Stuck on 3-digit number puzzle where reverse is 297 more

I’m revising fundamentals and trying to strengthen my number-puzzle reasoning. Here’s a riddle I’m stuck on:

Find the three-digit number with digits A, B, C (in that order) such that:
– The sum of the digits is 12.
– Reversing the digits makes a number that is exactly 297 larger than the original.
– The middle digit B is prime.
– All digits are different.

My attempt so far:
– Let N = 100A + 10B + C. Reversing gives 100C + 10B + A, and the difference is 297, so 99(C − A) = 297, hence C − A = 3, i.e., C = A + 3.
– Using the sum condition A + B + C = 12, we get A + B + (A + 3) = 12 ⇒ 2A + B = 9 ⇒ B = 9 − 2A.
– Since B must be a prime digit, B ∈ {2, 3, 5, 7}, and A is a nonzero digit with C = A + 3 ≤ 9. I tried a couple of A values: one seems to give a valid-looking triple and another runs into a repeated digit, but I’m not confident I’m checking the constraints in the cleanest way.

Question: Is my setup correct, and is there a neat way to finish this logically (without just brute-forcing A) to pin down the unique solution? Also, any tips for spotting these reverse-difference patterns faster while I’m revising?

Any help appreciated!

3 Responses

  1. Your setup is exactly right: from 99(C−A)=297 we get C=A+3, and with A+B+C=12 we have B=9−2A; requiring B ∈ {2,3,5,7} and all digits distinct leaves only A=1, B=7, C=4, so the number is 174. When I first learned these, the “reversal ⇒ multiple of 99” trick made them click-I’d go straight to C−A = (difference)/99 and then test the few prime options for B; a nice walkthrough is here: https://www.purplemath.com/modules/digits.htm

  2. Nice work-your setup is perfect: from 99(C−A)=297 we get C=A+3, and with A+B+C=12 we get B=9−2A; testing prime-digit B with distinct digits leaves only A=1 (A=2 gives B=C=5, A=3 gives A=B, A=4 makes B=1 not prime, larger A makes B negative). Worked example: A=1 ⇒ B=7, C=4, so the number is 174, and indeed 471−174=297.

  3. Your setup is right: from 99(C−A)=297 we get C=A+3 and 2A+B=9, and checking A=1..6 with B prime leaves only A=1 → B=7 → C=4 with distinct digits, so the number is 174 (since 471−174=297). I’m pretty sure it’s unique because A=2,3 give repeated digits and A≥4 makes B non-prime or negative.

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