2002 AIME II Problem 1

Below is the professionally curated solution for Problem 1 of the 2002 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AIME II solutions, or check the answer key.

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Concepts:place valuedigits

Difficulty rating: 1790

1.

Given that

(1) xx and yy are both integers between 100100 and 999,999, inclusive;

(2) yy is the number formed by reversing the digits of x;x; and

(3) z=xy.z = |x - y|.

How many distinct values of zz are possible?

Solution:

Write x=100h+10t+ux = 100h + 10t + u with digits h,h, t,t, u.u. Then y=100u+10t+h,y = 100u + 10t + h, so z=xy=99hu.z = |x - y| = 99\,|h - u|.

Since both xx and yy are three-digit numbers, both hh and uu run from 11 to 9,9, so hu|h - u| can be any of 0,1,,8.0, 1, \ldots, 8. Each choice gives a different multiple of 99,99, so there are 99 distinct values of z.z.

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