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HMMT 十一月 2015 · GEN 赛 · 第 1 题

HMMT November 2015 — GEN Round — Problem 1

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Find the number of triples ( a, b, c ) of positive integers such that a + ab + abc = 11 .
解析
  1. Find the number of triples ( a, b, c ) of positive integers such that a + ab + abc = 11 . Proposed by: Yang Liu Answer: 3 We can write a + ab + abc = a (1 + b + bc ). Since 11 is prime, a = 11 or a = 1. But since b, c are both positive integers, we cannot have a = 11, and so a = 1. Then 1 + b + bc = 11 = ⇒ b + bc = 10 = ⇒ b ( c + 1) = 10, and since c is a positive integer, only b = 1 , 2 , 5 are possible. This gives the 3 triples ( a, b, c ) = (1 , 1 , 9) , (1 , 2 , 4) , (1 , 5 , 1).