- Can there exist a number, consisting of only distinct digits, all of whose "rotations" (including the number itself) are exact multiples of each of its digits?
- Can there exist a number, consisting of at least 2 distinct digits (repetition of digits allowed), all of whose "rotations" (including the number itself) are exact multiples of its distinct digits?

Source: Reader Sudipta Das.

Solutions were received from Joseph DeVincentis, Jozef Hanenberg, Jeremy Galvagni, Algirdas Rascius, Kenneth Suprin, Graeme McRae, Claudio Baiocchi, Jimmy Chng Gim Hong, John Hewson, Philippe Fondanaiche, and Ross Millikan. The first question has no solutions, and the second has many solutions (within certain restrictions).

Joseph DeVincentis' solution is representative of most solutions:

Part 1: Can there exist a (multi-digit) number, consisting of only distinct digits, all of whose "rotations" (including the number itself) are exact multiples of each of its digits? Such a number cannot contain 0, since all multiples of 0 are 0. Such a number cannot contain 5, since all multiples of 5 end in 5 or 0, and we already know the number cannot contain 0. If such a number contains any even digit, then all the digits in the number must be even. So, we are left with combinations of 2, 4, 6, 8 and combinations of 1, 3, 7, 9. If the number contains 4 or 8, then it cannot contain 2 or 6, because at some point in the number a 2 or 6 must follow a 4 or 8, and numbers that end in 42, 46, 82, or 86 are not multiples of 4. And likewise 48 and 84 do not work because 84 is not a multiple of 8. The numbers 26 and 62 do not work because they are not multiples of 6. If the number contains 3 or 9, it cannot contain 1 or 7 because multiples of 3 always have digits which add to a multiple of 3, but the combinations of 1, 3, 7, and 9 which contain a 1 or 7 have digit sums which = 1 or 2 mod 3. The numbers 39 and 93 do not work because they are not multiples of 9. Thus, no such number exists. Part 2: Can there exist a number, consisting of at least 2 distinct digits (repetition of digits allowed), all of whose "rotations" (including the number itself) are exact multiples of its distinct digits? Some of the reasoning from above still works, but there are now a few cases where we can create a number with this property: 2226, and other combinations of 2s and 6s with a multiple of three 2s. 1113, and other combinations of 1s and 3s with a multiple of three 1s. 3339, 111339, 1111111119, and other combinations of 1s, 3s, and 9s with a digit sum that is a multiple of 9, including cases with no 1s and no 3s. Some combinations of 7s with other odd digits also work. For example, keeping in mind that 1001 = 7 * 11 * 13, numbers like 117117 and 177177 work.

Here is my (Ken's) solution I drew up when Sudipta Das sent me the puzzle:

For the second part:

2226 (or any number with 3k twos and at least one six.)

A quick summary of some things as I see them:

First consider the even case:

Such a number cannot contain both 4s and [2s or 6s] (because 4 only divides into a number ending in [2 or 6] if the 10s digit is odd.) So 4 could ONLY be in a number with 8s, but 8 only divides into a number ending in 4 if the 10s digit is odd or 2 or 6.

--> So 4 cannot be in the number.

Such a number cannot contain both 8s and [2s or 6s] (because 8 only divides into a number ending in [2 or 6] if the 10s digit is odd.)

--> So 8 cannot be in the number.

So, such a number can only contain 2s and 6s. A two-digit example is not possible, but a 4-digit exampe is (as above.)

Now the odd case:

1113 (or any number with 3k ones and at least one three.)

3339 (or any number with 3k threes and at least one nine.)

1111111119 (or any number with 9k ones ant at least one nine.)

111339 (basically if using (1,3,9) and the first digits sum to 9, the
rotations will all work out.)

373737 (any 6k digit number divisible by 7 will rotate to a number
divisible by 7.)

171717

373779

717913 (uses all possible odds.)

As you've pointed out, 5 cannot be in the number.

7 can only be used in a 6k digit number, so repetition of digits is
required if a 7 is involved.

This leaves only 1, 3, and 9. We can't use them in a 3-digit number
(the sum of the digits shows we won't get a number divisible by 9),
and we can see the two-digit numbers won't work. So repetition is
required and some examples are above.

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