1997 AIME Problem 1

Attempt Problem 1 of the 1997 AIME below, then check your answer against the professionally curated solution 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 problems are used with official legal permission of the Mathematical Association of America (MAA).

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?

Answer: 750
Concepts:difference of squaresparity
Difficulty rating: 1890
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 Exam

Problem 1 in Other Years