 3x - 6.

Is there anything else to simplify? Let me check if there are any more like terms. The terms are x³, 3x, and -6. There are no other like terms. So that should be the simplified form.

Wait, but let me go back through each step again to make sure I didn't make a mistake.

Starting with the original expression:

x[ x(1 + x) + 2x ] - 3(x² - x + 2)

First, inside the brackets: x(1 + x) + 2x.

x(1 + x) is x + x². Then adding 2x gives x + x² + 2x = x² + 3x. Correct.

Then multiplying by x: x*(x² + 3x) = x³ + 3x². Correct.

Then subtracting 3*(x² - x + 2). Distribute the -3: -3x² + 3x -6. Correct.

Combine with previous terms: x³ + 3x² -3x² + 3x -6. The 3x² -3x² cancels. So x³ + 3x -6.

Yes, that seems right. Let me check if I can factor anything or if there's a different way to write it, but the question says to expand and simplify completely. Since there are no like terms left, this should be the final answer.

Alternatively, maybe I can check by plugging in a value for x. Let me test with x = 1.

Original expression:

First compute the original expression at x=1:

1*(1*(1 + 1) + 2*1) - 3*(1² -1 + 2)

Compute step by step:

Inside the first bracket: 1*(1*(2) + 2) = 1*(2 + 2) = 1*4 = 4

Then subtract 3*(1 -1 + 2) = 3*(0 + 2) = 3*2 = 6

So total expression is 4 - 6 = -2

Now compute my simplified expression x³ + 3x -6 at x=1:

1³ + 3*1 -6 = 1 + 3 -6 = -2. Same result. Good.

Let me try another value, say x=2.

Original expression:

2*(2*(1 + 2) + 2*2) - 3*(4 - 2 + 2)

Compute step by step:

Inside first bracket: 2*(2*3 + 4) = 2*(6 + 4) = 2*10 = 20

Subtract 3*(4 -2 +2) = 3*(4) = 12

Total expression: 20 - 12 = 8

Simplified expression: 2³ + 3*2 -6 = 8 + 6 -6 = 8. Correct again.

Another test, x=0.

Original expression:

0*(0*(1 + 0) + 2*0) - 3*(0 - 0 + 2) = 0*(0 + 0) - 3*(2) = 0 -6 = -6

Simplified expression: 0³ + 3*0 -6 = 0 + 0 -6 = -6. Correct.

Okay, so seems like my simplified expression is correct. Therefore, the answer is x³ + 3x -6.

**Final Answer**
The simplified expression is \boxed{x^3 + 3x - 6}.

To simplify the expression:

$$
x\left(x(1+x)+2x\right)-3(x^2-x+2)
$$

we proceed step by step, focusing on expanding and combining like terms.

---

### Step 1: Simplify the inner expression

Start with the innermost parentheses:

$$
x(1 + x) = x + x^2
$$

Now add the remaining term inside the brackets:

$$
x(1 + x) + 2x = (x + x^2) + 2x = x^2 + 3x
$$

So the expression becomes:

$$
x(x^2 + 3x) - 3(x^2 - x + 2)
$$

---

### Step 2: Expand each term

**First term:**

$$
x(x^2 + 3x) = x \cdot x^2 + x \cdot 3x = x^3 + 3x^2
$$

**Second term:**

$$
-3(x^2 - x + 2) = -3x^2 + 3x - 6
$$

---

### Step 3: Combine all terms

Now combine the expanded parts:

$$
x^3 + 3x^2 - 3x^2 + 3x - 6
$$

Group like terms:

- $ x^3 $ remains as is.
- $ 3x^2 - 3x^2 = 0 $
- $ 3x $ remains as is.
- $ -6 $ remains as is.

So the simplified expression is:

$$
x^3 + 3x - 6
$$

---

### Final Answer

$$
\boxed{x^3 + 3x - 6}
$$

[Assistant]
Explanation:
Okay, so I need to expand and simplify this algebraic expression: x times (x times (1 + x) plus 2x) minus 3 times (x squared minus x plus 2). Let me take it step by step. First, I should probably start by looking at the innermost parentheses and work my way out.

Starting with the first part: x(1 + x). Wait, no, actually, the expression is x multiplied by [x(1 + x) + 2x]. So maybe I should first handle the part inside the brackets. Let me write that down.

So inside the brackets, there's x(1 + x) + 2x. Let me expand x(1 + x) first. Multiplying x by 1 gives x, and multiplying x by x gives x squared. So that part becomes x + x². Then we add 2x to that. So combining like terms, x + 2x is 3x. Therefore, the entire expression inside the brackets simplifies to x² + 3x.

Wait, let me check that again. Original inside the brackets: x(1 + x) + 2x. Expanding x(1 + x) is x*1 + x*x = x + x². Then adding 2x gives x + x² + 2x. Combining x and 2x gives 3x, so yes, it's x² + 3x. Okay, that seems right.

So now the original expression becomes x multiplied by (x² + 3x) minus 3 times (x² - x + 2). Let me write that down:

x(x² + 3x) - 3(x² - x + 2)

Now I need to expand both of these terms. Let me start with the first term: x multiplied by (x² + 3x). Distribute the x to each term inside the parentheses.

Multiplying x by x² gives x³, and multiplying x by 3x gives 3x². So the first part becomes x³ + 3x².

Now the second part is -3 multiplied by (x² - x + 2). Let me distribute the -3 to each term inside the parentheses.

First term: -3 * x² = -3x²

Second term: -3 * (-x) = +3x

Third term: -3 * 2 = -6

So putting that together, the second part is -3x² + 3x - 6.

Now, combining both parts: the first part was x³ + 3x² and the second part is -3x² + 3x - 6. So adding them together:

x³ + 3x² - 3x² + 3x - 6

Now let me combine like terms. The 3x² and -3x² cancel each other out because 3x² - 3x² is 0. So that leaves us with x³ + 3x - 6.

Wait, is that all? Let me check again.

Original expression after expansion:

First part: x³ + 3x²

Second part: -3x² + 3x - 6

Adding them together:

x³ + 3x² - 3x² + 3x - 6

Yes, the 3x² and -3x² cancel. So we have x³ + 3x - 6.

Is there anything else to simplify? Let me check if there are any more like terms. The terms are x³, 3x, and -6. There are no other like terms. So that should be the simplified form.

Wait, but let me go back through each step again to make sure I didn't make a mistake.

Starting with the original expression:

x[ x(1 + x) + 2x ] - 3(x² - x + 2)

First, inside the brackets: x(1 + x) + 2x.

x(1 + x) is x + x². Then adding 2x gives x + x² + 2x = x² + 3x. Correct.

Then multiplying by x: x*(x² + 3x) = x³ + 3x². Correct.

Then subtracting 3*(x² - x + 2). Distribute the -3: -3x² + 3x -6. Correct.

Combine with previous terms: x³ + 3x² -3x² + 3x -6. The 3x² -3x² cancels. So x³ + 3x -6.

Yes, that seems right. Let me check if I can factor anything or if there's a different way to write it, but the question says to expand and simplify completely. Since there are no like terms left, this should be the final answer.

Alternatively, maybe I can check by plugging in a value for x. Let me test with x = 1.

Original expression:

First compute the original expression at x=1:

1*(1*(1 + 1) + 2*1) - 3*(1² -1 + 2)

Compute step by step:

Inside the first bracket: 1*(1*(2) + 2) = 1*(2 + 2) = 1*4 = 4

Then subtract 3*(1 -1 + 2) = 3*(0 + 2) = 3*2 = 6

So total expression is 4 - 6 = -2

Now compute my simplified expression x³ + 3x -6 at x=1:

1³ + 3*1 -6 = 1 + 3 -6 = -2. Same result. Good.

Let me try another value, say x=2.

Original expression:

2*(2*(1 + 2) + 2*2) - 3*(4 - 2 + 2)

Compute step by step:

Inside first bracket: 2*(2*3 + 4) = 2*(6 + 4) = 2*10 = 20

Subtract 3*(4 -2 +2) = 3*(4) = 12

Total expression: 20 - 12 = 8

Simplified expression: 2³ + 3*2 -6 = 8 + 6 -6 = 8. Correct again.

Another test, x=0.

Original expression:

0*(0*(1 + 0) + 2*0) - 3*(0 - 0 + 2) = 0*(0 + 0) - 3*(2) = 0 -6 = -6

Simplified expression: 0³ + 3*0 -6 = 0 + 0 -6 = -6. Correct.

Okay, so seems like my simplified expression is correct. Therefore, the answer is x³ + 3x -6.

**Final Answer**
The simplified expression is \boxed{x^3 + 3x - 6}.

To simplify the expression:

$$
x\left(x(1+x)+2x\right)-3(x^2-x+2)
$$

we proceed step by step, focusing on expanding and combining like terms.

---

### Step 1: Simplify the inner expression

Start with the innermost parentheses:

$$
x(1 + x) = x + x^2
$$

Now add the remaining term inside the brackets:

$$
x(1 + x) + 2x = (x + x^2) + 2x = x^2 + 3x
$$

So the expression becomes:

$$
x(x^2 + 3x) - 3(x^2 - x + 2)
$$

---

### Step 2: Expand each term

**First term:**

$$
x(x^2 + 3x) = x \cdot x^2 + x \cdot 3x = x^3 + 3x^2
$$

**Second term:**

$$
-3(x^2 - x + 2) = -3x^2 + 3x - 6
$$

---

### Step 3: Combine all terms

Now combine the expanded parts:

$$
x^3 + 3x^2 - 3x^2 + 3x - 6
$$

Group like terms:

- $ x^3 $ remains as is.
- $ 3x^2 - 3x^2 = 0 $
- $ 3x $ remains as is.
- $ -6 $ remains as is.

So the simplified expression is:

$$
x^3 + 3x - 6
$$

---

### Final Answer

$$
\boxed{x^3 + 3x - 6}
$$

Solution:
Begin by distributing in the innermost parentheses: \begin{align*}
&\ \ \ \ x\left(x(1+x)+2x\right)-3(x^2-x+2) \\&= x(x+x^2+2x) - 3(x^2-x+2)
\end{align*} Now, distribute again: \begin{align*}
x^2+x^3+2x^2-3x^2+3x-6
\end{align*} Finally, combine like terms to get \begin{align*}
\boxed{x^3+3x-6}
\end{align*}


Explanation:
Okay, let's see. I need to figure out how many treeks weigh the same as one squig. The problem gives me two equations, so I should probably set up some variables and solve them. Let me start by assigning variables to each of the creatures. Let me call the weight of a treek "T", a squig "S", and a goolee "G".

The first statement says: Ten treeks weigh as much as three squigs and one goolee. So in equation form, that would be 10T = 3S + G. Got that.

The second statement says: Two treeks and one goolee are equal in weight to one squig. So that would be 2T + G = S.

So now I have two equations:

1) 10T = 3S + G

2) 2T + G = S

And the question is asking: The combined weight of how many treeks equals the weight of one squig? So they want to know what number multiplied by T equals S. In other words, find x such that xT = S. So x is the number of treeks equivalent to one squig.

Alright, so I need to solve these equations to find S in terms of T, or maybe find the ratio between S and T. Let me see. Since equation 2 gives me S in terms of T and G, maybe I can substitute that into equation 1.

From equation 2: S = 2T + G. So I can replace S in equation 1 with that expression. Let me do that.

Substituting into equation 1: 10T = 3*(2T + G) + G. Let me compute the right side.

First, 3*(2T + G) is 6T + 3G. Then adding the G from the original equation gives 6T + 3G + G = 6T + 4G.

So equation 1 becomes: 10T = 6T + 4G.

Now, subtract 6T from both sides: 10T - 6T = 4G => 4T = 4G.

Divide both sides by 4: T = G.

Oh! So a treek weighs the same as a goolee? Interesting. So T = G.

Now, going back to equation 2: S = 2T + G. But since G = T, substitute that in: S = 2T + T = 3T.

So S = 3T. Therefore, one squig weighs the same as three treeks.

Wait, so the combined weight of three treeks equals the weight of one squig. Therefore, the answer is 3?

Let me check my steps again to make sure I didn't make a mistake.

Starting with equations:

1) 10T = 3S + G

2) 2T + G = S

Substituted equation 2 into equation 1:

10T = 3*(2T + G) + G => 10T = 6T + 3G + G => 10T = 6T + 4G => 4T = 4G => T = G.

Then equation 2: S = 2T + G = 2T + T = 3T.

Yes, that seems correct. So S = 3T. Therefore, one squig is equal to three treeks.

But let me verify with the original equations. Let's check if these values satisfy both equations.

First equation: 10T = 3S + G. If S = 3T and G = T, then the right side is 3*(3T) + T = 9T + T = 10T. Which equals the left side. Good.

Second equation: 2T + G = S. Substituting, 2T + T = 3T = S. Which is correct.

So both equations are satisfied. Therefore, the answer is 3.

But wait, the problem says "the combined weight of how many treeks equals the weight of one squig?" So that would be 3 treeks.

But let me think again. Maybe I misread the problem? Let me check again.

Original problem:

Ten treeks weigh as much as three squigs and one goolee.

So 10T = 3S + G.

Two treeks and one goolee are equal in weight to one squig.

So 2T + G = S.

Yes, that's what I used.

So solving gives S = 3T. Therefore, 3 treeks equal one squig.

Therefore, the answer is 3.

But just to make sure, let me try plugging in numbers. Let me assign a value to T. Let's say T = 1 (treek weighs 1 unit). Then G = T = 1. Then S = 3T = 3.

Check first equation: 10T = 10*1 = 10.

3S + G = 3*3 + 1 = 9 + 1 = 10. Correct.

Second equation: 2T + G = 2*1 + 1 = 3, which is equal to S. Correct.

So yes, that works. Therefore, the answer is 3.

I think that's solid. So the combined weight of three treeks equals one squig.

**Final Answer**
The combined weight of \boxed{3} treeks equals the weight of one squig.

To determine how many treeks weigh the same as one squig, we begin by translating the given information into algebraic equations.

---

### Step 1: Define Variables

Let:
- $ T $ = weight of one treek
- $ S $ = weight of one squig
- $ G $ = weight of one goolee

---

### Step 2: Translate the Problem into Equations

From the problem, we are given two key relationships:

1. **Ten treeks weigh as much as three squigs and one goolee:**
   $$
   10T = 3S + G \tag{1}
   $$

2. **Two treeks and one goolee are equal in weight to one squig:**
   $$
   2T + G = S \tag{2}
   $$

---

### Step 3: Solve the System of Equations

From equation (2), solve for $ S $ in terms of $ T $ and $ G $:
$$
S = 2T + G \tag{3}
$$

Now substitute equation (3) into equation (1):
$$
10T = 3(2T + G) + G
$$
$$
10T = 6T + 3G + G
$$
$$
10T = 6T + 4G
$$
$$
4T = 4G \Rightarrow T = G \tag{4}
$$

Now substitute $ T = G $ into equation (3):
$$
S = 2T + T = 3T
$$

---

### Step 4: Interpret the Result

We now know:
$$
S = 3T
$$

This means **one squig weighs the same as three treeks**.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, let's see. I need to figure out how many treeks weigh the same as one squig. The problem gives me two equations, so I should probably set up some variables and solve them. Let me start by assigning variables to each of the creatures. Let me call the weight of a treek "T", a squig "S", and a goolee "G".

The first statement says: Ten treeks weigh as much as three squigs and one goolee. So in equation form, that would be 10T = 3S + G. Got that.

The second statement says: Two treeks and one goolee are equal in weight to one squig. So that would be 2T + G = S.

So now I have two equations:

1) 10T = 3S + G

2) 2T + G = S

And the question is asking: The combined weight of how many treeks equals the weight of one squig? So they want to know what number multiplied by T equals S. In other words, find x such that xT = S. So x is the number of treeks equivalent to one squig.

Alright, so I need to solve these equations to find S in terms of T, or maybe find the ratio between S and T. Let me see. Since equation 2 gives me S in terms of T and G, maybe I can substitute that into equation 1.

From equation 2: S = 2T + G. So I can replace S in equation 1 with that expression. Let me do that.

Substituting into equation 1: 10T = 3*(2T + G) + G. Let me compute the right side.

First, 3*(2T + G) is 6T + 3G. Then adding the G from the original equation gives 6T + 3G + G = 6T + 4G.

So equation 1 becomes: 10T = 6T + 4G.

Now, subtract 6T from both sides: 10T - 6T = 4G => 4T = 4G.

Divide both sides by 4: T = G.

Oh! So a treek weighs the same as a goolee? Interesting. So T = G.

Now, going back to equation 2: S = 2T + G. But since G = T, substitute that in: S = 2T + T = 3T.

So S = 3T. Therefore, one squig weighs the same as three treeks.

Wait, so the combined weight of three treeks equals the weight of one squig. Therefore, the answer is 3?

Let me check my steps again to make sure I didn't make a mistake.

Starting with equations:

1) 10T = 3S + G

2) 2T + G = S

Substituted equation 2 into equation 1:

10T = 3*(2T + G) + G => 10T = 6T + 3G + G => 10T = 6T + 4G => 4T = 4G => T = G.

Then equation 2: S = 2T + G = 2T + T = 3T.

Yes, that seems correct. So S = 3T. Therefore, one squig is equal to three treeks.

But let me verify with the original equations. Let's check if these values satisfy both equations.

First equation: 10T = 3S + G. If S = 3T and G = T, then the right side is 3*(3T) + T = 9T + T = 10T. Which equals the left side. Good.

Second equation: 2T + G = S. Substituting, 2T + T = 3T = S. Which is correct.

So both equations are satisfied. Therefore, the answer is 3.

But wait, the problem says "the combined weight of how many treeks equals the weight of one squig?" So that would be 3 treeks.

But let me think again. Maybe I misread the problem? Let me check again.

Original problem:

Ten treeks weigh as much as three squigs and one goolee.

So 10T = 3S + G.

Two treeks and one goolee are equal in weight to one squig.

So 2T + G = S.

Yes, that's what I used.

So solving gives S = 3T. Therefore, 3 treeks equal one squig.

Therefore, the answer is 3.

But just to make sure, let me try plugging in numbers. Let me assign a value to T. Let's say T = 1 (treek weighs 1 unit). Then G = T = 1. Then S = 3T = 3.

Check first equation: 10T = 10*1 = 10.

3S + G = 3*3 + 1 = 9 + 1 = 10. Correct.

Second equation: 2T + G = 2*1 + 1 = 3, which is equal to S. Correct.

So yes, that works. Therefore, the answer is 3.

I think that's solid. So the combined weight of three treeks equals one squig.

**Final Answer**
The combined weight of \boxed{3} treeks equals the weight of one squig.

To determine how many treeks weigh the same as one squig, we begin by translating the given information into algebraic equations.

---

### Step 1: Define Variables

Let:
- $ T $ = weight of one treek
- $ S $ = weight of one squig
- $ G $ = weight of one goolee

---

### Step 2: Translate the Problem into Equations

From the problem, we are given two key relationships:

1. **Ten treeks weigh as much as three squigs and one goolee:**
   $$
   10T = 3S + G \tag{1}
   $$

2. **Two treeks and one goolee are equal in weight to one squig:**
   $$
   2T + G = S \tag{2}
   $$

---

### Step 3: Solve the System of Equations

From equation (2), solve for $ S $ in terms of $ T $ and $ G $:
$$
S = 2T + G \tag{3}
$$

Now substitute equation (3) into equation (1):
$$
10T = 3(2T + G) + G
$$
$$
10T = 6T + 3G + G
$$
$$
10T = 6T + 4G
$$
$$
4T = 4G \Rightarrow T = G \tag{4}
$$

Now substitute $ T = G $ into equation (3):
$$
S = 2T + T = 3T
$$

---

### Step 4: Interpret the Result

We now know:
$$
S = 3T
$$

This means **one squig weighs the same as three treeks**.

---

### Final Answer

$$
\boxed{3}
$$

Solution:
Let $t,s,g$ be the weight of one treek, the weight of one squig, and the weight of one goolee, respectively. Then the given information tells us \begin{align*}
10t &=3s+g\\
2t +g &= s.
\end{align*} Since we would like to solve for $s$ in terms of $t$, we want to eliminate $g$. Add the two equations to obtain \begin{align*}
10t+2t+g &= 3s+g+s\\
\Rightarrow 10t+2t &= 3s+s\\
\Rightarrow 4s &= 12t\\
\Rightarrow s &=3t.
\end{align*} So one squig weighs $\boxed{3}$ treeks.