<Question>
What is the average of the two smallest positive integer solutions to the congruence $$14u \equiv 46 \pmod{100}~?$$
</Question>

<think>
<Plan_1>
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 start by recalling how to solve linear congruences. 

First, a linear congruence of the form ax ≡ b mod m has solutions if and only if the greatest common divisor (gcd) of a and m divides b. So, I should check the gcd of 14 and 100. Let me compute that.
The prime factors of 14 are 2 and 7, and the prime factors of 100 are 2^2 and 5^2. So, the gcd is 2. 

Since 2 divides 46 (because 46 divided by 2 is 23), there should be solutions. The number of solutions modulo 100 will be equal to the gcd, which is 2. That means there are two distinct solutions modulo 100. So, the solutions will be two numbers between 1 and 100, and then every 100/2 = 50 numbers after that. Wait, but since we're looking for the two smallest positive integer solutions, they should be within the first 100 numbers, right?

Let me confirm. To solve 14u ≡ 46 mod 100, since gcd(14, 100) = 2, I can divide the entire congruence by 2 to simplify it. Let's do that: 

Divide both sides by 2: 7u ≡ 23 mod 50.

Now, the congruence is 7u ≡ 23 mod 50. Here, gcd(7, 50) is 1 because 7 is prime and doesn't divide 50. So, 7 has an inverse modulo 50. To find the inverse of 7 mod 50, I need a number x such that 7x ≡ 1 mod 50.

Let me find that inverse. Let's try to solve 7x ≡ 1 mod 50. Testing small numbers:

<prompt_1.1> Find the modular inverse of 7 modulo 50 by testing small numbers. </prompt_1.1>
-----

So the solutions are u ≡ 29 mod 50. But since the original modulus was 100, do I need to adjust these solutions?

<prompt_1.2> Determine the solutions to the original congruence 14u ≡ 46 mod 100. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
7*1=7 → 7
7*2=14 →14
7*3=21→21
7*4=28→28
7*5=35→35
7*6=42→42
7*7=49→49
7*8=56→56 mod50 is 6
7*9=63→13
7*10=70→20
7*11=77→27
7*12=84→34
7*13=91→41
7*14=98→48
7*15=105→5
</execution_1.1>

<execution_1.2>
Wait, when we divided by 2, we went from modulus 100 to modulus 50. So, the solutions modulo 50 are 29, and then adding 50 gives the next solution modulo 100. Therefore, the solutions modulo 100 are 29 and 79. 

Therefore, the two smallest positive integer solutions are 29 and 79. Their average is (29 + 79)/2 = 108/2 = 54.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
None of these are 1. Let me try 7*11=77→77-50=27, 27. 7*13=91-2*50=91-100=-9, which is 41 mod50. Hmm. Maybe I need a better method than trial and error.

Using the extended Euclidean algorithm to find x such that 7x ≡1 mod50. 

The extended Euclidean algorithm finds integers x and y such that 7x + 50y = 1.

<prompt_2.1> Use the extended Euclidean algorithm to find the inverse of 7 modulo 50. </prompt_2.1>
-----

Based on execution_1.2:
Wait, let me verify that. Let's check if 29 and 79 satisfy the original congruence 14u ≡46 mod100.

<prompt_2.2> Verify that 29 and 79 are solutions to the original congruence 14u ≡ 46 mod 100. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let me apply the algorithm:

50 divided by 7 is 7, remainder 1. So, 50 = 7*7 + 1.

Then, 7 divided by 1 is 7, remainder 0. So, gcd is 1. Now, backtracking:

1 = 50 - 7*7.

Therefore, x = -7 and y = 1 is a solution. So, the inverse of 7 mod50 is -7 mod50, which is 43 because 50 -7=43. Let me check: 7*43=301. 301 divided by 50 is 6*50=300, so 301-300=1. Yes, 7*43≡1 mod50. So, the inverse is 43.
</execution_2.1>

<execution_2.2>
For u=29: 14*29. Let's compute 14*30=420, subtract 14 to get 14*29=420-14=406. Now, 406 mod100 is 6. But 6 is not 46. Wait, that can't be right. Did I make a mistake here? 

Wait, wait, wait. Hold on. If 7u ≡23 mod50, and u=29, then 7*29=203. 203 mod50 is 203 - 4*50=203-200=3. But 3 is not 23. So, that means I messed up somewhere.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Therefore, multiplying both sides of 7u ≡23 mod50 by 43: u ≡23*43 mod50.

<prompt_3.1> Solve the congruence 7u ≡ 23 mod 50 using the modular inverse. </prompt_3.1>
-----

Based on execution_2.2:
Let me check my steps again.

<prompt_3.2> Re-evaluate the steps to verify the solution of the congruence. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let me compute 23*43. 20*43=860, 3*43=129. So, 860+129=989. Now, 989 mod50. 50*19=950, so 989-950=39. Therefore, u ≡39 mod50. Wait, so u ≡39 mod50? But earlier I thought it was 29. Hmm, maybe I made a mistake earlier. Let me check.
</execution_3.1>

<execution_3.2>
Original congruence: 14u ≡46 mod100.

Divide by gcd(14,100)=2: 7u≡23 mod50. Correct. Then, inverse of 7 mod50 is 43. 7*43=301≡1 mod50. Correct. So, u≡23*43 mod50. Let's compute 23*43. 20*43=860, 3*43=129. So, 860+129=989. 989 mod50: 50*19=950, 989-950=39. So, u≡39 mod50. Wait, so u≡39 mod50. Then, the solutions modulo100 would be 39 and 39+50=89. Then, the two smallest positive solutions are 39 and 89. Then, their average is (39+89)/2=128/2=64.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Wait, so going back, after dividing by 2, we had 7u ≡23 mod50. Then, multiplying both sides by 43 gives u ≡23*43 mod50. So, 23*43: Let me compute this again. 20*43=860, 3*43=129. 860+129=989. 989 divided by 50: 50*19=950, 989-950=39. So, yes, 989≡39 mod50. Therefore, u≡39 mod50. 

Therefore, the solutions to the congruence 7u ≡23 mod50 are all integers congruent to 39 modulo50. Therefore, the solutions mod100 (since we divided by 2 earlier) would be u≡39 mod50. But since modulus 100, the solutions would be 39 and 39+50=89.
-----

Based on execution_3.2:
But wait, this contradicts my previous result. So, I must have made a mistake earlier when I thought the inverse was 15. Let's see. 

Wait, when I first tried to find the inverse of 7 mod50, I said 7*15=105≡5 mod50, which is not 1. Then, I tried to use the extended Euclidean algorithm and found x=-7, which is 43 mod50. So, that's correct. Then, 23*43=989≡39 mod50. Therefore, u≡39 mod50. 

Therefore, the solutions modulo100 are 39 and 89. Hence, the two smallest positive integer solutions are 39 and 89. Their average is (39+89)/2=128/2=64. 

Wait, so where did I get 29 earlier? Let me check. Maybe I confused the inverse? Let me verify 7*43: 7*40=280, 7*3=21, so 280+21=301. 301-6*50=301-300=1. So, 7*43=1 mod50. Correct. So, 23*43=989. 989 mod50: 50*19=950, 989-950=39. So, 39. 

So, the mistake earlier was when I thought the inverse was 15, but actually, the inverse is 43. Therefore, the correct solution is u≡39 mod50. Therefore, the two solutions are 39 and 89. 

But let me check 14*39. 14*40=560, subtract 14 to get 14*39=560-14=546. 546 mod100 is 46. Correct. Similarly, 14*89=14*(90-1)=1260-14=1246. 1246 mod100 is 46. So, both 39 and 89 are solutions.

<prompt_4.1> Verify the solutions 39 and 89 for the congruence 14u ≡ 46 mod 100. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
So, the two smallest positive integer solutions are 39 and 89. Therefore, their average is (39 +89)/2=128/2=64. 

Wait, so why did I initially get 29 and 79? Because I must have miscalculated the inverse. Let me check again. If the inverse is 43, not 15. So, 23*43=989≡39 mod50. So, 39 is the solution mod50. Then, mod100, adding 50 gives 89. So, 39 and 89. 

So, the answer should be 64.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But let me check once more. Let's see, if the original congruence is 14u ≡46 mod100.

<prompt_5.1> Re-confirm the solution and verify the calculations for the congruence 14u ≡ 46 mod 100. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Divided by 2: 7u≡23 mod50. Solution u≡39 mod50. Then, mod100, solutions are 39 and 89. 

Yes, that's correct. Therefore, the average is 64. 

But let me check another way. Let's solve 14u ≡46 mod100. We can write this as 14u = 46 + 100k for some integer k. Then, 14u = 46 + 100k. Dividing both sides by 2: 7u = 23 + 50k. So, 7u ≡23 mod50. 

Solving 7u ≡23 mod50. As before, inverse of 7 mod50 is 43. So, u≡23*43 mod50. 23*43=989. 989 divided by 50: 50*19=950, 989-950=39. So, u≡39 mod50. Therefore, solutions mod100 are 39 and 89. 

Therefore, the two smallest positive solutions are 39 and 89. Average is (39 +89)/2=128/2=64. 

Therefore, the answer is 64.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Let me check once more with another method. Let's solve 14u ≡46 mod100.

<prompt_6.1> Confirm the solution using another method for the congruence 14u ≡ 46 mod 100. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
We can write this as 14u -100v =46 for some integer v. But since we are dealing with congruence, v is an integer. Let's write it as 14u + (-100)v =46. But solving this Diophantine equation. Let's find particular solutions. 

First, find gcd(14,100)=2. Since 2 divides 46, solutions exist. Divide equation by 2: 7u -50v=23. Now, solve 7u -50v=23. 

We can use the extended Euclidean algorithm to find integers u and v such that 7u -50v=23. 

First, find the Bezout coefficients for 7 and 50. 

50 =7*7 +1

7=1*7 +0

So, gcd=1. But we need to express 1 as a combination of 7 and 50:

1=50 -7*7

Therefore, multiplying both sides by 23: 23=23*(50 -7*7)=23*50 -23*7*7

So, one particular solution is u=-161, v=-23. 

Therefore, the general solution is given by:

u= -161 + (50/2)t= -161 +25t

v= -23 + (7/2)t= but since t must be even? Wait, no. Wait, the general solution for equations of the form ax + by = c is x = x0 + (b/d)t, y = y0 - (a/d)t, where d = gcd(a,b). 

Wait, in our case, the equation is 7u -50v=23. So, a=7, b=-50, c=23. The general solution is:

u = u0 + (50/d)t = u0 +50t

v = v0 + (7/d)t = v0 +7t

Wait, but in the equation 7u -50v=23, the coefficients are 7 and -50. So, the standard form is 7u + (-50)v=23. Therefore, the general solution is:

u = u0 + (50/d)t

v = v0 + (7/d)t

Where d=gcd(7,50)=1. Therefore, u= u0 +50t

v= v0 +7t

From our particular solution, u0=-161, v0=-23. Therefore, the general solution is:

u= -161 +50t

v= -23 +7t

But we are interested in positive integer solutions u. So, find t such that u>0. 

u= -161 +50t >0

50t >161

t>161/50≈3.22. So, t≥4.

Therefore, the smallest positive solution is when t=4:

u= -161 +50*4= -161 +200=39

Then, next solution is t=5: u= -161 +50*5= -161 +250=89

t=6: u= -161 +300=139, etc. 

Therefore, the two smallest positive solutions are 39 and89. Hence, their average is (39+89)/2=64. 

Therefore, the answer is 64. 

I think this confirms it. So, the average is 64. I must have made a mistake earlier when I thought the inverse was 15, leading to the 29 solution, but actually, the inverse is 43, leading to 39. Therefore, the correct average is 64.
</execution_6.1>
</Execution_6>

<Final_answer>
Therefore, the average of the two smallest positive integer solutions to the congruence $14u \equiv 46 \pmod{100}$ is $\boxed{64}$.
</Final_answer>
</think>