HMMT 二月 2004 · GEN1 赛 · 第 6 题
HMMT February 2004 — GEN1 Round — Problem 6
题目详情
- Find the ordered quadruple of digits ( A, B, C, D ) , with A > B > C > D, such that ABCD − DCBA = BDAC.
解析
- Find the ordered quadruple of digits ( A, B, C, D ) , with A > B > C > D, such that ABCD − DCBA = BDAC. Solution: (7 , 6 , 4 , 1) Since D < A , when A is subtracted from D in the ones’ column, there will be a borrow from C in the tens’ column. Thus, D + 10 − A = C . Next, consider the subtraction in the tens’ column, ( C − 1) − B . Since C < B , there will be a borrow from the hundreds’ column, so ( C − 1 + 10) − B = A . In the hundreds’ column, B − 1 ≥ C , so we do not need to borrow from A in the thousands’ column. Thus, ( B − 1) − C = D and A − D = B . We thus have a system of four equations in four variables A, B, C, D , and solving by standard methods (e.g. substitution) produces ( A, B, C, D ) = (7 , 6 , 4 , 1).