2013 AIME II Problem 6

Below is the professionally curated solution for Problem 6 of the 2013 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AIME II solutions, or check the answer key.

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Concepts:perfect squareinequalitybounding to limit cases

Difficulty rating: 2430

6.

Find the least positive integer NN such that the set of 10001000 consecutive integers beginning with 1000N1000 \cdot N contains no square of an integer.

Solution:

The block {1000N,,1000N+999}\{1000N, \ldots, 1000N + 999\} misses all squares exactly when some consecutive squares x2x^2 and (x+1)2(x+1)^2 jump over it, which requires (x+1)2x2=2x+1>1000,(x+1)^2 - x^2 = 2x + 1 \gt 1000, so x500.x \ge 500. In particular every block below 5002=250000500^2 = 250000 contains a square, so we search from there.

Write x=500+ax = 500 + a with a0.a \ge 0. Then x2=1000(250+a)+a2,x^2 = 1000(250 + a) + a^2, so as long as a2<1000a^2 \lt 1000 (that is, a31a \le 31), the square x2x^2 lies in block 250+a;250 + a; these cover blocks 250250 through 281.281. Block 251+a251 + a is skipped exactly when (x+1)2=1000(250+a)+a2+2a+10011000(252+a),(x+1)^2 = 1000(250 + a) + a^2 + 2a + 1001 \ge 1000(252 + a), that is, a2+2a999.a^2 + 2a \ge 999. For a30a \le 30 this fails (so (x+1)2(x+1)^2 lands in block 251+a251 + a), and it first holds at a=31,a = 31, since 961+62=1023.961 + 62 = 1023.

Indeed 5312=281961531^2 = 281961 and 5322=283024532^2 = 283024 straddle the block starting at 282000.282000. The least such NN is 251+31=282.251 + 31 = 282.

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