HMMT 十一月 2012 · 冲刺赛 · 第 27 题

HMMT November 2012 — Guts Round — Problem 27

[ 13 ] Find the number of ordered triples of positive integers ( a, b, c ) satisfying a − b + ac − bc = 2012. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND 2 ab + b

解析

[ 13 ] Answer: 1755 We write this as ( a − b )( a + b ) + ( a − b )( c ) = ( a − b )( a + b + c ) = 2012. Since a, b, c are positive integers, a − b < a + b + c . So, we have three possibilities: a − b = 1 and a + b + c = 2012, a − b = 2 and a + b + c = 1006, and a − b = 4 and a + b + c = 503. The first solution gives a = b + 1 and c = 2011 − 2 b , so b can range from 1 through 1005, which determines a and c completely. Similarly, the second solution gives a = b + 2 and c = 1004 − 2 b , so b can range from 1 through 501. Finally, the third solution gives a = b + 4 and c = 499 − 2 b , so b can range from 1 through 249. Hence, the total number of solutions is 1005 + 501 + 249 = 1755.