Math CentralQuandaries & Queries


Question from Dr.:
Many years ago I discovered that there are 33 numbers which, when multiplied by 0123456789, simply rearrange the digits. Surprisingly (to me at least) is the fact that the digits of all of those 33 numbers add up
to one of the 6 "cardinal" numbers 1, 2, 4, 5, 7, and 8. For example 6 + 2 = 8 and 62 X 0123456789 = 7654320918 etc.

Interesting! Yes, I think this is probably known - but it is a very respectable rediscovery. The reasons for it are interesting, and have a lot to do with the theory of repeating decimals.

First, note that 1/9 = 0.11111.... This is not a coincidence: in base B, 1/(1-B) always equals 0.11111.....

Next, note that 11*11 = 121, 111*111 = 12321, etc. (the pattern breaks holds up to 111111111*111111111.) This is not a coincidence either; it comes from the fact that numbers in place value notation act like polynomials until you have to carry. With ones, there isn't a lot of carrying.

From these, you expect to see that 1/81 = 0.012345679. It's not quite so clear that it repeats immediately:

1/81 = 0.012345679012345679012345679....

but it does.

Thus, multiplying 0123456789 by N is essentially the same as finding N/81, except for the last couple digits. It's common to have patterns like this in repeating decimals - try finding 1/7, 2/7, 3/7... Why does this happen? Well, 1/81 has a loop of period 9. N/81 must also loop with a pure period of 9, or else with period 3 or 1.

1/27 loops with period 3, so if N is divisible by 3, N/81 loops with period 3. (Your "cardinal" numbers are just those not divisible by 3 - if a number is divisible by 3 so is the sum of its digits!) This will certainly bust the pattern early on.

If N is not divisible by 3, N/81 will have 9 of the digits (not always the same 9) in its circulant. This means that most of the digits of 012345679 * N will be different; and most of these these agree with the digits of 0123456789 * N. Note that (if N is not divisible by 3) the first six (seven?) digits are always different!

Finally, we note that 0+1+2+3+4+5+6+7+8+9 is divisible by 9. Thus 0123456789*N is divisible by 9, so its digits must also add to a multiple of 9. But if all but the last digit are different, the last one must be the missing one! Thus there are only a couple decimal places where the pattern can break down.

A final thought - a lot of this will work in other number bases.

Good Hunting!

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