1997 AIME Problem 1

Below is the professionally curated solution for Problem 1 of the 1997 AIME, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1997 AIME 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:difference of squaresparity

Difficulty rating: 1890

1.

How many of the integers between 11 and 1000,1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

Solution:

Write n=a2b2=(ab)(a+b).n = a^2 - b^2 = (a - b)(a + b). The factors aba - b and a+ba + b differ by the even number 2b,2b, so they have the same parity. If both are odd, nn is odd; if both are even, 4n.4 \mid n. Hence no integer n2(mod4)n \equiv 2 \pmod 4 is a difference of two squares.

Conversely, every odd number 2k+12k + 1 equals (k+1)2k2,(k+1)^2 - k^2, and every multiple of 4,4, say 4k,4k, equals (k+1)2(k1)2(k+1)^2 - (k-1)^2 (with k10k - 1 \ge 0 since k1k \ge 1).

Between 11 and 10001000 there are 500500 odd numbers and 250250 multiples of 4,4, for a total of 500+250=750.500 + 250 = 750.

Full ExamProblem 2

Problem 1 in Other Years