2005 AIME I Problem 4

Below is the professionally curated solution for Problem 4 of the 2005 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AIME I solutions, or check the answer key.

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Concepts:Diophantine Equationcompleting the squaredifference of squares

Difficulty rating: 2230

4.

The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 55 members left over. The director finds that if they are arranged in a rectangular formation with 77 more rows than columns, the desired result can be obtained. Find the maximum number of members this band can have.

Solution:

Let the band have nn members, with n=s2+5n = s^2 + 5 for the square formation and n=x(x+7)n = x(x + 7) for the rectangular formation with xx columns. Multiplying x2+7x=s2+5x^2 + 7x = s^2 + 5 by 44 and completing the square gives (2x+7)2(2s)2=69,(2x + 7)^2 - (2s)^2 = 69, so (2x+72s)(2x+7+2s)=69.(2x + 7 - 2s)(2x + 7 + 2s) = 69.

Writing 69=169=32369 = 1 \cdot 69 = 3 \cdot 23 with the larger factor second: from 1691 \cdot 69 we get 2x+7=352x + 7 = 35 and 2s=34,2s = 34, so x=14,x = 14, s=17,s = 17, and n=172+5=294.n = 17^2 + 5 = 294. From 3233 \cdot 23 we get 2x+7=132x + 7 = 13 and 2s=10,2s = 10, so x=3,x = 3, s=5,s = 5, and n=30.n = 30.

The maximum is 294,294, achieved by a 21×1421 \times 14 rectangle.

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