2003 AIME II Problem 2

Below is the professionally curated solution for Problem 2 of the 2003 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AIME II solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:divisibilitydigits

Difficulty rating: 1970

2.

Let NN be the greatest integer multiple of 8,8, no two of whose digits are the same. What is the remainder when NN is divided by 1000?1000?

Solution:

An integer is divisible by 88 exactly when the number formed by its last three digits is. To make NN as large as possible, use all ten digits once each and put the largest digits first: the leading digits are 9876543,9876543, and the final three digits are some arrangement of 0,0, 1,1, 22 — provided one of those arrangements is a multiple of 8.8.

Checking 012,012, 021,021, 102,102, 120,120, 201,201, 210,210, the only multiple of 88 is 120.120. So N=9,876,543,120,N = 9{,}876{,}543{,}120, and the remainder upon division by 10001000 is 120.120.

← Problem 1Full ExamProblem 3

Problem 2 in Other Years