Talk:2022 AIME I Problems/Problem 2

seems the most beautiful answers using modulus and inequality are flawed, and need minor revisions to fit the rigor of math

(modulus) 8c mod 71 can be 1 if c=9 instead of 8c

(inequality) 99(a-b) can be negative, and 4(2c-7b) can have a few negative values to get them equal.

Which solutions are you referring to?

All possible remainders modulo $71$ are $\{0,1,2,\ldots,70\}.$

Although $72$ and $143$ are not in the set, it is still true that $72\equiv143\pmod{71}.$ Recall that the numbers $a$ and $b$ are congruent modulo $71$ if and only if $a-b\equiv0\pmod{71}.$

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