2013 AIME I Problem 2

Below is the professionally curated solution for Problem 2 of the 2013 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AIME I solutions, or check the answer key.

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Concepts:digitsdivisibilitymultiplication principle

Difficulty rating: 2020

2.

Find the number of five-digit positive integers, n,n, that satisfy the following conditions:

• the number nn is divisible by 5,5,

• the first and last digits of nn are equal, and

• the sum of the digits of nn is divisible by 5.5.

Solution:

Since nn is divisible by 5,5, its last digit is 00 or 5;5; since the first digit equals the last digit and cannot be 0,0, both are 5.5. The outer digits contribute 1010 to the digit sum, so the three middle digits must also sum to a multiple of 5.5.

Choose the second and third digits freely, in 1010=10010 \cdot 10 = 100 ways. Whatever their sum is, the fourth digit must land in a prescribed residue class modulo 5,5, and exactly 22 of the digits 00 through 99 lie in each class. The count is 10102=200.10 \cdot 10 \cdot 2 = 200.

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