HMMT 二月 2006 · GEN1 赛 · 第 10 题

HMMT February 2006 — GEN1 Round — Problem 10

A positive integer n is called “flippant” if n does not end in 0 (when written in decimal notation) and, moreover, n and the number obtained by reversing the digits of n are both divisible by 7. How many flippant integers are there between 10 and 1000?

解析

A positive integer n is called “flippant” if n does not end in 0 (when written in decimal notation) and, moreover, n and the number obtained by reversing the digits of n are both divisible by 7. How many flippant integers are there between 10 and 1000? Answer: 17 Solution: We use the notation “ | ” to mean “divides.” There is only one flippant 2-digit number, namely 77. Indeed, if 10 a + b is flippant (where a, b are integers 1–9), then 7 | 10 a + b and 7 | 10 b + a . Thus, 7 | 3(10 a + b ) − (10 b + a ) = 29 a − 7 b = a + 7(4 a − b ) , so that 7 | a , and similarly 7 | b , so we’d better have a = b = 7. 3 There are 16 flippant 3-digit numbers. First consider the 12 palindromic ones (ones where the hundreds and units digits are the same): 161, 252, 343, 434, 525, 595, 616, 686, 707, 777, 868, and 959. Now consider the general case: suppose 100 a + 10 b + c is flippant, where a, b, c are integers 1–9. Then 7 | 100 a + 10 b + c and 7 | 100 c + 10 b + a , so 7 | (100 a + 10 b + c ) − (100 c + 10 b + a ) = 99( a − c ), and so 7 | a − c . In order for this not to result in a palindromic integer, we must have a − c = ± 7 and, moreover, both 100 a + 10 b + a and 100 c + 10 b + c must be palindromic flippant integers. Consulting our list above, we find 4 more flippant integers: 168, 259, 861, and 952. 4