Solution of the Week #474 - The Search for Seven

For the GCD to be 7, a necessary condition is that the two numbers are divisible by 7. This of course means that not only are abc and cba divisible by 7, but so is abc-cba.

abc = 100a+10b+c and cba = 100c+10b+a, so their difference is 99a-99c, or 99(a-c).

Since 99 is not a multiple of 7, (a-c) must be, so a and c must differ by 7 (they can’t be identical, since the GCD of aba and aba is aba).

Since they are single digits, they could only be: (0,7), (1,8) or (2,9).

For each case we can find the middle digit or digits b that makes abc (and therefore also cba) a multiple of 7.

This leads us to 007,700; 077,770; 168,861; and 259,952.

We can rule out 007,700 and 077,770 on the basis that 007 and 077 are not three-digit numbers.

The third pair 168 and 861 are also both divisible by 3 and therefore have a GCD of 21.

The final pair of 259 and 952 are therefore the numbers we are looking for. 259 = 7x37 and 952 = 7x8x17.