## #math trivia #248 solution

In the current example, 2, 4 and 8 are all powers of 2.

To see what the others might be, let’s start with the possible “bases” that everything else must be a power of.

If the base is 0, then the other digits must be 0 or 1:

100, 101, 110

(Recall that three-digit numbers can’t start with 0, and that 1 is a power of 0 and every other base.)

If the base is 1, all the digits must be 1:

111

If it’s 2, the other digits must be 1, 2, 4 or 8. Given that each digit can have four possible values, there are 64 three-digit numbers whose digits are all powers of 2.

If it’s 3, the other digits must be 1, 3 or 9 — three possible values for each digit — for a total of 27 numbers.

If the base is 4, 5, 6, 7, 8, or 9, the other digits must be either the base or 1 — two possible values for each digit — for a total of 8 numbers for each base.

Note, however, that there is some double-counting here. The lists for bases 2 through 9 all include 111, which was already counted for base 1. The list for base 4 includes 114, 141, 144, 411, 414, 441 and 444 which were also counted for base 2 (as well as 111, which is also on the base 2 list, but has already been noted).  Similarly, the list for 8 includes seven numbers also counted for base 2, and the list for 9 includes seven numbers also counted for base 3.

So the total across the lists enumerated here is:

1 + 3 + 64 + 27 + 6*8 – 7 – 7 – 7 – 7 = 115 numbers.