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Today Puzzle #657

Puzzle No. 657– 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

Click here for the answer

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.

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