Alright, let's tackle this problem. The question is asking for the average of the two smallest positive integer solutions to the congruence equation 14u ≡ 46 mod 100. Hmm, okay. So, first, I need to solve the congruence equation for u, find the two smallest positive solutions, and then average them. Let me break this down step by step.

Starting with the congruence 14u ≡ 46 mod 100. To solve for u, I remember that solving a linear congruence like ax ≡ b mod m involves finding the modular inverse of a modulo m, provided that a and m are coprime. But first, I need to check if 14 and 100 are coprime. The greatest common divisor (GCD) of 14 and 100 is... let's see, 14 factors into 2 and 7, and 100 factors into 2² and 5². So the GCD is 2. Since the GCD is not 1, that means 14 and 100 are not coprime, so the inverse of 14 mod 100 doesn't exist. Wait, but does that mean the congruence has no solutions? No, actually, even if a and m are not coprime, the congruence can still have solutions if the GCD divides the constant term b. In this case, the GCD is 2, and 2 divides 46 (since 46 ÷ 2 = 23). So, the congruence does have solutions. The number of solutions modulo 100 will be equal to the GCD, which is 2. So, there are two distinct solutions modulo 100.

To solve 14u ≡ 46 mod 100, I can simplify the equation by dividing all terms by the GCD, which is 2. That gives 7u ≡ 23 mod 50. Now, since 7 and 50 are coprime (their GCD is 1), the inverse of 7 mod 50 exists. Let's find that inverse. The inverse of 7 mod 50 is a number x such that 7x ≡ 1 mod 50. Let me try to find x by testing multiples of 7:

7*1 = 7 mod 50 = 7
7*2 = 14 mod 50 =14
7*3=21
7*4=28
7*5=35
7*6=42
7*7=49
7*8=56 mod50=6
7*9=63 mod50=13
7*10=70 mod50=20
7*11=77 mod50=27
7*12=84 mod50=34
7*13=91 mod50=41
7*14=98 mod50=48
7*15=105 mod50=5
7*16=112 mod50=12
7*17=119 mod50=19
7*18=126 mod50=26
7*19=133 mod50=33
7*20=140 mod50=40
7*21=147 mod50=47
7*22=154 mod50=4
7*23=161 mod50=11
7*24=168 mod50=18
7*25=175 mod50=25
7*26=182 mod50=32
7*27=189 mod50=39
7*28=196 mod50=46
7*29=203 mod50=3
7*30=210 mod50=10
7*31=217 mod50=17
7*32=224 mod50=24
7*33=231 mod50=31
7*34=238 mod50=38
7*35=245 mod50=45
7*36=252 mod50=2
7*37=259 mod50=9
7*38=266 mod50=16
7*39=273 mod50=23
7*40=280 mod50=30
7*41=287 mod50=37
7*42=294 mod50=44
7*43=301 mod50=1
Ah, there we go! 7*43 = 301, which is 1 mod50. So, the inverse of 7 mod50 is 43. Therefore, multiplying both sides of the simplified congruence 7u ≡23 mod50 by 43 gives:

u ≡23*43 mod50

Calculating 23*43: 20*43=860, 3*43=129, so total is 860+129=989. Then 989 mod50: 50*19=950, 989-950=39. So, u ≡39 mod50. That means the solutions to the original congruence 14u ≡46 mod100 are u ≡39 mod50. Wait, but since we divided by 2 earlier, the modulus becomes 50. So, the solutions are u ≡39 mod50. But since the original modulus was 100, there are two solutions separated by 50. So, the solutions are u ≡39 mod50 and u ≡39 +50=89 mod50, but since modulus is 100, adding 50 to 39 gives 89, which is 39 +50=89. So, the two smallest positive solutions are 39 and 89. Therefore, the average of 39 and 89 is (39 +89)/2 =128/2=64. So, the answer should be 64. Wait, let me verify that.

Wait, hold on. Let me confirm the steps again to make sure I didn't make a mistake. Starting with 14u ≡46 mod100. GCD(14,100)=2. Since 2 divides 46, there are two solutions modulo100. Divide both sides by 2: 7u ≡23 mod50. Then find inverse of7 mod50, which is43. Then u ≡23*43 mod50. 23*43=989. 989 mod50. 50*19=950, 989-950=39. So, u≡39 mod50. Therefore, solutions modulo100 are u≡39 and u≡39+50=89. Then average is (39+89)/2=64. Seems correct.

But let me check if 14*39 ≡46 mod100. 14*39=546. 546 mod100 is 46. Correct. 14*89=1246. 1246 mod100 is 46. Correct. So, both solutions work. Therefore, the two smallest positive integer solutions are 39 and 89, whose average is 64. Therefore, the answer is 64. 

But to make sure I didn't skip any steps or make any calculation errors, let me go through the inverse calculation again. The inverse of7 mod50. We found that 7*43=301, which is 1 mod50. Let's verify 7*43=301. 7*40=280, 7*3=21, so 280+21=301. 301 divided by50 is6 with remainder1. Yes, 50*6=300, 301-300=1. So, correct. Therefore, inverse is indeed43. So, 23*43=989. 989 divided by50 is19*50=950, remainder39. Correct. So, u=39 mod50. So, solutions are39 and89 mod100. So, average is64. Seems solid.

Alternatively, maybe there's another way to approach this problem? For example, solving the congruence equation step by step without reducing the modulus. Let me try that.

Original equation:14u ≡46 mod100. So, 14u -46 is divisible by100. So, 14u -46 =100k for some integerk. Then, 14u =100k +46. Divide both sides by2:7u=50k +23. Then, 7u=50k +23. So, 7u ≡23 mod50. Which is the same as before. So, same result. Therefore, regardless of the approach, we end up with the same congruence. So, the solutions are u≡39 mod50, leading to solutions39 and89 mod100. Therefore, average is64. So, confident that the answer is64.

**Final Answer**
The average of the two smallest positive integer solutions is \boxed{64}.
Therefore, the average of the two smallest positive integer solutions to the congruence $$14u \equiv 46 \pmod{100}~$$ is $\boxed{64}$.
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