1971 AMC 12 Problem 12

Attempt Problem 12 of the 1971 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1971 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

For each integer N>1,N\gt1, define positive integers to be congruent if they leave the same nonnegative remainder when divided by N.N. If 69,69, 90,90, and 125125 are congruent in one such system, then in that same system, 8181 is congruent to:
33
44
55
77
88
Answer: B
Concepts:modular arithmeticgreatest common divisordivisibility
Small hint:
The modulus divides the difference of any two congruent integers
Big hint:
Use the differences 2121 and 3535
Solution:
The modulus NN divides both 90−69=2190-69=21 and 125−90=35.125-90=35. Since N>1,N\gt1, this forces N=7.N=7. Then 81≡4(mod7).81\equiv4\pmod7.
Therefore, the correct answer is B.
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