# Today Puzzle #657

## Puzzle No. 657â€“ Monday 20 January

"Two squared" is four, "two cubed" is eight and so on. Then when we get to "two to the 29" we find that it is a nine digit number with all of the digits zero to nine appearing once only - except for one digit which doesn't appear at all. What is the missing digit?

Today’s #PuzzleForToday has been set by Hugh Hunt, Reader in Engineering Dynamics and Vibration at Trinity College, Cambridge

4

2^29 is only divisible by 2, 4, 8 etc. In particular it isn't divisible by 3 or 9. For any number we know that the sum of its digits is divisible by 3 or 9 if the number itself is divisible by 3 or 9. The digits 0 to 9 sum to 45 so we know that 0, 3, 6 or 9 can't be missing because the sum of digits would still be divisible by 3.

But there's another useful property of nines, which is that the remainder of a number when divided by nine is also equal to the remainder of the sum of digits when divided by nine. And there is a repeating pattern for powers of 2 when divided by nine:

2 2<-- repeating pattern

4 4

8 8

16 7

32 5

64 1

128 2 <-- repeating pattern

256 4

...

So you can deduce from this sequence that the remainder of 2^29 when divided by nine will be 5. The sum of digits is therefore 41 and so 4 is the missing digit.