HMMT 十一月 2012 · 冲刺赛 · 第 27 题
HMMT November 2012 — Guts Round — Problem 27
题目详情
- [ 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 ROUNDab + b 2
英文原题
[ 13 ] Find the number of ordered triples of positive integers ( a, b, c ) satisfying a 2 − b 2 + ac − bc = 2012.
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HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND
解析
英文解析
- [ 13 ]
Answer: 1755 We write this as ( a − b )( a + b ) + ( a − b )( c ) = ( a − b )( a + b + c ) = 2012. Since a, b, care 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, whichdetermines 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.