HMMT 十一月 2016 · 冲刺赛 · 第 12 题

HMMT November 2016 — Guts Round — Problem 12

[ 8 ] A positive integer ABC , where A , B , C are digits, satisfies C ABC = B − A Find ABC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2016, November 12, 2016 — GUTS ROUND Organization Team Team ID# 3

解析

[ 8 ] A positive integer ABC , where A , B , C are digits, satisfies C ABC = B − A Find ABC . Proposed by: Henrik Boecken C C The equation is equivalent to 100 A +10 B + C = B − A . Suppose A = 0, so that we get 10 B + C = B . Reducing mod B , we find that C must be divisible by B . C 6 = 0, since otherwise 10 B = 1, contradiction, B B so C ≥ B . Thus 10 B + C ≥ B for digits B, C . For B ≥ 4, we have 100 > 10 B + C ≥ B > 100, a contradiction, so B = 1 , 2 , 3. We can easily test that these do not yield solutions, so there are no solutions when A = 0. C Thus A ≥ 1, and so 100 ≤ 100 A + 10 B + C ≤ 1000, and thus 100 ≤ B − A ≤ 1000. 1 ≤ A ≤ 10, C so we have 101 ≤ B ≤ 1010. We can test that the only pairs ( B, C ) that satisfy this condition are (2 , 7) , (2 , 8) , (2 , 9) , (3 , 5) , (3 , 6) , (4 , 4) , (5 , 3) , (6 , 3) , (7 , 3) , (8 , 3) , (9 , 3). Of these pairs, only (2 , 7) yields a solution to the original equation, namely A = 1 , B = 2 , C = 7. Thus ABC = 127. 3