2017 AIME I Problem 2

Attempt Problem 2 of the 2017 AIME I 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 2017 AIME I solutions, or check the answer key.

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

2.

When each of 702,702, 787,787, and 855855 is divided by the positive integer m,m, the remainder is always the positive integer r.r. When each of 412,412, 722,722, and 815815 is divided by the positive integer n,n, the remainder is always the positive integer sr.s \neq r. Find m+n+r+s.m + n + r + s.

Answer: 62
Concepts:greatest common divisormodular arithmeticdivisibility
Difficulty rating: 2070
Solution:

Numbers leaving equal remainders upon division by mm differ by multiples of m,m, so mm divides both 787702=85787 - 702 = 85 and 855787=68.855 - 787 = 68. Since gcd(85,68)=17\gcd(85, 68) = 17 and mm must exceed the positive remainder r,r, we get m=17,m = 17, and r=7024117=5.r = 702 - 41 \cdot 17 = 5.

Similarly nn divides both 722412=310722 - 412 = 310 and 815722=93,815 - 722 = 93, and gcd(310,93)=31,\gcd(310, 93) = 31, so n=31n = 31 and s=4121331=9,s = 412 - 13 \cdot 31 = 9, which indeed differs from r.r.

The requested sum is 17+31+5+9=62.17 + 31 + 5 + 9 = 62.

← Problem 1#1
Full Exam

Problem 2 in Other Years