HMMT 十一月 2016 · 冲刺赛 · 第 12 题

英文解析

12. [ 8 ] A positive integer ABC , where A , B , C are digits, satisfies
    ABC = B − AC
    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 Bso 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 nosolutions when A = 0.
    Thus A ≥ 1, and so 100 ≤ 100 A + 10 B + C ≤ 1000, and thus 100 ≤ B − A ≤ 1000. 1 ≤ A ≤ 10,Cso we have 101 ≤ B ≤ 1010. We can test that the only pairs ( B, C ) that satisfy this condition are c
    (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 asolution to the original equation, namely A = 1 , B = 2 , C = 7. Thus ABC = 127. 3