Explanation:
Okay, so I need to find the inverse of 17 modulo 83. That means I need to find some number x such that when I multiply 17 by x and then divide by 83, the remainder is 1. In other words, 17x ≡ 1 mod 83. Hmm, how do I approach this?

The problem also mentions that 17 times 5 is 85, which is 83 + 2. Maybe that's a hint? Let me think. Since 85 is congruent to 2 mod 83, that means 17*5 ≡ 2 mod 83. So if I can use this somehow to find the inverse...

Let me recall that to find the modular inverse, one method is the Extended Euclidean Algorithm. Maybe I should try that? Let me remember how that works. The Extended Euclidean Algorithm not only finds the greatest common divisor (gcd) of two numbers but also finds integers x and y such that ax + by = gcd(a, b). In this case, since 17 and 83 are coprime (because 83 is a prime number and 17 is less than 83 and not a multiple of it), their gcd should be 1. Therefore, there exist integers x and y such that 17x + 83y = 1. The x here would be the inverse of 17 modulo 83.

Alright, let me try applying the Extended Euclidean Algorithm step by step.

First, divide 83 by 17:

83 divided by 17 is 4 with a remainder. Let me compute 17*4 = 68. Then 83 - 68 = 15. So 83 = 17*4 + 15.

Next step: divide 17 by 15.

17 divided by 15 is 1 with a remainder of 2. So 17 = 15*1 + 2.

Then divide 15 by 2:

15 divided by 2 is 7 with a remainder of 1. So 15 = 2*7 + 1.

Finally, divide 2 by 1:

2 divided by 1 is 2 with a remainder of 0. So 2 = 1*2 + 0.

Since the last non-zero remainder is 1, the gcd is 1, which confirms that the inverse exists.

Now, working backwards to express 1 as a combination of 17 and 83.

Starting from the equation 1 = 15 - 2*7.

But 2 is from the previous step: 2 = 17 - 15*1.

Substitute that into the equation:

1 = 15 - (17 - 15*1)*7
= 15 - 17*7 + 15*7
= 15*8 - 17*7

Now, 15 is from the first step: 15 = 83 - 17*4.

Substitute that in:

1 = (83 - 17*4)*8 - 17*7
= 83*8 - 17*32 - 17*7
= 83*8 - 17*(32 + 7)
= 83*8 - 17*39

So, rearranged, this is:

-17*39 + 83*8 = 1

Which means that -39*17 ≡ 1 mod 83. Therefore, the inverse of 17 modulo 83 is -39. But we need a positive residue between 0 and 82. To convert -39 to a positive number modulo 83, we can add 83 to it:

-39 + 83 = 44.

So, 44 is the inverse? Let me check that. Let me compute 17*44 mod 83.

First, 17*44. Let me compute 17*40 = 680, and 17*4 = 68, so total is 680 + 68 = 748. Now divide 748 by 83 and find the remainder.

Let me see how many times 83 goes into 748. Let's compute 83*9 = 747. Oh! So 83*9 = 747, so 748 - 747 = 1. Therefore, 17*44 ≡ 1 mod 83. Perfect! That works. So 44 is the inverse. But wait, the problem mentioned that 17*5 = 85, which is 2 mod 83. Maybe there's another way to get there?

Alternatively, since 17*5 ≡ 2 mod 83, maybe I can use that to find the inverse? Let me think.

If 17*5 ≡ 2 mod 83, then multiplying both sides by something might get me to 1. For instance, if I can find a number x such that 2x ≡ 1 mod 83, then multiplying both sides by x would give me 17*5x ≡ 1 mod 83, so 17*(5x) ≡ 1 mod 83, so 5x would be the inverse. So first, find the inverse of 2 mod 83, then multiply by 5?

Let me try that. Let's find the inverse of 2 mod 83. That's easier, maybe? Let me think.

We need y such that 2y ≡ 1 mod 83. Which means 2y - 1 is divisible by 83, so 2y = 83k + 1 for some integer k. Trying small k:

k=1: 83 +1 = 84. 84 /2=42. So y=42. Let me check: 2*42=84. 84 mod 83 is 1. Yes! So inverse of 2 mod 83 is 42. Therefore, going back, the inverse of 17 is 5*42 mod 83. Let me compute that.

5*42 = 210. Now compute 210 mod 83. Let me divide 210 by 83. 83*2=166. 210-166=44. So 210 mod 83 is 44. Which matches the previous result. So that's another way to get 44. Therefore, 44 is indeed the inverse.

But maybe there's another approach? Let me think. Since 17*5 = 85 ≡ 2 mod 83, then 17*5*42 ≡ 2*42 ≡ 1 mod 83, so that's another way. But it seems that both methods lead to 44. So maybe 44 is the correct answer. Let me check once more.

Alternatively, maybe I can use Fermat's little theorem? Since 83 is prime, then for any a not divisible by 83, a^(82) ≡ 1 mod 83. Therefore, the inverse of a is a^(81) mod 83. So maybe compute 17^81 mod 83? But that seems computationally intensive. But maybe there's a smarter way?

Alternatively, maybe using exponentiation by squaring? But since 81 is a large exponent, maybe not necessary here. But let me see if I can do it.

But since I already have 44 as the inverse through two different methods, maybe it's safe. However, perhaps I should verify once more by multiplying 17 and 44.

So 17*44. As before, 17*40=680, 17*4=68. 680+68=748. Now divide 748 by 83. Let me do this division properly. 83*9=747. So 748-747=1. So 748=83*9 +1. Therefore 748 mod 83 is 1. Therefore 17*44 ≡ 1 mod 83. Perfect. So that's definitely correct.

But just to make sure I didn't make a mistake earlier, let me check another way. Maybe using the Extended Euclidean Algorithm steps again.

So we had:

83 = 17*4 + 15

17 = 15*1 + 2

15 = 2*7 + 1

2 = 1*2 + 0

Then working backwards:

1 = 15 - 2*7

But 2 = 17 - 15*1, so substitute:

1 = 15 - (17 - 15*1)*7 = 15 -17*7 +15*7 = 15*8 -17*7

Then 15 = 83 -17*4, substitute:

1 = (83 -17*4)*8 -17*7 = 83*8 -17*32 -17*7 = 83*8 -17*(32+7) = 83*8 -17*39

Therefore, -39*17 +83*8 =1, so -39*17 ≡1 mod 83. Hence, inverse is -39. Then -39 mod 83 is 83-39=44. Correct. So that's consistent again.

Alternatively, maybe there's a way to use the given hint of 17*5=85≡2 mod 83. So if I can write 17*5 ≡2, then 17*5*42≡2*42≡84≡1 mod83, so 17*(5*42)=17*210≡1 mod83. Then 210 mod83 is 210-2*83=210-166=44. So that's the same as before.

Alternatively, maybe we can solve the equation 17x ≡1 mod83. Let me try to solve it directly. Let me think about solving 17x ≡1 mod83. Which is equivalent to 17x -83y=1. So again, same as the Extended Euclidean Algorithm. But perhaps I can do some trial and error? Though with numbers up to 83, it's a bit tedious, but maybe?

Alternatively, maybe using the fact that 17x ≡1 mod83 means that 17x -1 is divisible by 83. So 17x -1 =83k, so 17x -83k=1. So solving for x and k. But again, same as the Diophantine equation. Maybe not helpful.

Alternatively, perhaps using some patterns. For example, since 17*5=85≡2, then if I can find a multiple of 2 that gives 1 mod83, then multiply by 5. Which is the same as before. So inverse of 2 is 42, so inverse of 17 is 5*42=210≡44. So that's again the same.

Alternatively, maybe using some other relations. Let me think. Suppose we have 17x ≡1 mod83. Let me try to find x by successive substitutions or something. But since 17 is relatively prime to 83, and there are 82 possible inverses, maybe not the best approach.

Alternatively, maybe using the fact that 17* something = 1 mod83. Let me think of 17*? Let me see, trying numbers:

But maybe there's a smarter way. Let me think of 17x ≡1 mod83. Let me write x as (1 + 83k)/17. So we need 1 + 83k divisible by 17. Let me find k such that 1 +83k ≡0 mod17. Which is equivalent to 83k ≡ -1 mod17. Let me compute 83 mod17. 17*4=68, 83-68=15. So 83≡15 mod17. Therefore, equation is 15k ≡-1 mod17. Which is 15k ≡16 mod17. Now, need to solve for k.

So find k such that 15k ≡16 mod17. Let me find inverse of 15 mod17. Since 15 and 17 are coprime. Let me find the inverse of 15 mod17. Let me use Extended Euclidean Algorithm again for 15 and 17.

Divide 17 by 15: 1 with remainder 2. 17=15*1+2

Divide 15 by 2: 7 with remainder 1. 15=2*7 +1

Divide 2 by 1: 2 with remainder 0. So gcd is 1.

Working backwards:

1 =15 -2*7

But 2=17 -15*1, so substitute:

1=15 - (17 -15*1)*7 =15 -17*7 +15*7 =15*8 -17*7

Therefore, 1=15*8 -17*7, so 15*8 ≡1 mod17. Therefore inverse of 15 mod17 is 8. Therefore, going back to the equation 15k ≡16 mod17. Multiply both sides by 8:

k ≡16*8 mod17. 16*8=128. 128 divided by 17: 17*7=119, 128-119=9. So k≡9 mod17. Therefore, k=9 +17m for some integer m. Then x=(1 +83k)/17. Let me take k=9. Then x=(1 +83*9)/17. Compute 83*9: 80*9=720, 3*9=27, total 747. Then 747 +1=748. Then 748 divided by 17. Let me check: 17*44=748, as before. So x=44. Therefore, that's the same result. So this method also gives 44.

So regardless of the method, it seems that 44 is the inverse. Therefore, I think I can be confident that the answer is 44.

But just to make sure, let me check another value. For example, let me compute 17*44 mod83. As before, 17*44=748. Now divide 748 by 83. Let me do this division step by step. 83*8=664. 748-664=84. 84 is larger than 83, so subtract 83: 84-83=1. Therefore, 748=83*8 +84=83*8 +83 +1=83*9 +1. Therefore, 748 mod83 is 1. Perfect. So that's correct.

Therefore, the inverse of 17 modulo 83 is indeed 44. So the answer is 44. Therefore, I think that's solid.

But just to recap all steps once again:

1. Recognized that finding the inverse of 17 mod83 requires solving 17x ≡1 mod83.
2. Used the Extended Euclidean Algorithm to find that -39*17 ≡1 mod83, so -39 mod83 is 44.
3. Verified using the given hint that 17*5≡2 mod83, then found inverse of 2 mod83 is 42, so 5*42=210≡44 mod83.
4. Tried solving the equation via another method by expressing x as (1 +83k)/17 and found k=9 leading to x=44.
5. Verified by direct computation that 17*44=748≡1 mod83.

All methods converge to 44. Therefore, 44 is the correct answer.

**Final Answer**
The inverse of $17$ modulo $83$ is $\boxed{44}$.

To find the modular inverse of $ 17 \mod 83 $, we want a residue $ x $ such that:

$$
17x \equiv 1 \pmod{83}
$$

This means we are looking for an integer $ x $ in the range $ 0 \leq x < 83 $ such that when $ 17x $ is divided by 83, the remainder is 1. We will solve this using the **Extended Euclidean Algorithm**, which is a reliable and structured method for computing modular inverses.

---

### Step 1: Apply the Extended Euclidean Algorithm

We begin by dividing 83 by 17:

$$
83 = 17 \cdot 4 + 15
$$

$$
17 = 15 \cdot 1 + 2
$$

$$
15 = 2 \cdot 7 + 1
$$

$$
2 = 1 \cdot 2 + 0
$$

The last nonzero remainder is 1, so the gcd of 17 and 83 is 1 (which confirms that 17 has an inverse modulo 83).

---

### Step 2: Work Backwards to Express 1 as a Linear Combination

We use the previous equations to express 1 as a combination of 17 and 83:

1. From the third equation:
   $$
   1 = 15 - 2 \cdot 7
   $$

2. From the second equation:
   $$
   2 = 17 - 15 \cdot 1
   $$

   Substituting into the previous step:
   $$
   1 = 15 - (17 - 15) \cdot 7 = 15 - 17 \cdot 7 + 15 \cdot 7 = 15 \cdot 8 - 17 \cdot 7
   $$

3. From the first equation:
   $$
   15 = 83 - 17 \cdot 4
   $$

   Substitute into the previous expression:
   $$
   1 = (83 - 17 \cdot 4) \cdot 8 - 17 \cdot 7 = 83 \cdot 8 - 17 \cdot 32 - 17 \cdot 7 = 83 \cdot 8 - 17 \cdot 39
   $$

Thus, we have:

$$
-39 \cdot 17 + 83 \cdot 8 = 1
$$

This tells us that the modular inverse of 17 modulo 83 is:

$$
-39 \mod 83
$$

To convert $ -39 $ to a positive integer between 0 and 82:

$$
-39 + 83 = 44
$$

---

### Step 3: Verification

We verify that $ 17 \cdot 44 \equiv 1 \mod 83 $:

$$
17 \cdot 44 = 748
$$

$$
748 \mod 83 = 748 - 83 \cdot 9 = 748 - 747 = 1
$$

So, the modular inverse is correct.

---

### Optional Alternative: Use the Given Hint

We are given that $ 17 \cdot 5 = 85 \equiv 2 \mod 83 $. Then, we can use this to find the inverse:

$$
17 \cdot 5 \equiv 2 \mod 83 \Rightarrow 17 \cdot 5 \cdot x \equiv 1 \mod 83 \Rightarrow x \equiv \frac{1}{2} \mod 83
$$

So, we need the inverse of 2 modulo 83. It's easy to find that:

$$
2 \cdot 42 = 84 \Rightarrow 84 \equiv 1 \mod 83 \Rightarrow \text{Inverse of } 2 \mod 83 = 42
$$

So, the inverse of 17 is:

$$
5 \cdot 42 = 210 \Rightarrow 210 \mod 83 = 44
$$

This confirms our previous result.

---

### ✅ Final Answer

$$
\boxed{44}
$$

[Assistant]
Explanation:
Okay, so I need to find the inverse of 17 modulo 83. That means I need to find some number x such that when I multiply 17 by x and then divide by 83, the remainder is 1. In other words, 17x ≡ 1 mod 83. Hmm, how do I approach this?

The problem also mentions that 17 times 5 is 85, which is 83 + 2. Maybe that's a hint? Let me think. Since 85 is congruent to 2 mod 83, that means 17*5 ≡ 2 mod 83. So if I can use this somehow to find the inverse...

Let me recall that to find the modular inverse, one method is the Extended Euclidean Algorithm. Maybe I should try that? Let me remember how that works. The Extended Euclidean Algorithm not only finds the greatest common divisor (gcd) of two numbers but also finds integers x and y such that ax + by = gcd(a, b). In this case, since 17 and 83 are coprime (because 83 is a prime number and 17 is less than 83 and not a multiple of it), their gcd should be 1. Therefore, there exist integers x and y such that 17x + 83y = 1. The x here would be the inverse of 17 modulo 83.

Alright, let me try applying the Extended Euclidean Algorithm step by step.

First, divide 83 by 17:

83 divided by 17 is 4 with a remainder. Let me compute 17*4 = 68. Then 83 - 68 = 15. So 83 = 17*4 + 15.

Next step: divide 17 by 15.

17 divided by 15 is 1 with a remainder of 2. So 17 = 15*1 + 2.

Then divide 15 by 2:

15 divided by 2 is 7 with a remainder of 1. So 15 = 2*7 + 1.

Finally, divide 2 by 1:

2 divided by 1 is 2 with a remainder of 0. So 2 = 1*2 + 0.

Since the last non-zero remainder is 1, the gcd is 1, which confirms that the inverse exists.

Now, working backwards to express 1 as a combination of 17 and 83.

Starting from the equation 1 = 15 - 2*7.

But 2 is from the previous step: 2 = 17 - 15*1.

Substitute that into the equation:

1 = 15 - (17 - 15*1)*7
= 15 - 17*7 + 15*7
= 15*8 - 17*7

Now, 15 is from the first step: 15 = 83 - 17*4.

Substitute that in:

1 = (83 - 17*4)*8 - 17*7
= 83*8 - 17*32 - 17*7
= 83*8 - 17*(32 + 7)
= 83*8 - 17*39

So, rearranged, this is:

-17*39 + 83*8 = 1

Which means that -39*17 ≡ 1 mod 83. Therefore, the inverse of 17 modulo 83 is -39. But we need a positive residue between 0 and 82. To convert -39 to a positive number modulo 83, we can add 83 to it:

-39 + 83 = 44.

So, 44 is the inverse? Let me check that. Let me compute 17*44 mod 83.

First, 17*44. Let me compute 17*40 = 680, and 17*4 = 68, so total is 680 + 68 = 748. Now divide 748 by 83 and find the remainder.

Let me see how many times 83 goes into 748. Let's compute 83*9 = 747. Oh! So 83*9 = 747, so 748 - 747 = 1. Therefore, 17*44 ≡ 1 mod 83. Perfect! That works. So 44 is the inverse. But wait, the problem mentioned that 17*5 = 85, which is 2 mod 83. Maybe there's another way to get there?

Alternatively, since 17*5 ≡ 2 mod 83, maybe I can use that to find the inverse? Let me think.

If 17*5 ≡ 2 mod 83, then multiplying both sides by something might get me to 1. For instance, if I can find a number x such that 2x ≡ 1 mod 83, then multiplying both sides by x would give me 17*5x ≡ 1 mod 83, so 17*(5x) ≡ 1 mod 83, so 5x would be the inverse. So first, find the inverse of 2 mod 83, then multiply by 5?

Let me try that. Let's find the inverse of 2 mod 83. That's easier, maybe? Let me think.

We need y such that 2y ≡ 1 mod 83. Which means 2y - 1 is divisible by 83, so 2y = 83k + 1 for some integer k. Trying small k:

k=1: 83 +1 = 84. 84 /2=42. So y=42. Let me check: 2*42=84. 84 mod 83 is 1. Yes! So inverse of 2 mod 83 is 42. Therefore, going back, the inverse of 17 is 5*42 mod 83. Let me compute that.

5*42 = 210. Now compute 210 mod 83. Let me divide 210 by 83. 83*2=166. 210-166=44. So 210 mod 83 is 44. Which matches the previous result. So that's another way to get 44. Therefore, 44 is indeed the inverse.

But maybe there's another approach? Let me think. Since 17*5 = 85 ≡ 2 mod 83, then 17*5*42 ≡ 2*42 ≡ 1 mod 83, so that's another way. But it seems that both methods lead to 44. So maybe 44 is the correct answer. Let me check once more.

Alternatively, maybe I can use Fermat's little theorem? Since 83 is prime, then for any a not divisible by 83, a^(82) ≡ 1 mod 83. Therefore, the inverse of a is a^(81) mod 83. So maybe compute 17^81 mod 83? But that seems computationally intensive. But maybe there's a smarter way?

Alternatively, maybe using exponentiation by squaring? But since 81 is a large exponent, maybe not necessary here. But let me see if I can do it.

But since I already have 44 as the inverse through two different methods, maybe it's safe. However, perhaps I should verify once more by multiplying 17 and 44.

So 17*44. As before, 17*40=680, 17*4=68. 680+68=748. Now divide 748 by 83. Let me do this division properly. 83*9=747. So 748-747=1. So 748=83*9 +1. Therefore 748 mod 83 is 1. Therefore 17*44 ≡ 1 mod 83. Perfect. So that's definitely correct.

But just to make sure I didn't make a mistake earlier, let me check another way. Maybe using the Extended Euclidean Algorithm steps again.

So we had:

83 = 17*4 + 15

17 = 15*1 + 2

15 = 2*7 + 1

2 = 1*2 + 0

Then working backwards:

1 = 15 - 2*7

But 2 = 17 - 15*1, so substitute:

1 = 15 - (17 - 15*1)*7 = 15 -17*7 +15*7 = 15*8 -17*7

Then 15 = 83 -17*4, substitute:

1 = (83 -17*4)*8 -17*7 = 83*8 -17*32 -17*7 = 83*8 -17*(32+7) = 83*8 -17*39

Therefore, -39*17 +83*8 =1, so -39*17 ≡1 mod 83. Hence, inverse is -39. Then -39 mod 83 is 83-39=44. Correct. So that's consistent again.

Alternatively, maybe there's a way to use the given hint of 17*5=85≡2 mod 83. So if I can write 17*5 ≡2, then 17*5*42≡2*42≡84≡1 mod83, so 17*(5*42)=17*210≡1 mod83. Then 210 mod83 is 210-2*83=210-166=44. So that's the same as before.

Alternatively, maybe we can solve the equation 17x ≡1 mod83. Let me try to solve it directly. Let me think about solving 17x ≡1 mod83. Which is equivalent to 17x -83y=1. So again, same as the Extended Euclidean Algorithm. But perhaps I can do some trial and error? Though with numbers up to 83, it's a bit tedious, but maybe?

Alternatively, maybe using the fact that 17x ≡1 mod83 means that 17x -1 is divisible by 83. So 17x -1 =83k, so 17x -83k=1. So solving for x and k. But again, same as the Diophantine equation. Maybe not helpful.

Alternatively, perhaps using some patterns. For example, since 17*5=85≡2, then if I can find a multiple of 2 that gives 1 mod83, then multiply by 5. Which is the same as before. So inverse of 2 is 42, so inverse of 17 is 5*42=210≡44. So that's again the same.

Alternatively, maybe using some other relations. Let me think. Suppose we have 17x ≡1 mod83. Let me try to find x by successive substitutions or something. But since 17 is relatively prime to 83, and there are 82 possible inverses, maybe not the best approach.

Alternatively, maybe using the fact that 17* something = 1 mod83. Let me think of 17*? Let me see, trying numbers:

But maybe there's a smarter way. Let me think of 17x ≡1 mod83. Let me write x as (1 + 83k)/17. So we need 1 + 83k divisible by 17. Let me find k such that 1 +83k ≡0 mod17. Which is equivalent to 83k ≡ -1 mod17. Let me compute 83 mod17. 17*4=68, 83-68=15. So 83≡15 mod17. Therefore, equation is 15k ≡-1 mod17. Which is 15k ≡16 mod17. Now, need to solve for k.

So find k such that 15k ≡16 mod17. Let me find inverse of 15 mod17. Since 15 and 17 are coprime. Let me find the inverse of 15 mod17. Let me use Extended Euclidean Algorithm again for 15 and 17.

Divide 17 by 15: 1 with remainder 2. 17=15*1+2

Divide 15 by 2: 7 with remainder 1. 15=2*7 +1

Divide 2 by 1: 2 with remainder 0. So gcd is 1.

Working backwards:

1 =15 -2*7

But 2=17 -15*1, so substitute:

1=15 - (17 -15*1)*7 =15 -17*7 +15*7 =15*8 -17*7

Therefore, 1=15*8 -17*7, so 15*8 ≡1 mod17. Therefore inverse of 15 mod17 is 8. Therefore, going back to the equation 15k ≡16 mod17. Multiply both sides by 8:

k ≡16*8 mod17. 16*8=128. 128 divided by 17: 17*7=119, 128-119=9. So k≡9 mod17. Therefore, k=9 +17m for some integer m. Then x=(1 +83k)/17. Let me take k=9. Then x=(1 +83*9)/17. Compute 83*9: 80*9=720, 3*9=27, total 747. Then 747 +1=748. Then 748 divided by 17. Let me check: 17*44=748, as before. So x=44. Therefore, that's the same result. So this method also gives 44.

So regardless of the method, it seems that 44 is the inverse. Therefore, I think I can be confident that the answer is 44.

But just to make sure, let me check another value. For example, let me compute 17*44 mod83. As before, 17*44=748. Now divide 748 by 83. Let me do this division step by step. 83*8=664. 748-664=84. 84 is larger than 83, so subtract 83: 84-83=1. Therefore, 748=83*8 +84=83*8 +83 +1=83*9 +1. Therefore, 748 mod83 is 1. Perfect. So that's correct.

Therefore, the inverse of 17 modulo 83 is indeed 44. So the answer is 44. Therefore, I think that's solid.

But just to recap all steps once again:

1. Recognized that finding the inverse of 17 mod83 requires solving 17x ≡1 mod83.
2. Used the Extended Euclidean Algorithm to find that -39*17 ≡1 mod83, so -39 mod83 is 44.
3. Verified using the given hint that 17*5≡2 mod83, then found inverse of 2 mod83 is 42, so 5*42=210≡44 mod83.
4. Tried solving the equation via another method by expressing x as (1 +83k)/17 and found k=9 leading to x=44.
5. Verified by direct computation that 17*44=748≡1 mod83.

All methods converge to 44. Therefore, 44 is the correct answer.

**Final Answer**
The inverse of $17$ modulo $83$ is $\boxed{44}$.

To find the modular inverse of $ 17 \mod 83 $, we want a residue $ x $ such that:

$$
17x \equiv 1 \pmod{83}
$$

This means we are looking for an integer $ x $ in the range $ 0 \leq x < 83 $ such that when $ 17x $ is divided by 83, the remainder is 1. We will solve this using the **Extended Euclidean Algorithm**, which is a reliable and structured method for computing modular inverses.

---

### Step 1: Apply the Extended Euclidean Algorithm

We begin by dividing 83 by 17:

$$
83 = 17 \cdot 4 + 15
$$

$$
17 = 15 \cdot 1 + 2
$$

$$
15 = 2 \cdot 7 + 1
$$

$$
2 = 1 \cdot 2 + 0
$$

The last nonzero remainder is 1, so the gcd of 17 and 83 is 1 (which confirms that 17 has an inverse modulo 83).

---

### Step 2: Work Backwards to Express 1 as a Linear Combination

We use the previous equations to express 1 as a combination of 17 and 83:

1. From the third equation:
   $$
   1 = 15 - 2 \cdot 7
   $$

2. From the second equation:
   $$
   2 = 17 - 15 \cdot 1
   $$

   Substituting into the previous step:
   $$
   1 = 15 - (17 - 15) \cdot 7 = 15 - 17 \cdot 7 + 15 \cdot 7 = 15 \cdot 8 - 17 \cdot 7
   $$

3. From the first equation:
   $$
   15 = 83 - 17 \cdot 4
   $$

   Substitute into the previous expression:
   $$
   1 = (83 - 17 \cdot 4) \cdot 8 - 17 \cdot 7 = 83 \cdot 8 - 17 \cdot 32 - 17 \cdot 7 = 83 \cdot 8 - 17 \cdot 39
   $$

Thus, we have:

$$
-39 \cdot 17 + 83 \cdot 8 = 1
$$

This tells us that the modular inverse of 17 modulo 83 is:

$$
-39 \mod 83
$$

To convert $ -39 $ to a positive integer between 0 and 82:

$$
-39 + 83 = 44
$$

---

### Step 3: Verification

We verify that $ 17 \cdot 44 \equiv 1 \mod 83 $:

$$
17 \cdot 44 = 748
$$

$$
748 \mod 83 = 748 - 83 \cdot 9 = 748 - 747 = 1
$$

So, the modular inverse is correct.

---

### Optional Alternative: Use the Given Hint

We are given that $ 17 \cdot 5 = 85 \equiv 2 \mod 83 $. Then, we can use this to find the inverse:

$$
17 \cdot 5 \equiv 2 \mod 83 \Rightarrow 17 \cdot 5 \cdot x \equiv 1 \mod 83 \Rightarrow x \equiv \frac{1}{2} \mod 83
$$

So, we need the inverse of 2 modulo 83. It's easy to find that:

$$
2 \cdot 42 = 84 \Rightarrow 84 \equiv 1 \mod 83 \Rightarrow \text{Inverse of } 2 \mod 83 = 42
$$

So, the inverse of 17 is:

$$
5 \cdot 42 = 210 \Rightarrow 210 \mod 83 = 44
$$

This confirms our previous result.

---

### ✅ Final Answer

$$
\boxed{44}
$$

Solution:
We may begin by noting that $17\cdot 5=85\equiv 2\pmod{83}$. However, we are looking for $n$ such that $17\cdot n\equiv 1\pmod{83}$.

Note that $2\cdot 42=84\equiv 1\pmod{83}$. Therefore, \begin{align*}
17\cdot 5\cdot 42 &\equiv 2\cdot 42 \\
&\equiv 1\pmod{83},
\end{align*}which tells us that $17$ and $5\cdot 42$ are each other's inverses modulo $83$. We can evaluate $5\cdot 42=210$, but this is not in the range $0$ to $82$, so we take its residue $\pmod{83}$, which is $44$.

Therefore, $17^{-1}\equiv \boxed{44}\pmod{83}$.

We may check our answer: $17\cdot 44 = 748 = 9\cdot 83+1 \equiv 1\pmod{83}$, so our answer is correct.


Explanation:
Okay, so I need to find the average of the two smallest positive integer solutions to the congruence 14u ≡ 46 mod 100. Hmm, let me think about how to approach this.

First, I remember that solving a linear congruence like ax ≡ b mod m involves finding integers x that satisfy the equation. The general method is to find the modular inverse of a modulo m, provided that a and m are coprime. But wait, before that, I should check if a and m have a common divisor. If they do, then there might be solutions only if that divisor also divides b.

So, let me check the gcd of 14 and 100. Let me compute that. The prime factors of 14 are 2 and 7, and for 100, it's 2² * 5². So the gcd(14,100) is 2. Okay, so since 2 divides 46? Let me check: 46 divided by 2 is 23, so yes. That means there are solutions. The number of solutions modulo 100 should be equal to gcd(14,100) which is 2. Therefore, there are two solutions modulo 100. Wait, but the question is asking for the two smallest positive integer solutions. So I need to find all solutions and then take the two smallest ones and average them.

But how do I find those solutions? Let me recall the method. When solving ax ≡ b mod m, if d = gcd(a, m), then we can divide the entire congruence by d. So first step is to divide the congruence by d.

So, let me do that. Here, d = 2. Therefore, dividing both sides by 2, we get:

(14/2)u ≡ 46/2 mod 100/2

Which simplifies to:

7u ≡ 23 mod 50

Okay, so now the congruence is 7u ≡ 23 mod 50. Now, since 7 and 50 are coprime (gcd(7,50)=1), there's a unique solution modulo 50. But since the original modulus was 100, and we divided by d=2, the solutions will be spaced every 50. Therefore, once I find the solution to 7u ≡ 23 mod 50, the general solution to the original congruence 14u ≡ 46 mod 100 would be u ≡ solution mod 50, and then adding 50 to get the next solution.

So let me first solve 7u ≡ 23 mod 50. To solve for u, I need the inverse of 7 modulo 50. Let me compute that.

What's the inverse of 7 mod 50? That is, find an integer x such that 7x ≡ 1 mod 50.

To find this, I can use the Extended Euclidean Algorithm. Let me apply that to 7 and 50.

First, divide 50 by 7: 50 = 7*7 + 1. Then 7 = 1*7 + 0. So the gcd is 1, which we already know. Now, working backwards:

1 = 50 - 7*7.

Therefore, 1 ≡ -7*7 mod 50. Which means that -7*7 ≡ 1 mod 50. Therefore, multiplying both sides by 1: -49 ≡ 1 mod 50. Wait, but -49 mod 50 is 1, yes. Therefore, the inverse of 7 mod 50 is -7. But -7 mod 50 is equal to 50 -7 = 43. Let me check: 7*43 = 301. Now, 301 divided by 50 is 6*50=300, so 301 - 300 = 1. Yes, 7*43 ≡ 1 mod 50. Perfect.

Therefore, the inverse of 7 mod 50 is 43. So, multiplying both sides of 7u ≡ 23 mod 50 by 43 gives:

u ≡ 23*43 mod 50.

Let me compute 23*43. Let me break it down:

First, 20*43 = 860, 3*43 = 129, so total is 860 + 129 = 989. Now compute 989 mod 50. Since 50*19 = 950, 989 - 950 = 39. Therefore, u ≡ 39 mod 50.

So the general solution to the congruence 7u ≡ 23 mod 50 is u = 39 + 50k, where k is integer. Therefore, going back to the original congruence 14u ≡ 46 mod 100, the solutions are u = 39 + 50k. Wait, but why? Because when we divided the original congruence by 2, we transformed it into 7u ≡ 23 mod 50, so solutions to that are u = 39 + 50k, so plugging back into original congruence, they should satisfy 14u ≡ 46 mod 100.

But let me check if that's correct. Let me take u = 39. Then 14*39 = 546. Let me compute 546 mod 100. 100*5 = 500, 546 - 500 = 46. So 546 ≡ 46 mod 100. Yes, that works.

Then the next solution would be u = 39 + 50 = 89. Let me check that. 14*89 = 1246. 1246 divided by 100 is 12*100 = 1200, 1246 - 1200 = 46. So 1246 ≡ 46 mod 100. Perfect.

So the solutions are 39, 89, 139, 189,... and so on. Therefore, the two smallest positive integer solutions are 39 and 89. Therefore, the average of these two is (39 + 89)/2. Let me compute that: 39 + 89 is 128, divided by 2 is 64. Therefore, the average is 64.

Wait, but before I conclude, let me check if there are smaller solutions. Because maybe there's a solution less than 39? Because sometimes when you divide congruences, you might have other solutions. Wait, no. Since we had divided by d = 2, the solutions are spaced every 50, so once we found the minimal solution mod 50, which is 39, then the next one is 39 + 50. So 39 is the smallest positive solution. But let me check if there is any smaller solution.

For example, maybe u=... Let me think. Let me check u=39 is the first positive solution. Let me check u= -11? Wait, negative numbers aren't positive, so we need positive integers.

But to be thorough, let me check if there's another solution between 1 and 39. For example, suppose I didn't do the method correctly. Let me try solving 14u ≡ 46 mod 100.

Alternatively, maybe I can think of 14u - 46 = 100k, for some integer k. Then, 14u = 100k + 46, so u = (100k + 46)/14. Let me compute this for k such that u is integer.

We can write u = (50k + 23)/7. So for u to be integer, 50k + 23 must be divisible by 7. So 50k + 23 ≡ 0 mod 7. Let me compute 50 mod 7. 7*7=49, so 50 ≡ 1 mod 7. Similarly, 23 mod 7 is 2, since 7*3=21, so 23 - 21 = 2. Therefore, the congruence becomes:

1*k + 2 ≡ 0 mod 7 => k ≡ -2 mod 7 => k ≡ 5 mod 7. Therefore, k = 5 + 7m, where m is integer. Therefore, substituting back into u:

u = [50*(5 + 7m) + 23]/7 = [250 + 350m + 23]/7 = [273 + 350m]/7 = 273/7 + 350m/7 = 39 + 50m.

So this gives the same solutions: u = 39 + 50m. Therefore, the solutions are 39, 89, 139,... as before. So, indeed, the smallest positive solution is 39, then 89, so the average is (39 + 89)/2 = 64.

But let me check with another approach. Suppose I didn't divide by gcd(14,100)=2. Let me try solving 14u ≡ 46 mod 100. Let me try to solve this directly.

We can write 14u ≡ 46 mod 100. Let me subtract 46 from both sides: 14u - 46 = 100k, which is the same as before. But maybe trying to find u by checking numbers? Though that might take time, but for verification.

Alternatively, since we have 14u ≡ 46 mod 100, we can reduce this congruence. Let me divide both sides by 2, which gives 7u ≡ 23 mod 50, which is what we did earlier. So the same steps. Therefore, leading to the same solutions.

Alternatively, maybe I can write 14u ≡ 46 mod 100 as 14u ≡ 46 mod 100. Since 14 and 100 have gcd 2, as we saw, so solutions exist.

Alternatively, think of it as looking for u where 14u - 46 is divisible by 100. So 14u = 46 + 100k. Then u = (46 + 100k)/14. Let me write that as u = (23 + 50k)/7. For u to be integer, 23 + 50k must be divisible by 7. Let me compute 23 mod 7 is 2, 50 mod 7 is 1. So 2 + k ≡ 0 mod 7 => k ≡ -2 mod 7 => k ≡ 5 mod 7. Therefore k = 5 + 7m, then u = (23 + 50*(5 + 7m))/7 = (23 + 250 + 350m)/7 = 273/7 + 50m = 39 + 50m. Same result.

So again, same conclusion. Therefore, the solutions are 39, 89, 139,... So the two smallest positive solutions are 39 and 89, average is 64.

But just to be 100% sure, let me check if there's a smaller solution. For example, let me check u= 39-50 = -11, which is negative, so not positive. So 39 is the first positive one. Then adding 50 gives 89. So there are no smaller positive solutions.

But just to check, maybe there's another solution between, say, 1 and 39? Let me check for u from 1 to 39.

Let me try u= 39 is the first one. Let me check maybe some smaller numbers. Let me compute 14u mod 100 for u=1: 14. 2:28, 3:42, 4:56, 5:70, 6:84, 7:98, 8: 14*8=112 → 12 mod 100, 9: 126 → 26, 10:140→40, 11:154→54, 12: 168→68, 13:182→82, 14:196→96, 15:210→10 mod 100, 16: 224→24, 17:238→38, 18:252→52, 19:266→66, 20:280→80, 21:294→94, 22: 308→08, 23:322→22, 24:336→36, 25:350→50, 26:364→64, 27:378→78, 28:392→92, 29:406→06, 30:420→20, 31:434→34, 32:448→48, 33:462→62, 34:476→76, 35:490→90, 36:504→04, 37:518→18, 38:532→32, 39:546→46.

There we go! At u=39, we get 46 mod 100. So indeed, u=39 is the first positive solution. Then the next one is u=39 + 50=89. Let me check u=89, 14*89=1246, mod 100 is 46. Yep.

So no solutions between 1 and 39. Therefore, the two smallest are 39 and 89. Their average is (39 + 89)/2 = 128/2 = 64.

But just to be thorough, let me check u= 39 - 50 = -11, which is negative. So not allowed.

Alternatively, maybe there is another solution in another range? Wait, no. Because the modulus is 100, and we have solutions spaced by 50. So once we have the minimal positive solution, the next is 50 more, then 100 more, etc. So in positive integers, the solutions are 39, 89, 139, 189,... So the two smallest positive solutions are 39 and 89.

Therefore, their average is 64.

But just to make sure that the original problem didn't have some trick or something. For example, maybe I misapplied the division by gcd? Let me check again.

Original congruence: 14u ≡ 46 mod 100.

gcd(14,100)=2, divides 46? Yes, 46 / 2 =23. Therefore, divide congruence by 2: 7u ≡23 mod 50. Then solve that.

Since 7 and 50 are coprime, solution is u ≡23*7^{-1} mod 50. Which we found inverse of 7 mod 50 is 43. Therefore, 23*43 mod 50. Which we calculated as 989 mod 50.

But let me check again: 23*43. Let me compute 23*40=920, 23*3=69. 920+69=989. Then 989 divided by 50 is 19*50=950, 989-950=39. So yes, 39.

Therefore, the solution is u ≡39 mod 50. Hence, solutions are 39, 89, 139,...

Therefore, the two smallest positive solutions are 39 and 89. Average is 64.

Alternatively, maybe there is a different approach? Let me think. Let me try solving 14u ≡ 46 mod 100.

Another method could be to write this as 14u = 46 + 100k. Then, rearranged as 14u - 100k = 46. This is a linear Diophantine equation. The solutions can be found using the extended Euclidean algorithm.

Let me try solving 14u - 100k = 46.

First, let me find integers u and k such that this equation holds.

First, we can write this as 14u - 100k = 46.

But since gcd(14,100)=2, and 2 divides 46, solutions exist.

First, divide the entire equation by 2:

7u - 50k = 23.

So now, solving 7u - 50k = 23.

Let me rearrange it as 7u + (-50)k = 23.

Let me find one particular solution.

Let me use the extended Euclidean algorithm for 7 and 50.

We already did this earlier. The gcd is 1.

From before, we had 1 = 50 - 7*7. Therefore, rearranged:

-7*7 + 1*50 = 1.

Multiply both sides by 23:

-7*161 + 1*1150 = 23.

Therefore, one particular solution is u = -161, k = -1150? Wait, no. Wait:

Original equation after dividing by 2 is 7u -50k =23.

From above: -7*161 + 1*1150 = 23. Which can be written as 7*(-161) + (-50)*(-1150) = 23? Wait, no. Wait, let's check:

Original equation: 7u -50k =23.

From the equation -7*7 + 1*50 = 1. Multiply by 23: -7*161 + 50*23 =23. Therefore, rearranged:

7*(-161) + 50*23 =23.

Therefore, comparing with 7u -50k =23. Let me write it as 7u + (-50)k =23. Therefore, if I have 7*(-161) + (-50)*(-23) = 23? Wait:

Wait, let me check again. Let me take the equation:

From -7*7 + 1*50 = 1, multiplying by 23: -7*(7*23) + 1*(50*23) = 23. So that's -7*161 + 50*23 = 23.

Therefore, if we write this as 7*(-161) + 50*23 = 23.

But our equation is 7u - 50k =23. So, comparing:

7u -50k =23.

Which can be written as 7u + (-50)k =23.

Therefore, if we set u = -161 and k = -23, then 7*(-161) + (-50)*(-23) = -1127 + 1150 = 23. So yes. Therefore, u = -161, k = -23 is a particular solution.

But we need positive solutions. Therefore, the general solution can be found by adding multiples of 50/gcd(7,50) and 7/gcd(7,50). Wait, the general solution for equations like ax + by = c is x = x0 + (b/d)*t, y = y0 - (a/d)*t, where d = gcd(a,b).

So here, our equation is 7u -50k =23. Let me write it as 7u + (-50)k = 23. So here, a=7, b=-50, c=23. Then the general solution is u = u0 + (-50)/d * t, k = k0 + 7/d * t, where d = gcd(7,50)=1.

Wait, more precisely, the general solution for u is u = u0 + (b/d)*t, k = k0 - (a/d)*t? Wait, let me recall. When solving ax + by = c, once you have a particular solution (x0, y0), the general solution is x = x0 + (b/d)*t, y = y0 - (a/d)*t, where t is an integer.

So here, our equation is 7u + (-50)k =23. So a =7, b=-50. So d=1. Therefore, general solution is u = u0 + (-50)/1 * t = u0 -50t, and k = k0 - (7)/1 * t = k0 -7t.

Wait, but let me check. Let me think of it as 7u + (-50)k = 23. So if I have a solution (u0, k0), then adding t*(b/d) to u and subtracting t*(a/d) from k? Wait, maybe I need to think of it in terms of homogeneous solutions.

The homogeneous equation is 7u + (-50)k =0. The solutions to this are multiples of (50, 7), since 7*50 = 350 = -50*(-7). Therefore, the general solution is the particular solution plus multiples of (50, -7). Therefore, u = u0 +50t and k = k0 -7t.

Wait, but the original equation is 7u -50k =23. So if we have a particular solution (u0, k0), then the general solution is u = u0 +50t and k = k0 +7t. Wait, let me check. Suppose we take t as a parameter.

If I take u = u0 + 50t, then plugging into the equation: 7(u0 +50t) -50k =23. Which is 7u0 -50k + 7*50t =23. Since 7u0 -50k =23, then we have 23 + 350t =23. Therefore, this holds for any t. Therefore, k must be adjusted accordingly.

But if we set u = u0 +50t, then we can solve for k. From the equation 7u -50k =23, rearranged: k = (7u -23)/50. Therefore, if u = u0 +50t, then k = [7(u0 +50t) -23]/50 = [7u0 -23 + 350t]/50 = k0 +7t.

Therefore, general solution is u = u0 +50t, k =k0 +7t.

So in our case, a particular solution is u0 = -161, k0 = -23. Therefore, general solution is u = -161 +50t, k = -23 +7t.

Therefore, to find positive solutions, we need u>0. So -161 +50t >0 => 50t >161 => t >161/50 =3.22. Therefore, t≥4.

For t=4: u= -161 +50*4 = -161 +200=39. Which matches our previous result. And k= -23 +7*4 = -23 +28=5.

So that's the minimal positive solution u=39. Then for t=5, u= -161 +50*5= -161 +250=89, which is the next solution.

Therefore, again, confirming that 39 and 89 are the two smallest positive solutions.

Therefore, their average is indeed 64.

Therefore, I think this confirms that my answer is correct.

But just to make sure, let me check another value. Let's say t= -3: u= -161 +50*(-3)= -161 -150= -311, which is negative. Not allowed. t=3: u= -161 +150= -11, still negative. t=4: 39, as before. So yes, 39 is the first positive solution.

Therefore, the answer is 64.

**Final Answer**
The average of the two smallest positive integer solutions is \boxed{64}.

To solve the congruence
$$
14u \equiv 46 \pmod{100},
$$
we want to find the two smallest **positive integer solutions** to this congruence and then compute their **average**.

---

### Step 1: Analyze the Congruence

We are given the linear congruence:
$$
14u \equiv 46 \pmod{100}.
$$

Let us analyze the **greatest common divisor** (gcd) of the coefficient of $ u $ (which is 14) and the modulus (which is 100):
$$
\gcd(14, 100) = 2.
$$

Since this gcd divides the right-hand side (46), the congruence has **solutions**.

---

### Step 2: Reduce the Congruence

We divide the entire congruence by the gcd:
$$
\frac{14}{2} u \equiv \frac{46}{2} \pmod{\frac{100}{2}} \quad \Rightarrow \quad 7u \equiv 23 \pmod{50}.
$$

Now we are solving a simpler congruence:
$$
7u \equiv 23 \pmod{50}.
$$

---

### Step 3: Solve the Simplified Congruence

We need the **modular inverse** of 7 modulo 50. That is, find an integer $ x $ such that:
$$
7x \equiv 1 \pmod{50}.
$$

Using the Extended Euclidean Algorithm:
$$
50 = 7 \times 7 + 1, \quad 7 = 1 \times 7 + 0,
$$
backwards:
$$
1 = 50 - 7 \times 7.
$$
So:
$$
1 \equiv -7 \times 7 \pmod{50} \quad \Rightarrow \quad -7 \equiv 43 \pmod{50}.
$$

Thus, the inverse of 7 mod 50 is **43**. Now multiply both sides of the simplified congruence by this inverse:
$$
u \equiv 23 \times 43 \pmod{50}.
$$

Compute:
$$
23 \times 43 = 989, \quad 989 \mod 50 = 39.
$$

So:
$$
u \equiv 39 \pmod{50}.
$$

---

### Step 4: Find All Positive Integer Solutions

The general solution to the simplified congruence $ 7u \equiv 23 \pmod{50} $ is:
$$
u = 39 + 50k, \quad \text{for integer } k.
$$

Now, recall the original modulus was 100, and we divided by 2, so solutions to the original $ 14u \equiv 46 \pmod{100} $ are also the same:
$$
u = 39 + 50k, \quad \text{for } k \in \mathbb{Z}.
$$

So the **positive** solutions are:
- When $ k = 0 $: $ u = 39 $
- When $ k = 1 $: $ u = 89 $
- When $ k = 2 $: $ u = 139 $, and so on.

Thus, the **two smallest positive solutions** are **39** and **89**.

---

### Step 5: Compute the Average

$$
\text{Average} = \frac{39 + 89}{2} = \frac{128}{2} = \boxed{64}.
$$

---

### ✅ Final Answer

$$
\boxed{64}
$$

[Assistant]
Explanation:
Okay, so I need to find the average of the two smallest positive integer solutions to the congruence 14u ≡ 46 mod 100. Hmm, let me think about how to approach this.

First, I remember that solving a linear congruence like ax ≡ b mod m involves finding integers x that satisfy the equation. The general method is to find the modular inverse of a modulo m, provided that a and m are coprime. But wait, before that, I should check if a and m have a common divisor. If they do, then there might be solutions only if that divisor also divides b.

So, let me check the gcd of 14 and 100. Let me compute that. The prime factors of 14 are 2 and 7, and for 100, it's 2² * 5². So the gcd(14,100) is 2. Okay, so since 2 divides 46? Let me check: 46 divided by 2 is 23, so yes. That means there are solutions. The number of solutions modulo 100 should be equal to gcd(14,100) which is 2. Therefore, there are two solutions modulo 100. Wait, but the question is asking for the two smallest positive integer solutions. So I need to find all solutions and then take the two smallest ones and average them.

But how do I find those solutions? Let me recall the method. When solving ax ≡ b mod m, if d = gcd(a, m), then we can divide the entire congruence by d. So first step is to divide the congruence by d.

So, let me do that. Here, d = 2. Therefore, dividing both sides by 2, we get:

(14/2)u ≡ 46/2 mod 100/2

Which simplifies to:

7u ≡ 23 mod 50

Okay, so now the congruence is 7u ≡ 23 mod 50. Now, since 7 and 50 are coprime (gcd(7,50)=1), there's a unique solution modulo 50. But since the original modulus was 100, and we divided by d=2, the solutions will be spaced every 50. Therefore, once I find the solution to 7u ≡ 23 mod 50, the general solution to the original congruence 14u ≡ 46 mod 100 would be u ≡ solution mod 50, and then adding 50 to get the next solution.

So let me first solve 7u ≡ 23 mod 50. To solve for u, I need the inverse of 7 modulo 50. Let me compute that.

What's the inverse of 7 mod 50? That is, find an integer x such that 7x ≡ 1 mod 50.

To find this, I can use the Extended Euclidean Algorithm. Let me apply that to 7 and 50.

First, divide 50 by 7: 50 = 7*7 + 1. Then 7 = 1*7 + 0. So the gcd is 1, which we already know. Now, working backwards:

1 = 50 - 7*7.

Therefore, 1 ≡ -7*7 mod 50. Which means that -7*7 ≡ 1 mod 50. Therefore, multiplying both sides by 1: -49 ≡ 1 mod 50. Wait, but -49 mod 50 is 1, yes. Therefore, the inverse of 7 mod 50 is -7. But -7 mod 50 is equal to 50 -7 = 43. Let me check: 7*43 = 301. Now, 301 divided by 50 is 6*50=300, so 301 - 300 = 1. Yes, 7*43 ≡ 1 mod 50. Perfect.

Therefore, the inverse of 7 mod 50 is 43. So, multiplying both sides of 7u ≡ 23 mod 50 by 43 gives:

u ≡ 23*43 mod 50.

Let me compute 23*43. Let me break it down:

First, 20*43 = 860, 3*43 = 129, so total is 860 + 129 = 989. Now compute 989 mod 50. Since 50*19 = 950, 989 - 950 = 39. Therefore, u ≡ 39 mod 50.

So the general solution to the congruence 7u ≡ 23 mod 50 is u = 39 + 50k, where k is integer. Therefore, going back to the original congruence 14u ≡ 46 mod 100, the solutions are u = 39 + 50k. Wait, but why? Because when we divided the original congruence by 2, we transformed it into 7u ≡ 23 mod 50, so solutions to that are u = 39 + 50k, so plugging back into original congruence, they should satisfy 14u ≡ 46 mod 100.

But let me check if that's correct. Let me take u = 39. Then 14*39 = 546. Let me compute 546 mod 100. 100*5 = 500, 546 - 500 = 46. So 546 ≡ 46 mod 100. Yes, that works.

Then the next solution would be u = 39 + 50 = 89. Let me check that. 14*89 = 1246. 1246 divided by 100 is 12*100 = 1200, 1246 - 1200 = 46. So 1246 ≡ 46 mod 100. Perfect.

So the solutions are 39, 89, 139, 189,... and so on. Therefore, the two smallest positive integer solutions are 39 and 89. Therefore, the average of these two is (39 + 89)/2. Let me compute that: 39 + 89 is 128, divided by 2 is 64. Therefore, the average is 64.

Wait, but before I conclude, let me check if there are smaller solutions. Because maybe there's a solution less than 39? Because sometimes when you divide congruences, you might have other solutions. Wait, no. Since we had divided by d = 2, the solutions are spaced every 50, so once we found the minimal solution mod 50, which is 39, then the next one is 39 + 50. So 39 is the smallest positive solution. But let me check if there is any smaller solution.

For example, maybe u=... Let me think. Let me check u=39 is the first positive solution. Let me check u= -11? Wait, negative numbers aren't positive, so we need positive integers.

But to be thorough, let me check if there's another solution between 1 and 39. For example, suppose I didn't do the method correctly. Let me try solving 14u ≡ 46 mod 100.

Alternatively, maybe I can think of 14u - 46 = 100k, for some integer k. Then, 14u = 100k + 46, so u = (100k + 46)/14. Let me compute this for k such that u is integer.

We can write u = (50k + 23)/7. So for u to be integer, 50k + 23 must be divisible by 7. So 50k + 23 ≡ 0 mod 7. Let me compute 50 mod 7. 7*7=49, so 50 ≡ 1 mod 7. Similarly, 23 mod 7 is 2, since 7*3=21, so 23 - 21 = 2. Therefore, the congruence becomes:

1*k + 2 ≡ 0 mod 7 => k ≡ -2 mod 7 => k ≡ 5 mod 7. Therefore, k = 5 + 7m, where m is integer. Therefore, substituting back into u:

u = [50*(5 + 7m) + 23]/7 = [250 + 350m + 23]/7 = [273 + 350m]/7 = 273/7 + 350m/7 = 39 + 50m.

So this gives the same solutions: u = 39 + 50m. Therefore, the solutions are 39, 89, 139,... as before. So, indeed, the smallest positive solution is 39, then 89, so the average is (39 + 89)/2 = 64.

But let me check with another approach. Suppose I didn't divide by gcd(14,100)=2. Let me try solving 14u ≡ 46 mod 100. Let me try to solve this directly.

We can write 14u ≡ 46 mod 100. Let me subtract 46 from both sides: 14u - 46 = 100k, which is the same as before. But maybe trying to find u by checking numbers? Though that might take time, but for verification.

Alternatively, since we have 14u ≡ 46 mod 100, we can reduce this congruence. Let me divide both sides by 2, which gives 7u ≡ 23 mod 50, which is what we did earlier. So the same steps. Therefore, leading to the same solutions.

Alternatively, maybe I can write 14u ≡ 46 mod 100 as 14u ≡ 46 mod 100. Since 14 and 100 have gcd 2, as we saw, so solutions exist.

Alternatively, think of it as looking for u where 14u - 46 is divisible by 100. So 14u = 46 + 100k. Then u = (46 + 100k)/14. Let me write that as u = (23 + 50k)/7. For u to be integer, 23 + 50k must be divisible by 7. Let me compute 23 mod 7 is 2, 50 mod 7 is 1. So 2 + k ≡ 0 mod 7 => k ≡ -2 mod 7 => k ≡ 5 mod 7. Therefore k = 5 + 7m, then u = (23 + 50*(5 + 7m))/7 = (23 + 250 + 350m)/7 = 273/7 + 50m = 39 + 50m. Same result.

So again, same conclusion. Therefore, the solutions are 39, 89, 139,... So the two smallest positive solutions are 39 and 89, average is 64.

But just to be 100% sure, let me check if there's a smaller solution. For example, let me check u= 39-50 = -11, which is negative, so not positive. So 39 is the first positive one. Then adding 50 gives 89. So there are no smaller positive solutions.

But just to check, maybe there's another solution between, say, 1 and 39? Let me check for u from 1 to 39.

Let me try u= 39 is the first one. Let me check maybe some smaller numbers. Let me compute 14u mod 100 for u=1: 14. 2:28, 3:42, 4:56, 5:70, 6:84, 7:98, 8: 14*8=112 → 12 mod 100, 9: 126 → 26, 10:140→40, 11:154→54, 12: 168→68, 13:182→82, 14:196→96, 15:210→10 mod 100, 16: 224→24, 17:238→38, 18:252→52, 19:266→66, 20:280→80, 21:294→94, 22: 308→08, 23:322→22, 24:336→36, 25:350→50, 26:364→64, 27:378→78, 28:392→92, 29:406→06, 30:420→20, 31:434→34, 32:448→48, 33:462→62, 34:476→76, 35:490→90, 36:504→04, 37:518→18, 38:532→32, 39:546→46.

There we go! At u=39, we get 46 mod 100. So indeed, u=39 is the first positive solution. Then the next one is u=39 + 50=89. Let me check u=89, 14*89=1246, mod 100 is 46. Yep.

So no solutions between 1 and 39. Therefore, the two smallest are 39 and 89. Their average is (39 + 89)/2 = 128/2 = 64.

But just to be thorough, let me check u= 39 - 50 = -11, which is negative. So not allowed.

Alternatively, maybe there is another solution in another range? Wait, no. Because the modulus is 100, and we have solutions spaced by 50. So once we have the minimal positive solution, the next is 50 more, then 100 more, etc. So in positive integers, the solutions are 39, 89, 139, 189,... So the two smallest positive solutions are 39 and 89.

Therefore, their average is 64.

But just to make sure that the original problem didn't have some trick or something. For example, maybe I misapplied the division by gcd? Let me check again.

Original congruence: 14u ≡ 46 mod 100.

gcd(14,100)=2, divides 46? Yes, 46 / 2 =23. Therefore, divide congruence by 2: 7u ≡23 mod 50. Then solve that.

Since 7 and 50 are coprime, solution is u ≡23*7^{-1} mod 50. Which we found inverse of 7 mod 50 is 43. Therefore, 23*43 mod 50. Which we calculated as 989 mod 50.

But let me check again: 23*43. Let me compute 23*40=920, 23*3=69. 920+69=989. Then 989 divided by 50 is 19*50=950, 989-950=39. So yes, 39.

Therefore, the solution is u ≡39 mod 50. Hence, solutions are 39, 89, 139,...

Therefore, the two smallest positive solutions are 39 and 89. Average is 64.

Alternatively, maybe there is a different approach? Let me think. Let me try solving 14u ≡ 46 mod 100.

Another method could be to write this as 14u = 46 + 100k. Then, rearranged as 14u - 100k = 46. This is a linear Diophantine equation. The solutions can be found using the extended Euclidean algorithm.

Let me try solving 14u - 100k = 46.

First, let me find integers u and k such that this equation holds.

First, we can write this as 14u - 100k = 46.

But since gcd(14,100)=2, and 2 divides 46, solutions exist.

First, divide the entire equation by 2:

7u - 50k = 23.

So now, solving 7u - 50k = 23.

Let me rearrange it as 7u + (-50)k = 23.

Let me find one particular solution.

Let me use the extended Euclidean algorithm for 7 and 50.

We already did this earlier. The gcd is 1.

From before, we had 1 = 50 - 7*7. Therefore, rearranged:

-7*7 + 1*50 = 1.

Multiply both sides by 23:

-7*161 + 1*1150 = 23.

Therefore, one particular solution is u = -161, k = -1150? Wait, no. Wait:

Original equation after dividing by 2 is 7u -50k =23.

From above: -7*161 + 1*1150 = 23. Which can be written as 7*(-161) + (-50)*(-1150) = 23? Wait, no. Wait, let's check:

Original equation: 7u -50k =23.

From the equation -7*7 + 1*50 = 1. Multiply by 23: -7*161 + 50*23 =23. Therefore, rearranged:

7*(-161) + 50*23 =23.

Therefore, comparing with 7u -50k =23. Let me write it as 7u + (-50)k =23. Therefore, if I have 7*(-161) + (-50)*(-23) = 23? Wait:

Wait, let me check again. Let me take the equation:

From -7*7 + 1*50 = 1, multiplying by 23: -7*(7*23) + 1*(50*23) = 23. So that's -7*161 + 50*23 = 23.

Therefore, if we write this as 7*(-161) + 50*23 = 23.

But our equation is 7u - 50k =23. So, comparing:

7u -50k =23.

Which can be written as 7u + (-50)k =23.

Therefore, if we set u = -161 and k = -23, then 7*(-161) + (-50)*(-23) = -1127 + 1150 = 23. So yes. Therefore, u = -161, k = -23 is a particular solution.

But we need positive solutions. Therefore, the general solution can be found by adding multiples of 50/gcd(7,50) and 7/gcd(7,50). Wait, the general solution for equations like ax + by = c is x = x0 + (b/d)*t, y = y0 - (a/d)*t, where d = gcd(a,b).

So here, our equation is 7u -50k =23. Let me write it as 7u + (-50)k = 23. So here, a=7, b=-50, c=23. Then the general solution is u = u0 + (-50)/d * t, k = k0 + 7/d * t, where d = gcd(7,50)=1.

Wait, more precisely, the general solution for u is u = u0 + (b/d)*t, k = k0 - (a/d)*t? Wait, let me recall. When solving ax + by = c, once you have a particular solution (x0, y0), the general solution is x = x0 + (b/d)*t, y = y0 - (a/d)*t, where t is an integer.

So here, our equation is 7u + (-50)k =23. So a =7, b=-50. So d=1. Therefore, general solution is u = u0 + (-50)/1 * t = u0 -50t, and k = k0 - (7)/1 * t = k0 -7t.

Wait, but let me check. Let me think of it as 7u + (-50)k = 23. So if I have a solution (u0, k0), then adding t*(b/d) to u and subtracting t*(a/d) from k? Wait, maybe I need to think of it in terms of homogeneous solutions.

The homogeneous equation is 7u + (-50)k =0. The solutions to this are multiples of (50, 7), since 7*50 = 350 = -50*(-7). Therefore, the general solution is the particular solution plus multiples of (50, -7). Therefore, u = u0 +50t and k = k0 -7t.

Wait, but the original equation is 7u -50k =23. So if we have a particular solution (u0, k0), then the general solution is u = u0 +50t and k = k0 +7t. Wait, let me check. Suppose we take t as a parameter.

If I take u = u0 + 50t, then plugging into the equation: 7(u0 +50t) -50k =23. Which is 7u0 -50k + 7*50t =23. Since 7u0 -50k =23, then we have 23 + 350t =23. Therefore, this holds for any t. Therefore, k must be adjusted accordingly.

But if we set u = u0 +50t, then we can solve for k. From the equation 7u -50k =23, rearranged: k = (7u -23)/50. Therefore, if u = u0 +50t, then k = [7(u0 +50t) -23]/50 = [7u0 -23 + 350t]/50 = k0 +7t.

Therefore, general solution is u = u0 +50t, k =k0 +7t.

So in our case, a particular solution is u0 = -161, k0 = -23. Therefore, general solution is u = -161 +50t, k = -23 +7t.

Therefore, to find positive solutions, we need u>0. So -161 +50t >0 => 50t >161 => t >161/50 =3.22. Therefore, t≥4.

For t=4: u= -161 +50*4 = -161 +200=39. Which matches our previous result. And k= -23 +7*4 = -23 +28=5.

So that's the minimal positive solution u=39. Then for t=5, u= -161 +50*5= -161 +250=89, which is the next solution.

Therefore, again, confirming that 39 and 89 are the two smallest positive solutions.

Therefore, their average is indeed 64.

Therefore, I think this confirms that my answer is correct.

But just to make sure, let me check another value. Let's say t= -3: u= -161 +50*(-3)= -161 -150= -311, which is negative. Not allowed. t=3: u= -161 +150= -11, still negative. t=4: 39, as before. So yes, 39 is the first positive solution.

Therefore, the answer is 64.

**Final Answer**
The average of the two smallest positive integer solutions is \boxed{64}.

To solve the congruence
$$
14u \equiv 46 \pmod{100},
$$
we want to find the two smallest **positive integer solutions** to this congruence and then compute their **average**.

---

### Step 1: Analyze the Congruence

We are given the linear congruence:
$$
14u \equiv 46 \pmod{100}.
$$

Let us analyze the **greatest common divisor** (gcd) of the coefficient of $ u $ (which is 14) and the modulus (which is 100):
$$
\gcd(14, 100) = 2.
$$

Since this gcd divides the right-hand side (46), the congruence has **solutions**.

---

### Step 2: Reduce the Congruence

We divide the entire congruence by the gcd:
$$
\frac{14}{2} u \equiv \frac{46}{2} \pmod{\frac{100}{2}} \quad \Rightarrow \quad 7u \equiv 23 \pmod{50}.
$$

Now we are solving a simpler congruence:
$$
7u \equiv 23 \pmod{50}.
$$

---

### Step 3: Solve the Simplified Congruence

We need the **modular inverse** of 7 modulo 50. That is, find an integer $ x $ such that:
$$
7x \equiv 1 \pmod{50}.
$$

Using the Extended Euclidean Algorithm:
$$
50 = 7 \times 7 + 1, \quad 7 = 1 \times 7 + 0,
$$
backwards:
$$
1 = 50 - 7 \times 7.
$$
So:
$$
1 \equiv -7 \times 7 \pmod{50} \quad \Rightarrow \quad -7 \equiv 43 \pmod{50}.
$$

Thus, the inverse of 7 mod 50 is **43**. Now multiply both sides of the simplified congruence by this inverse:
$$
u \equiv 23 \times 43 \pmod{50}.
$$

Compute:
$$
23 \times 43 = 989, \quad 989 \mod 50 = 39.
$$

So:
$$
u \equiv 39 \pmod{50}.
$$

---

### Step 4: Find All Positive Integer Solutions

The general solution to the simplified congruence $ 7u \equiv 23 \pmod{50} $ is:
$$
u = 39 + 50k, \quad \text{for integer } k.
$$

Now, recall the original modulus was 100, and we divided by 2, so solutions to the original $ 14u \equiv 46 \pmod{100} $ are also the same:
$$
u = 39 + 50k, \quad \text{for } k \in \mathbb{Z}.
$$

So the **positive** solutions are:
- When $ k = 0 $: $ u = 39 $
- When $ k = 1 $: $ u = 89 $
- When $ k = 2 $: $ u = 139 $, and so on.

Thus, the **two smallest positive solutions** are **39** and **89**.

---

### Step 5: Compute the Average

$$
\text{Average} = \frac{39 + 89}{2} = \frac{128}{2} = \boxed{64}.
$$

---

### ✅ Final Answer

$$
\boxed{64}
$$

Solution:
Note that $14$, $46$, and $100$ all have a common factor of $2$, so we can divide it out: the solutions to $$14u \equiv 46 \pmod{100}$$ are identical to the solutions to $$7u \equiv 23 \pmod{50}.$$ Make sure you see why this is the case.

Now we can multiply both sides of the congruence by $7$ to obtain $$49u \equiv 161 \pmod{50},$$ which also has the same solutions as the previous congruence, since we could reverse the step above by multiplying both sides by $7^{-1}$. (We know that $7^{-1}$ exists modulo $50$ because $7$ and $50$ are relatively prime.)

Replacing each side of $49u\equiv 161$ by a $\pmod{50}$ equivalent, we have $$-u \equiv 11\pmod{50},$$ and thus $$u \equiv -11\pmod{50}.$$ This is the set of solutions to our original congruence. The two smallest positive solutions are $-11+50 = 39$ and $-11+2\cdot 50 = 89$. Their average is $\boxed{64}$.


Explanation:
Okay, so I need to find the image of the line y = 2x + 1 under the matrix M. But first, I don't know what matrix M is. However, I do know that M maps two specific vectors: the vector [2, -1]^T to [9, 3]^T, and [1, -3]^T to [7, -1]^T.

Hmm, so maybe I can figure out what matrix M is by using these two transformations. Since M is a 2x2 matrix, let me denote it as:

M = [ a  b ]
        [ c  d ]

So, when I multiply M by the first vector [2, -1]^T, I should get [9, 3]^T. Let me write that out as equations.

Multiplying the matrix by the vector:

a*2 + b*(-1) = 9
c*2 + d*(-1) = 3

Similarly, for the second vector [1, -3]^T mapped to [7, -1]^T:

a*1 + b*(-3) = 7
c*1 + d*(-3) = -1

So now I have a system of four equations:

1) 2a - b = 9
2) 2c - d = 3
3) a - 3b = 7
4) c - 3d = -1

Okay, so I can solve equations 1 and 3 for a and b, and equations 2 and 4 for c and d.

Starting with equations 1 and 3:

Equation 1: 2a - b = 9
Equation 3: a - 3b = 7

Maybe I can solve equation 1 for b and substitute into equation 3. Let me try that.

From equation 1: 2a - 9 = b, so b = 2a - 9.

Substituting into equation 3:

a - 3*(2a - 9) = 7
a - 6a + 27 = 7
-5a + 27 = 7
-5a = 7 - 27
-5a = -20
a = (-20)/(-5) = 4

Okay, so a = 4. Then plugging back into equation 1:

2*4 - b = 9 => 8 - b = 9 => -b = 1 => b = -1.

So, a = 4, b = -1.

Now moving on to equations 2 and 4 for c and d.

Equation 2: 2c - d = 3
Equation 4: c - 3d = -1

Similarly, let me solve equation 2 for d and substitute into equation 4.

From equation 2: 2c - 3 = d => d = 2c - 3.

Substituting into equation 4:

c - 3*(2c - 3) = -1
c - 6c + 9 = -1
-5c + 9 = -1
-5c = -1 - 9
-5c = -10
c = (-10)/(-5) = 2

So, c = 2. Then d = 2c - 3 = 2*2 - 3 = 4 - 3 = 1.

Therefore, c = 2, d = 1.

So putting it all together, the matrix M is:

[ 4  -1 ]
[ 2   1 ]

Let me check if this matrix works with the given vectors.

First vector [2, -1]^T:

M * [2]   = [4*2 + (-1)*(-1)] = [8 + 1] = [9]
         [-1]      [2*2 + 1*(-1)]   [4 -1]   [3]

Yes, that gives [9, 3]^T as desired.

Second vector [1, -3]^T:

M * [1]   = [4*1 + (-1)*(-3)] = [4 + 3] = [7]
         [-3]      [2*1 + 1*(-3)]   [2 - 3]   [-1]

Yes, that gives [7, -1]^T as required. Good, so matrix M is correct.

Now, the next part is to find the image of the line y = 2x + 1 under the matrix M.

So, the image of a line under a linear transformation is another line (assuming the transformation is invertible, which it is since determinant is 4*1 - (-1)*2 = 4 + 2 = 6 ≠ 0). So the image will be a line, and I need to express it in the form y = mx + b.

To find the image of the line, one approach is to take a general point (x, y) on the original line, apply the matrix M to get a new point (x', y'), and then find the relationship between x' and y'.

But since the original line is y = 2x + 1, we can parametrize the line or express it parametrically, then apply the transformation. Alternatively, we can find two points on the original line, apply the transformation to them, and then find the equation of the line through the transformed points.

Let me try both approaches and see which is easier.

First, maybe parametrize the line. Let me think.

Let me take a parameter t such that x = t, then y = 2t + 1. So any point on the line can be written as (t, 2t + 1). Then applying matrix M to this vector gives:

x' = 4*t + (-1)*(2t + 1) = 4t - 2t - 1 = 2t - 1

y' = 2*t + 1*(2t + 1) = 2t + 2t + 1 = 4t + 1

So now, the transformed coordinates are x' = 2t - 1 and y' = 4t + 1.

Now, we need to eliminate the parameter t to find the relationship between x' and y'.

From x' = 2t - 1, solve for t:

x' + 1 = 2t => t = (x' + 1)/2

Substitute into y':

y' = 4*( (x' + 1)/2 ) + 1 = 2*(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3

So, y' = 2x' + 3. Therefore, the image of the line under M is y = 2x + 3.

But let me verify this with another approach to be sure.

Alternatively, take two points on the original line, apply the matrix to them, then find the equation of the line through their images.

Take two points on y = 2x + 1. Let's choose two points.

First, when x = 0, y = 1. So point A is (0, 1).

Second, when x = 1, y = 2*1 + 1 = 3. So point B is (1, 3).

Now, apply matrix M to these points.

For point A (0, 1):

x' = 4*0 + (-1)*1 = 0 -1 = -1

y' = 2*0 + 1*1 = 0 + 1 = 1

So image of A is (-1, 1).

For point B (1, 3):

x' = 4*1 + (-1)*3 = 4 - 3 = 1

y' = 2*1 + 1*3 = 2 + 3 = 5

So image of B is (1, 5).

Now, find the equation of the line passing through (-1, 1) and (1, 5).

First, compute the slope m:

m = (5 - 1)/(1 - (-1)) = 4 / 2 = 2.

So slope is 2. Now, using point-slope form with point (-1, 1):

y - 1 = 2(x - (-1)) => y - 1 = 2(x + 1) => y = 2x + 2 + 1 => y = 2x + 3.

Same result as before! So that confirms that the image is y = 2x + 3.

Alternatively, let me check another point. Let's take point (2, 5) on the original line y=2x +1. Applying matrix M:

x' = 4*2 + (-1)*5 = 8 -5 = 3

y' = 2*2 + 1*5 = 4 + 5 = 9

So the transformed point is (3, 9). Let's check if it lies on y = 2x + 3:

2*3 + 3 = 6 + 3 = 9. Yes, that's correct. So (3,9) is indeed on y = 2x + 3.

Therefore, seems like the answer is y = 2x + 3.

Wait, but let me think again if there is another way someone might approach this, just to make sure.

Another approach: The line y = 2x + 1 can be represented as a set of vectors. The transformation M maps this line to another line. Since linear transformations map lines to lines (unless the line is mapped to a point, which would be if the line is in the kernel, but since M is invertible, the image of a line is a line).

Alternatively, we can find the equation of the image line by considering the transformation of the original line. Since the original line is y = 2x + 1, we can write it as y - 2x - 1 = 0. Let me consider the general equation of a line: ax + by + c = 0. But here, it's y - 2x -1 = 0.

But when we apply a linear transformation, how does it affect the equation? Hmm, not sure. Maybe using coordinates transformation? Let me think.

Alternatively, suppose (x, y) is on the original line. Then (x', y') = M*(x, y). So x' = 4x - y, y' = 2x + y. Wait, that's from the matrix M:

Original coordinates (x, y), then transformed coordinates:

x' = 4x - y

y' = 2x + y

So, we can express x and y in terms of x' and y', and then substitute into the original line equation.

But solving for x and y from x' and y':

We have:

x' = 4x - y

y' = 2x + y

We can solve this system for x and y. Let's write:

Equation 1: 4x - y = x'

Equation 2: 2x + y = y'

Adding equations 1 and 2:

(4x - y) + (2x + y) = x' + y'

=> 6x = x' + y'

So x = (x' + y') / 6

Similarly, to find y, subtract equation 1 and equation 2? Let me see:

Equation 1: 4x - y = x'

Equation 2: 2x + y = y'

Let me solve for y. Let me add equations 1 and 2 to get x as above. Alternatively, maybe express y from equation 2:

From equation 2: y = y' - 2x. Substitute into equation 1:

4x - (y' - 2x) = x'

=> 4x - y' + 2x = x'

=> 6x - y' = x'

=> 6x = x' + y'

Same as before. So x = (x' + y') / 6.

Then, substitute back into equation 2 to find y:

y = y' - 2x = y' - 2*(x' + y')/6 = y' - (x' + y')/3 = (3y' - x' - y')/3 = (2y' - x') / 3

So, y = (2y' - x') / 3.

Now, original line equation is y = 2x + 1. Substitute x and y in terms of x', y':

(2y' - x') / 3 = 2*( (x' + y') / 6 ) + 1

Simplify the right-hand side:

2*( (x' + y') / 6 ) + 1 = (x' + y') / 3 + 1

So, equation:

(2y' - x') / 3 = (x' + y') / 3 + 1

Multiply both sides by 3:

2y' - x' = x' + y' + 3

Bring all terms to the left side:

2y' - x' - x' - y' - 3 = 0 => (2y' - y') + (-x' - x') -3 = 0 => y' - 2x' - 3 = 0 => y' = 2x' + 3

Which gives y = 2x + 3. Same answer. So this method also confirms it. So definitely, the image is y = 2x + 3.

Therefore, the answer is y = 2x + 3.

But just to be thorough, let me check another point. Let's take the original point ( -1, -1) on the line y = 2x + 1. Wait, when x = -1, y = 2*(-1) + 1 = -2 + 1 = -1. So (-1, -1) is on the line.

Applying matrix M:

x' = 4*(-1) - (-1) = -4 + 1 = -3

y' = 2*(-1) + (-1) = -2 -1 = -3

So transformed point is (-3, -3). Let's check if that lies on y = 2x + 3:

2*(-3) + 3 = -6 + 3 = -3. Yes, y' = -3 matches. So that works.

Another point: Let me take (0.5, 2*0.5 +1)= (0.5, 2). Applying M:

x' = 4*(0.5) - 2 = 2 - 2 = 0

y' = 2*(0.5) + 2 = 1 + 2 = 3

So transformed point (0, 3). Check if it's on y = 2x + 3: 2*0 + 3 = 3. Correct.

Therefore, all these checks confirm that the image is y = 2x + 3.

Therefore, I can be confident the answer is y = 2x + 3.

**Final Answer**
The image of the line under matrix $\mathbf{M}$ is \boxed{y = 2x + 3}.

To solve the problem, we first determine the matrix $\mathbf{M}$ that satisfies the given mappings:

- $\mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \end{pmatrix}$
- $\mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \end{pmatrix}$

Let’s assume the matrix $\mathbf{M}$ has the form:

$$
\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
$$

Applying the matrix to the input vectors gives us four equations:

1. $2a - b = 9$
2. $2c - d = 3$
3. $a - 3b = 7$
4. $c - 3d = -1$

### Step 1: Solve for $a$, $b$, $c$, and $d$

**From equations (1) and (3):**

- From (1): $b = 2a - 9$
- Substitute into (3): $a - 3(2a - 9) = 7$
- Simplify: $a - 6a + 27 = 7 \Rightarrow -5a = -20 \Rightarrow a = 4$

Now compute $b$:

$$
b = 2a - 9 = 2(4) - 9 = 8 - 9 = -1
$$

**From equations (2) and (4):**

- From (2): $d = 2c - 3$
- Substitute into (4): $c - 3(2c - 3) = -1$
- Simplify: $c - 6c + 9 = -1 \Rightarrow -5c = -10 \Rightarrow c = 2$

Now compute $d$:

$$
d = 2c - 3 = 2(2) - 3 = 4 - 3 = 1
$$

So, the matrix is:

$$
\mathbf{M} = \begin{pmatrix} 4 & -1 \\ 2 & 1 \end{pmatrix}
$$

### Step 2: Find the image of the line $y = 2x + 1$

We find the image of the line under the transformation by applying $\mathbf{M}$ to a point on the line.

Let $\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ be a point on the line $y = 2x + 1$, so $y = 2x + 1$. The transformation is:

$$
\mathbf{M} \cdot \begin{pmatrix} x \\ 2x + 1 \end{pmatrix}
= \begin{pmatrix} 4x - (2x + 1) \\ 2x + (2x + 1) \end{pmatrix}
= \begin{pmatrix} 2x - 1 \\ 4x + 1 \end{pmatrix}
$$

Let the transformed point be $(x', y')$. Then:

$$
x' = 2x - 1 \quad \text{and} \quad y' = 4x + 1
$$

We eliminate the parameter $x$ to find the relation between $x'$ and $y'$. From the first equation:

$$
x = \frac{x' + 1}{2}
$$

Substitute into the second equation:

$$
y' = 4\left(\frac{x' + 1}{2}\right) + 1 = 2(x' + 1) + 1 = 2x' + 3
$$

Thus, the image of the line under the transformation is:

$$
y = 2x + 3
$$

---

### Final Answer

$$
\boxed{y = 2x + 3}
$$

[Assistant]
Explanation:
Okay, so I need to find the image of the line y = 2x + 1 under the matrix M. But first, I don't know what matrix M is. However, I do know that M maps two specific vectors: the vector [2, -1]^T to [9, 3]^T, and [1, -3]^T to [7, -1]^T.

Hmm, so maybe I can figure out what matrix M is by using these two transformations. Since M is a 2x2 matrix, let me denote it as:

M = [ a  b ]
        [ c  d ]

So, when I multiply M by the first vector [2, -1]^T, I should get [9, 3]^T. Let me write that out as equations.

Multiplying the matrix by the vector:

a*2 + b*(-1) = 9
c*2 + d*(-1) = 3

Similarly, for the second vector [1, -3]^T mapped to [7, -1]^T:

a*1 + b*(-3) = 7
c*1 + d*(-3) = -1

So now I have a system of four equations:

1) 2a - b = 9
2) 2c - d = 3
3) a - 3b = 7
4) c - 3d = -1

Okay, so I can solve equations 1 and 3 for a and b, and equations 2 and 4 for c and d.

Starting with equations 1 and 3:

Equation 1: 2a - b = 9
Equation 3: a - 3b = 7

Maybe I can solve equation 1 for b and substitute into equation 3. Let me try that.

From equation 1: 2a - 9 = b, so b = 2a - 9.

Substituting into equation 3:

a - 3*(2a - 9) = 7
a - 6a + 27 = 7
-5a + 27 = 7
-5a = 7 - 27
-5a = -20
a = (-20)/(-5) = 4

Okay, so a = 4. Then plugging back into equation 1:

2*4 - b = 9 => 8 - b = 9 => -b = 1 => b = -1.

So, a = 4, b = -1.

Now moving on to equations 2 and 4 for c and d.

Equation 2: 2c - d = 3
Equation 4: c - 3d = -1

Similarly, let me solve equation 2 for d and substitute into equation 4.

From equation 2: 2c - 3 = d => d = 2c - 3.

Substituting into equation 4:

c - 3*(2c - 3) = -1
c - 6c + 9 = -1
-5c + 9 = -1
-5c = -1 - 9
-5c = -10
c = (-10)/(-5) = 2

So, c = 2. Then d = 2c - 3 = 2*2 - 3 = 4 - 3 = 1.

Therefore, c = 2, d = 1.

So putting it all together, the matrix M is:

[ 4  -1 ]
[ 2   1 ]

Let me check if this matrix works with the given vectors.

First vector [2, -1]^T:

M * [2]   = [4*2 + (-1)*(-1)] = [8 + 1] = [9]
         [-1]      [2*2 + 1*(-1)]   [4 -1]   [3]

Yes, that gives [9, 3]^T as desired.

Second vector [1, -3]^T:

M * [1]   = [4*1 + (-1)*(-3)] = [4 + 3] = [7]
         [-3]      [2*1 + 1*(-3)]   [2 - 3]   [-1]

Yes, that gives [7, -1]^T as required. Good, so matrix M is correct.

Now, the next part is to find the image of the line y = 2x + 1 under the matrix M.

So, the image of a line under a linear transformation is another line (assuming the transformation is invertible, which it is since determinant is 4*1 - (-1)*2 = 4 + 2 = 6 ≠ 0). So the image will be a line, and I need to express it in the form y = mx + b.

To find the image of the line, one approach is to take a general point (x, y) on the original line, apply the matrix M to get a new point (x', y'), and then find the relationship between x' and y'.

But since the original line is y = 2x + 1, we can parametrize the line or express it parametrically, then apply the transformation. Alternatively, we can find two points on the original line, apply the transformation to them, and then find the equation of the line through the transformed points.

Let me try both approaches and see which is easier.

First, maybe parametrize the line. Let me think.

Let me take a parameter t such that x = t, then y = 2t + 1. So any point on the line can be written as (t, 2t + 1). Then applying matrix M to this vector gives:

x' = 4*t + (-1)*(2t + 1) = 4t - 2t - 1 = 2t - 1

y' = 2*t + 1*(2t + 1) = 2t + 2t + 1 = 4t + 1

So now, the transformed coordinates are x' = 2t - 1 and y' = 4t + 1.

Now, we need to eliminate the parameter t to find the relationship between x' and y'.

From x' = 2t - 1, solve for t:

x' + 1 = 2t => t = (x' + 1)/2

Substitute into y':

y' = 4*( (x' + 1)/2 ) + 1 = 2*(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3

So, y' = 2x' + 3. Therefore, the image of the line under M is y = 2x + 3.

But let me verify this with another approach to be sure.

Alternatively, take two points on the original line, apply the matrix to them, then find the equation of the line through their images.

Take two points on y = 2x + 1. Let's choose two points.

First, when x = 0, y = 1. So point A is (0, 1).

Second, when x = 1, y = 2*1 + 1 = 3. So point B is (1, 3).

Now, apply matrix M to these points.

For point A (0, 1):

x' = 4*0 + (-1)*1 = 0 -1 = -1

y' = 2*0 + 1*1 = 0 + 1 = 1

So image of A is (-1, 1).

For point B (1, 3):

x' = 4*1 + (-1)*3 = 4 - 3 = 1

y' = 2*1 + 1*3 = 2 + 3 = 5

So image of B is (1, 5).

Now, find the equation of the line passing through (-1, 1) and (1, 5).

First, compute the slope m:

m = (5 - 1)/(1 - (-1)) = 4 / 2 = 2.

So slope is 2. Now, using point-slope form with point (-1, 1):

y - 1 = 2(x - (-1)) => y - 1 = 2(x + 1) => y = 2x + 2 + 1 => y = 2x + 3.

Same result as before! So that confirms that the image is y = 2x + 3.

Alternatively, let me check another point. Let's take point (2, 5) on the original line y=2x +1. Applying matrix M:

x' = 4*2 + (-1)*5 = 8 -5 = 3

y' = 2*2 + 1*5 = 4 + 5 = 9

So the transformed point is (3, 9). Let's check if it lies on y = 2x + 3:

2*3 + 3 = 6 + 3 = 9. Yes, that's correct. So (3,9) is indeed on y = 2x + 3.

Therefore, seems like the answer is y = 2x + 3.

Wait, but let me think again if there is another way someone might approach this, just to make sure.

Another approach: The line y = 2x + 1 can be represented as a set of vectors. The transformation M maps this line to another line. Since linear transformations map lines to lines (unless the line is mapped to a point, which would be if the line is in the kernel, but since M is invertible, the image of a line is a line).

Alternatively, we can find the equation of the image line by considering the transformation of the original line. Since the original line is y = 2x + 1, we can write it as y - 2x - 1 = 0. Let me consider the general equation of a line: ax + by + c = 0. But here, it's y - 2x -1 = 0.

But when we apply a linear transformation, how does it affect the equation? Hmm, not sure. Maybe using coordinates transformation? Let me think.

Alternatively, suppose (x, y) is on the original line. Then (x', y') = M*(x, y). So x' = 4x - y, y' = 2x + y. Wait, that's from the matrix M:

Original coordinates (x, y), then transformed coordinates:

x' = 4x - y

y' = 2x + y

So, we can express x and y in terms of x' and y', and then substitute into the original line equation.

But solving for x and y from x' and y':

We have:

x' = 4x - y

y' = 2x + y

We can solve this system for x and y. Let's write:

Equation 1: 4x - y = x'

Equation 2: 2x + y = y'

Adding equations 1 and 2:

(4x - y) + (2x + y) = x' + y'

=> 6x = x' + y'

So x = (x' + y') / 6

Similarly, to find y, subtract equation 1 and equation 2? Let me see:

Equation 1: 4x - y = x'

Equation 2: 2x + y = y'

Let me solve for y. Let me add equations 1 and 2 to get x as above. Alternatively, maybe express y from equation 2:

From equation 2: y = y' - 2x. Substitute into equation 1:

4x - (y' - 2x) = x'

=> 4x - y' + 2x = x'

=> 6x - y' = x'

=> 6x = x' + y'

Same as before. So x = (x' + y') / 6.

Then, substitute back into equation 2 to find y:

y = y' - 2x = y' - 2*(x' + y')/6 = y' - (x' + y')/3 = (3y' - x' - y')/3 = (2y' - x') / 3

So, y = (2y' - x') / 3.

Now, original line equation is y = 2x + 1. Substitute x and y in terms of x', y':

(2y' - x') / 3 = 2*( (x' + y') / 6 ) + 1

Simplify the right-hand side:

2*( (x' + y') / 6 ) + 1 = (x' + y') / 3 + 1

So, equation:

(2y' - x') / 3 = (x' + y') / 3 + 1

Multiply both sides by 3:

2y' - x' = x' + y' + 3

Bring all terms to the left side:

2y' - x' - x' - y' - 3 = 0 => (2y' - y') + (-x' - x') -3 = 0 => y' - 2x' - 3 = 0 => y' = 2x' + 3

Which gives y = 2x + 3. Same answer. So this method also confirms it. So definitely, the image is y = 2x + 3.

Therefore, the answer is y = 2x + 3.

But just to be thorough, let me check another point. Let's take the original point ( -1, -1) on the line y = 2x + 1. Wait, when x = -1, y = 2*(-1) + 1 = -2 + 1 = -1. So (-1, -1) is on the line.

Applying matrix M:

x' = 4*(-1) - (-1) = -4 + 1 = -3

y' = 2*(-1) + (-1) = -2 -1 = -3

So transformed point is (-3, -3). Let's check if that lies on y = 2x + 3:

2*(-3) + 3 = -6 + 3 = -3. Yes, y' = -3 matches. So that works.

Another point: Let me take (0.5, 2*0.5 +1)= (0.5, 2). Applying M:

x' = 4*(0.5) - 2 = 2 - 2 = 0

y' = 2*(0.5) + 2 = 1 + 2 = 3

So transformed point (0, 3). Check if it's on y = 2x + 3: 2*0 + 3 = 3. Correct.

Therefore, all these checks confirm that the image is y = 2x + 3.

Therefore, I can be confident the answer is y = 2x + 3.

**Final Answer**
The image of the line under matrix $\mathbf{M}$ is \boxed{y = 2x + 3}.

To solve the problem, we first determine the matrix $\mathbf{M}$ that satisfies the given mappings:

- $\mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \end{pmatrix}$
- $\mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \end{pmatrix}$

Let’s assume the matrix $\mathbf{M}$ has the form:

$$
\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
$$

Applying the matrix to the input vectors gives us four equations:

1. $2a - b = 9$
2. $2c - d = 3$
3. $a - 3b = 7$
4. $c - 3d = -1$

### Step 1: Solve for $a$, $b$, $c$, and $d$

**From equations (1) and (3):**

- From (1): $b = 2a - 9$
- Substitute into (3): $a - 3(2a - 9) = 7$
- Simplify: $a - 6a + 27 = 7 \Rightarrow -5a = -20 \Rightarrow a = 4$

Now compute $b$:

$$
b = 2a - 9 = 2(4) - 9 = 8 - 9 = -1
$$

**From equations (2) and (4):**

- From (2): $d = 2c - 3$
- Substitute into (4): $c - 3(2c - 3) = -1$
- Simplify: $c - 6c + 9 = -1 \Rightarrow -5c = -10 \Rightarrow c = 2$

Now compute $d$:

$$
d = 2c - 3 = 2(2) - 3 = 4 - 3 = 1
$$

So, the matrix is:

$$
\mathbf{M} = \begin{pmatrix} 4 & -1 \\ 2 & 1 \end{pmatrix}
$$

### Step 2: Find the image of the line $y = 2x + 1$

We find the image of the line under the transformation by applying $\mathbf{M}$ to a point on the line.

Let $\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ be a point on the line $y = 2x + 1$, so $y = 2x + 1$. The transformation is:

$$
\mathbf{M} \cdot \begin{pmatrix} x \\ 2x + 1 \end{pmatrix}
= \begin{pmatrix} 4x - (2x + 1) \\ 2x + (2x + 1) \end{pmatrix}
= \begin{pmatrix} 2x - 1 \\ 4x + 1 \end{pmatrix}
$$

Let the transformed point be $(x', y')$. Then:

$$
x' = 2x - 1 \quad \text{and} \quad y' = 4x + 1
$$

We eliminate the parameter $x$ to find the relation between $x'$ and $y'$. From the first equation:

$$
x = \frac{x' + 1}{2}
$$

Substitute into the second equation:

$$
y' = 4\left(\frac{x' + 1}{2}\right) + 1 = 2(x' + 1) + 1 = 2x' + 3
$$

Thus, the image of the line under the transformation is:

$$
y = 2x + 3
$$

---

### Final Answer

$$
\boxed{y = 2x + 3}
$$

Solution:
We have that $\mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \end{pmatrix}$ and $\mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \end{pmatrix}.$  Then $\mathbf{M} \begin{pmatrix} 6 \\ -3 \end{pmatrix} = \begin{pmatrix} 27 \\ 9 \end{pmatrix},$ so
\[\mathbf{M} \begin{pmatrix} 6 \\ -3 \end{pmatrix} - \mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 27 \\ 9 \end{pmatrix} - \begin{pmatrix} 7 \\ -1 \end{pmatrix}.\]This gives us $\mathbf{M} \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 20 \\ 10 \end{pmatrix},$ so
\[\mathbf{M} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}.\]Then
\[\mathbf{M} \begin{pmatrix} 1 \\ 0 \end{pmatrix} - \mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} - \begin{pmatrix} 7 \\ -1 \end{pmatrix}.\]This gives us $\mathbf{M} \begin{pmatrix} 0 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 3 \end{pmatrix},$ so
\[\mathbf{M} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}.\]Finally,
\begin{align*}
\mathbf{M} \begin{pmatrix} 1 \\ 3 \end{pmatrix} &= \mathbf{M} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + 3 \mathbf{M} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\
&= \begin{pmatrix} 4 \\ 2 \end{pmatrix} + 3 \begin{pmatrix} -1 \\ 1 \end{pmatrix} \\
&= \begin{pmatrix} 1 \\ 5 \end{pmatrix}.
\end{align*}Since $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ lie on the line $y = 2x + 1,$ we want to compute the equation of the line through $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 5 \end{pmatrix}.$  The equation of this line is $\boxed{y = 2x + 3}.$


Explanation:
Okay, so I need to figure out how much a 3-inch cube would cost to make, given the costs for 1-inch and 2-inch cubes. The problem says that fake gold bricks are made by covering concrete cubes with gold paint. The cost of the paint is proportional to the surface area, and the cost of the concrete is proportional to the volume.

First, let me recall the formulas for surface area and volume of a cube. The surface area of a cube is 6 times the side length squared, right? So if the side length is s, then surface area A = 6s². The volume is s cubed, so V = s³.

So, the total cost of making a cube would be the sum of the cost of the paint (which is proportional to surface area) and the cost of the concrete (proportional to volume). Let me denote the cost of paint per unit surface area as k₁ and the cost of concrete per unit volume as k₂. Then, the total cost C for a cube with side length s would be:

C = k₁ * A + k₂ * V = k₁ * 6s² + k₂ * s³.

Given that, we have two data points: a 1-inch cube costs $1.30 and a 2-inch cube costs $6.80. Let me write those equations down.

For the 1-inch cube:

C₁ = k₁ * 6*(1)² + k₂*(1)³ = 6k₁ + k₂ = 1.30.

For the 2-inch cube:

C₂ = k₁ * 6*(2)² + k₂*(2)³ = 6k₁*4 + k₂*8 = 24k₁ + 8k₂ = 6.80.

So we have a system of two equations:

1) 6k₁ + k₂ = 1.30

2) 24k₁ + 8k₂ = 6.80

Now, I need to solve this system for k₁ and k₂. Let me use substitution or elimination. Let's try elimination.

If I multiply the first equation by 8, then:

1) 6k₁*8 + k₂*8 = 1.30*8

Which is 48k₁ + 8k₂ = 10.40

Now, subtract the second equation from this:

(48k₁ + 8k₂) - (24k₁ + 8k₂) = 10.40 - 6.80

Calculating left side: 48k₁ -24k₁ + 8k₂ -8k₂ = 24k₁

Right side: 10.40 - 6.80 = 3.60

So 24k₁ = 3.60 => k₁ = 3.60 / 24 = 0.15

So k₁ is $0.15 per unit surface area? Wait, units? Let me confirm. The cost of paint is proportional to surface area, so k₁ is the cost per unit surface area. Similarly, k₂ is cost per unit volume.

So k₁ = 0.15 dollars per square inch? Wait, actually, the units might be dollars per square inch for k₁ and dollars per cubic inch for k₂. But since we don't need to worry about units for the final answer, as long as we are consistent.

So once we have k₁ = 0.15, we can plug back into the first equation to find k₂.

From equation 1):

6k₁ + k₂ = 1.30 => 6*(0.15) + k₂ = 1.30 => 0.90 + k₂ = 1.30 => k₂ = 1.30 - 0.90 = 0.40

So k₂ = $0.40 per unit volume.

Now, let's check if these values work with the second equation.

24k₁ + 8k₂ = 24*(0.15) + 8*(0.40) = 3.60 + 3.20 = 6.80, which matches the given cost for the 2-inch cube. Good.

Now, we need to find the cost for a 3-inch cube. Using the formula:

C = 6k₁s² + k₂s³.

So plugging in s = 3 inches:

C = 6*(0.15)*(3)² + 0.40*(3)³.

Let me compute each term step by step.

First, compute 6*(0.15): that's 0.90.

Then multiply by 3² = 9: 0.90 * 9 = 8.10

Next term: 0.40*(27) since 3³ = 27. 0.40*27.

Let me calculate 0.4 * 27: 0.4*20 = 8.0, 0.4*7 = 2.8, total 10.80.

Therefore total cost is 8.10 + 10.80 = 18.90.

Wait, so the total cost for a 3-inch cube is $18.90?

But let me double-check my calculations to be sure.

First, k₁ = 0.15, k₂ = 0.40.

For 3-inch cube:

Surface area: 6*(3)^2 = 6*9 = 54 square inches.

Volume: 3^3 = 27 cubic inches.

So cost of paint: 0.15 * 54 = 8.10 dollars.

Cost of concrete: 0.40 * 27 = 10.80 dollars.

Total: 8.10 + 10.80 = 18.90. Yeah, same answer.

Hmm, but let me check if there's another way to approach this problem? Maybe using ratios?

Alternatively, since the cost is linear in terms of surface area and volume, perhaps we can think of the cost as a function of s. Let me think.

But maybe not necessary since the previous method worked. Wait, but just to check if there's an alternative way. Maybe think of the cost as C(s) = k₁*6s² + k₂*s³. We have two equations and solved for k₁ and k₂. So seems solid.

Alternatively, maybe the problem is expecting something else? Let me see. Maybe the cost is proportional to surface area and volume. But since the problem says "the cost of the paint is proportional to their surface area while the cost of the concrete is proportional to their volume." So total cost is sum of the two.

So our approach is correct.

Alternatively, maybe there's a different way to model it? Let me think. Suppose that the cost is proportional to surface area plus proportional to volume. So, C = a * surface area + b * volume. Which is exactly what we have.

So solving for a and b gives us the same equations. So yes, the answer is $18.90. Let me check if that seems reasonable.

Looking at the costs:

1-inch cube: surface area 6, volume 1. So cost is 6a + 1b = 1.30

2-inch cube: surface area 24, volume 8. So cost 24a + 8b = 6.80

Which is exactly what we had. Then solving gives a = 0.15, b = 0.40. So for 3-inch cube, surface area 54, volume 27.

So 54*0.15 + 27*0.40 = 8.10 + 10.80 = 18.90. So yes. Seems correct.

But maybe check if maybe the cost is supposed to be something else? For example, maybe the cost is a combination where the paint is proportional to surface area and concrete proportional to volume, but maybe they have different units?

Wait, but in the problem statement, it's just saying that the cost of the paint is proportional to surface area, and the cost of concrete is proportional to volume. So we can assume that there are constants k_p and k_c such that total cost is k_p * surface area + k_c * volume.

Therefore, the answer is $18.90. So in boxed form, it would be \boxed{18.90}

But let me check once again all steps for errors.

First, equations:

For 1 inch:

6k₁ + k₂ = 1.30.

For 2 inches:

24k₁ + 8k₂ = 6.80.

Multiply first equation by 8: 48k₁ + 8k₂ = 10.40.

Subtract second equation: 48k₁ +8k₂ -24k₁ -8k₂ = 10.40 -6.80 => 24k₁ = 3.60 => k₁= 3.60 /24 = 0.15. Correct.

Then k₂= 1.30 -6*0.15 = 1.30 - 0.90 = 0.40. Correct.

Then for 3 inches:

Surface area: 6*(3)^2 = 54. Volume: 27. So cost is 54*0.15 +27*0.40.

Calculating 54*0.15. Let me compute 50*0.15=7.5, 4*0.15=0.6, so total 8.1. Correct.

27*0.4: 27*0.4 = 10.8. Correct. 8.1 +10.8=18.9. So yes.

Therefore, the answer is \$18.90.

But maybe check if the question expects cents? But \$18.90 is correct. Maybe written as 18.9? But no, the given costs are in dollars with two decimal places. \$1.30 and \$6.80. So \$18.90 is correct.

Alternatively, maybe fractions? Let me check if 0.15 and 0.40 can be represented as fractions.

0.15 is 3/20, 0.40 is 2/5.

So for the 3-inch cube:

Surface area 54. 54*(3/20) = (54/20)*3 = (27/10)*3 = 81/10 = 8.1

Volume 27. 27*(2/5) = 54/5 = 10.8

So total is 8.1 +10.8= 18.9. Correct.

So yes, \$18.90.

Therefore, the answer is \boxed{18.90}.

**Final Answer**
The cost of a 3 inch cube is \boxed{18.90} dollars.

To determine the cost of a 3-inch cube, we analyze the given information about a 1-inch and 2-inch cube, and use the relationship between the cost, surface area, and volume of cubes.

---

### Step 1: Understand the Cost Structure

We are told that:

- The cost of the **paint** is **proportional** to the **surface area**.
- The cost of the **concrete** is **proportional** to the **volume**.

Let:

- $ k_1 $ = cost per unit surface area (dollars per square inch)
- $ k_2 $ = cost per unit volume (dollars per cubic inch)

For a cube with side length $ s $:

- **Surface Area** = $ 6s^2 $
- **Volume** = $ s^3 $

Thus, the total cost $ C $ of a cube is:

$$
C = k_1 \cdot 6s^2 + k_2 \cdot s^3
$$

---

### Step 2: Set Up Equations Based on Given Data

We are told:

- A **1-inch** cube costs \$1.30
- A **2-inch** cube costs \$6.80

Using the formula:

1. For $ s = 1 $:

$$
6k_1 + k_2 = 1.30 \quad \text{(Equation 1)}
$$

2. For $ s = 2 $:

$$
24k_1 + 8k_2 = 6.80 \quad \text{(Equation 2)}
$$

---

### Step 3: Solve the System of Equations

From Equation 1:

$$
6k_1 + k_2 = 1.30 \quad \Rightarrow \quad k_2 = 1.30 - 6k_1
$$

Substitute into Equation 2:

$$
24k_1 + 8(1.30 - 6k_1) = 6.80 \\
24k_1 + 10.40 - 48k_1 = 6.80 \\
-24k_1 + 10.40 = 6.80 \\
-24k_1 = -3.60 \\
k_1 = \frac{3.60}{24} = 0.15
$$

Now find $ k_2 $:

$$
k_2 = 1.30 - 6(0.15) = 1.30 - 0.90 = 0.40
$$

---

### Step 4: Calculate Cost for a 3-inch Cube

Now, calculate the cost for a cube with side length $ s = 3 $:

- Surface area = $ 6(3)^2 = 54 $
- Volume = $ 3^3 = 27 $

$$
C = 0.15 \cdot 54 + 0.40 \cdot 27 = 8.10 + 10.80 = 18.90
$$

---

### Final Answer

The cost of a 3-inch cube is:

$$
\boxed{18.90}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how much a 3-inch cube would cost to make, given the costs for 1-inch and 2-inch cubes. The problem says that fake gold bricks are made by covering concrete cubes with gold paint. The cost of the paint is proportional to the surface area, and the cost of the concrete is proportional to the volume.

First, let me recall the formulas for surface area and volume of a cube. The surface area of a cube is 6 times the side length squared, right? So if the side length is s, then surface area A = 6s². The volume is s cubed, so V = s³.

So, the total cost of making a cube would be the sum of the cost of the paint (which is proportional to surface area) and the cost of the concrete (proportional to volume). Let me denote the cost of paint per unit surface area as k₁ and the cost of concrete per unit volume as k₂. Then, the total cost C for a cube with side length s would be:

C = k₁ * A + k₂ * V = k₁ * 6s² + k₂ * s³.

Given that, we have two data points: a 1-inch cube costs $1.30 and a 2-inch cube costs $6.80. Let me write those equations down.

For the 1-inch cube:

C₁ = k₁ * 6*(1)² + k₂*(1)³ = 6k₁ + k₂ = 1.30.

For the 2-inch cube:

C₂ = k₁ * 6*(2)² + k₂*(2)³ = 6k₁*4 + k₂*8 = 24k₁ + 8k₂ = 6.80.

So we have a system of two equations:

1) 6k₁ + k₂ = 1.30

2) 24k₁ + 8k₂ = 6.80

Now, I need to solve this system for k₁ and k₂. Let me use substitution or elimination. Let's try elimination.

If I multiply the first equation by 8, then:

1) 6k₁*8 + k₂*8 = 1.30*8

Which is 48k₁ + 8k₂ = 10.40

Now, subtract the second equation from this:

(48k₁ + 8k₂) - (24k₁ + 8k₂) = 10.40 - 6.80

Calculating left side: 48k₁ -24k₁ + 8k₂ -8k₂ = 24k₁

Right side: 10.40 - 6.80 = 3.60

So 24k₁ = 3.60 => k₁ = 3.60 / 24 = 0.15

So k₁ is $0.15 per unit surface area? Wait, units? Let me confirm. The cost of paint is proportional to surface area, so k₁ is the cost per unit surface area. Similarly, k₂ is cost per unit volume.

So k₁ = 0.15 dollars per square inch? Wait, actually, the units might be dollars per square inch for k₁ and dollars per cubic inch for k₂. But since we don't need to worry about units for the final answer, as long as we are consistent.

So once we have k₁ = 0.15, we can plug back into the first equation to find k₂.

From equation 1):

6k₁ + k₂ = 1.30 => 6*(0.15) + k₂ = 1.30 => 0.90 + k₂ = 1.30 => k₂ = 1.30 - 0.90 = 0.40

So k₂ = $0.40 per unit volume.

Now, let's check if these values work with the second equation.

24k₁ + 8k₂ = 24*(0.15) + 8*(0.40) = 3.60 + 3.20 = 6.80, which matches the given cost for the 2-inch cube. Good.

Now, we need to find the cost for a 3-inch cube. Using the formula:

C = 6k₁s² + k₂s³.

So plugging in s = 3 inches:

C = 6*(0.15)*(3)² + 0.40*(3)³.

Let me compute each term step by step.

First, compute 6*(0.15): that's 0.90.

Then multiply by 3² = 9: 0.90 * 9 = 8.10

Next term: 0.40*(27) since 3³ = 27. 0.40*27.

Let me calculate 0.4 * 27: 0.4*20 = 8.0, 0.4*7 = 2.8, total 10.80.

Therefore total cost is 8.10 + 10.80 = 18.90.

Wait, so the total cost for a 3-inch cube is $18.90?

But let me double-check my calculations to be sure.

First, k₁ = 0.15, k₂ = 0.40.

For 3-inch cube:

Surface area: 6*(3)^2 = 6*9 = 54 square inches.

Volume: 3^3 = 27 cubic inches.

So cost of paint: 0.15 * 54 = 8.10 dollars.

Cost of concrete: 0.40 * 27 = 10.80 dollars.

Total: 8.10 + 10.80 = 18.90. Yeah, same answer.

Hmm, but let me check if there's another way to approach this problem? Maybe using ratios?

Alternatively, since the cost is linear in terms of surface area and volume, perhaps we can think of the cost as a function of s. Let me think.

But maybe not necessary since the previous method worked. Wait, but just to check if there's an alternative way. Maybe think of the cost as C(s) = k₁*6s² + k₂*s³. We have two equations and solved for k₁ and k₂. So seems solid.

Alternatively, maybe the problem is expecting something else? Let me see. Maybe the cost is proportional to surface area and volume. But since the problem says "the cost of the paint is proportional to their surface area while the cost of the concrete is proportional to their volume." So total cost is sum of the two.

So our approach is correct.

Alternatively, maybe there's a different way to model it? Let me think. Suppose that the cost is proportional to surface area plus proportional to volume. So, C = a * surface area + b * volume. Which is exactly what we have.

So solving for a and b gives us the same equations. So yes, the answer is $18.90. Let me check if that seems reasonable.

Looking at the costs:

1-inch cube: surface area 6, volume 1. So cost is 6a + 1b = 1.30

2-inch cube: surface area 24, volume 8. So cost 24a + 8b = 6.80

Which is exactly what we had. Then solving gives a = 0.15, b = 0.40. So for 3-inch cube, surface area 54, volume 27.

So 54*0.15 + 27*0.40 = 8.10 + 10.80 = 18.90. So yes. Seems correct.

But maybe check if maybe the cost is supposed to be something else? For example, maybe the cost is a combination where the paint is proportional to surface area and concrete proportional to volume, but maybe they have different units?

Wait, but in the problem statement, it's just saying that the cost of the paint is proportional to surface area, and the cost of concrete is proportional to volume. So we can assume that there are constants k_p and k_c such that total cost is k_p * surface area + k_c * volume.

Therefore, the answer is $18.90. So in boxed form, it would be \boxed{18.90}

But let me check once again all steps for errors.

First, equations:

For 1 inch:

6k₁ + k₂ = 1.30.

For 2 inches:

24k₁ + 8k₂ = 6.80.

Multiply first equation by 8: 48k₁ + 8k₂ = 10.40.

Subtract second equation: 48k₁ +8k₂ -24k₁ -8k₂ = 10.40 -6.80 => 24k₁ = 3.60 => k₁= 3.60 /24 = 0.15. Correct.

Then k₂= 1.30 -6*0.15 = 1.30 - 0.90 = 0.40. Correct.

Then for 3 inches:

Surface area: 6*(3)^2 = 54. Volume: 27. So cost is 54*0.15 +27*0.40.

Calculating 54*0.15. Let me compute 50*0.15=7.5, 4*0.15=0.6, so total 8.1. Correct.

27*0.4: 27*0.4 = 10.8. Correct. 8.1 +10.8=18.9. So yes.

Therefore, the answer is \$18.90.

But maybe check if the question expects cents? But \$18.90 is correct. Maybe written as 18.9? But no, the given costs are in dollars with two decimal places. \$1.30 and \$6.80. So \$18.90 is correct.

Alternatively, maybe fractions? Let me check if 0.15 and 0.40 can be represented as fractions.

0.15 is 3/20, 0.40 is 2/5.

So for the 3-inch cube:

Surface area 54. 54*(3/20) = (54/20)*3 = (27/10)*3 = 81/10 = 8.1

Volume 27. 27*(2/5) = 54/5 = 10.8

So total is 8.1 +10.8= 18.9. Correct.

So yes, \$18.90.

Therefore, the answer is \boxed{18.90}.

**Final Answer**
The cost of a 3 inch cube is \boxed{18.90} dollars.

To determine the cost of a 3-inch cube, we analyze the given information about a 1-inch and 2-inch cube, and use the relationship between the cost, surface area, and volume of cubes.

---

### Step 1: Understand the Cost Structure

We are told that:

- The cost of the **paint** is **proportional** to the **surface area**.
- The cost of the **concrete** is **proportional** to the **volume**.

Let:

- $ k_1 $ = cost per unit surface area (dollars per square inch)
- $ k_2 $ = cost per unit volume (dollars per cubic inch)

For a cube with side length $ s $:

- **Surface Area** = $ 6s^2 $
- **Volume** = $ s^3 $

Thus, the total cost $ C $ of a cube is:

$$
C = k_1 \cdot 6s^2 + k_2 \cdot s^3
$$

---

### Step 2: Set Up Equations Based on Given Data

We are told:

- A **1-inch** cube costs \$1.30
- A **2-inch** cube costs \$6.80

Using the formula:

1. For $ s = 1 $:

$$
6k_1 + k_2 = 1.30 \quad \text{(Equation 1)}
$$

2. For $ s = 2 $:

$$
24k_1 + 8k_2 = 6.80 \quad \text{(Equation 2)}
$$

---

### Step 3: Solve the System of Equations

From Equation 1:

$$
6k_1 + k_2 = 1.30 \quad \Rightarrow \quad k_2 = 1.30 - 6k_1
$$

Substitute into Equation 2:

$$
24k_1 + 8(1.30 - 6k_1) = 6.80 \\
24k_1 + 10.40 - 48k_1 = 6.80 \\
-24k_1 + 10.40 = 6.80 \\
-24k_1 = -3.60 \\
k_1 = \frac{3.60}{24} = 0.15
$$

Now find $ k_2 $:

$$
k_2 = 1.30 - 6(0.15) = 1.30 - 0.90 = 0.40
$$

---

### Step 4: Calculate Cost for a 3-inch Cube

Now, calculate the cost for a cube with side length $ s = 3 $:

- Surface area = $ 6(3)^2 = 54 $
- Volume = $ 3^3 = 27 $

$$
C = 0.15 \cdot 54 + 0.40 \cdot 27 = 8.10 + 10.80 = 18.90
$$

---

### Final Answer

The cost of a 3-inch cube is:

$$
\boxed{18.90}
$$

Solution: Let $x$ be the cost per square inch of gold paint, and let $y$ be the cost per cubic inch of concrete.  Since a 1 inch cube has surface area 6 $\text{in}^2$ and volume 1 $\text{in}^3$ its total price will be $6x+y$ dollars.  Similarly, a 2 inch cube has surface area 24 $\text{in}^2$ and volume 8 $\text{in}^3$, so its total price will be $24x+8y$ dollars.  We are given that \begin{align*} 6x+y &=\$1.30 \\ 24x+8y&= \$6.80 \end{align*} Subtracting 4 times the first equation from the second gives $4y=\$1.60$, so $y=\$0.40$.  Hence $6x=\$0.90$, so $x=\$0.15$.  Since a 3 inch cube has surface area 54 $\text{in}^2$ and volume 27 $\text{in}^3$, its total price will be a total of $54(\$0.15)+27(\$0.40)=\boxed{\$18.90}$.