Harvard-MIT Mathematics Tournament

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 $7 \mid 10a + b$ and $7 \mid 10b + a$. Thus,
$$7 \mid 3(10a + b) - (10b + a) = 29a - 7b = a + 7(4a - b)$$
so that $7 \mid a$, and similarly $7 \mid b$, 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 $7 \mid 100a + 10b + c$ and $7 \mid 100c + 10b + a$, so $7 \mid (100a + 10b + c) - (100c + 10b + a) = 99(a - c)$, and so $7 \mid a - c$. In order for this not to result in a palindromic integer, we must have $a - c = \pm 7$ 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 $952$.