Solution to The flippant number

Answer: 17
Solution: We use the notation “|” to mean “divides.”
There is only one flippant 2-digit number, namely 77. Indeed, if 10a + b is flippant
(where a, b are integers 1–9), then 7 | 10a + b and 7 | 10b + a. Thus,
7 | 3(10a + b) − (10b + a) = 29a − 7b = a + 7(4a − b),
so that 7 | a, and similarly 7 | b, so we’d better have a = b = 7

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 100a + 10b + c is
flippant, where a, b, c are integers 1–9. Then 7 | 100a + 10b + c and 7 | 100c + 10b + a,
so 7 | (100a + 10b + c) − (100c + 10b + 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
100a + 10b + a and 100c + 10b + c must be palindromic flippant integers. Consulting
our list above, we find 4 more flippant integers: 168, 259, 861, and 952.