2025 AMC 10A Problem 17

Attempt Problem 17 of the 2025 AMC 10A 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 2025 AMC 10A solutions, or check the answer key.

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

17.

Let NN be the unique positive integer such that dividing 273436273436 by NN leaves a remainder of 16,16, and dividing 272760272760 by NN leaves a remainder of 15.15. What is the tens digit of N?N?
00
11
22
33
44
Answer: E
Concepts:greatest common divisordivisibility
Small hint:
Both 273436−16273436 - 16 and 272760−15272760 - 15 are multiples of NN
Big hint:
So is their difference; reduce it by subtracting further multiples, and remember N>16N \gt 16 since the remainder is 1616
Solution:
Subtract off the remainders. Both 273420=273436−16273420 = 273436 - 16 and 272745=272760−15272745 = 272760 - 15 are multiples of N,N, so their difference 675675 is too. Now 272745=404⋅675+45,272745 = 404 \cdot 675 + 45, which makes 4545 a multiple of NN as well. The remainder 1616 means N>16,N \gt 16, and the only divisor of 4545 bigger than 1616 is 4545 itself. So N=45,N = 45, and its tens digit is 4.4. Thus, E is the correct answer.
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