Tuesday puzzle

Stumbled across it...

Bill Earl

Can't claim any analytical process. Too tired to write a program. Fiddled around with the HP-16C for a few and stumbled across a solution.

Subtract the third digit from the second to get the first.

Subtract the third and fourth from the second to get the last.

Alex Y

Correct.

I narrowed the choices a little by looking at the "period" over which each digit repeated in its powers. So the period of 5 is 1, since every power of 5 ends in 5, while the period of 7 is 4, so I only needed to test 7^1, 7^5, and 7^9.

Of course, doing it by hand, that ended up not helping much -- just cutting in half the number of powers of 4 I had to test before moving on to find the solution. (2 and 3 were rejected trivially based on the ninth power of each.)