# Solution to "The Digit Puzzles"

### Puzzle 1: What are the digits?

We have 10 digits to consider, and three of them must be rejected.

Since it is nonsensical to say that the number N is divisible by 0, we can immediately conclude that 0 cannot be a digit of N.

Now consider that we can only eliminate two more digits. There are 4 even digits remaining, so at least two will be included, and that means that N must be even.

So we ask, can 5 be a digit? If so, N must be divisible by 5, and hence its last digit must either be 5 or 0. But we're eliminated 0 already, and if the last digit is 5, the number is not even. Therefore, 5 cannot be a digit of N.

Now keep in mind that either 3 or 9 must be a digit (we can only eliminate one digit) and so the digital root of N (the sum of all the digits of N) must be divisible by 3 for sure, and if 9 is one of the digits, the digital root must actually be divisible by 9.

Since we only need to eliminate one digit, here are the cases:

```        xx234x6789 - digital root 39, not divisible by 9.
x1x34x6789 - digital root 38, not divisible by 3 (or 9).
x12x4x6789 - digital root 37, not divisible by 3 (or 9).
x123xx6789 - digital root 36,     divisible by 3 and 9.
x1234xx789 - digital root 34, not divisible by 3 (or 9).
x1234x6x89 - digital root 33, not divisible by 9.
x1234x67x9 - digital root 32, not divisible by 3 (or 9).
x1234x678x - digital root 31, not divisible by 3.
```

Thus, the only possibility for the digits is 1236789, not necessarily in that order, of course!

### Puzzle 2: What is an example of such a number?

This part is much harder! You can see several ways to start. For one thing, the number must be a multiple of the least common multiple of 1, 2, 3, 6, 7, 8 and 9 = 7 * 8 * 9 = 504. The multiplier of 504 has to be at least big enough to product a 7 digit number, so at least 1985. It can't product an 8 digit number, so the multiplier is no bigger than 19841.

Divisibility by 2 implies the last digit is 2, 6 or 8.

We also must have divisibility by 4 (even though there is no 4 in the number) and 100 is divisible by 4, so the last two digits must be:
12, 32, 72, 92,
16, 36, 76, 96,
28, or 68.

Divisibility by 8 means the last 3 digits are divisible by 8 (since 1000 is divisible by 8), so we must have one of:
312, 712, 912,
632, 832,
672, 872,
192, 392, 792,
216, 816,
136, 736, 936,
176, 376, 976,
296, 896,
128, 328, 728, 928,
168, 368, 768, or 968.

But ultimately, we ended up with a brute force calculation which got us 1,687,392 = 504 * 3348.

Last revised on 21 February 2008