On March 14, 2016, Pizza Hut held a online math competition in honor of Pi Day, offering three questions posed by Princeton mathematician John H. Conway. As luck would have it, years ago, I had actually heard of the first question before from a colleague who had heard it from Conway himself:

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first *n* of them is divisible by *n* for each *n* from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. So far in this series, I described why the solution must have one of the following four forms:

O , 4 O 2 , 5 8 O , 6 O 0,

O , 6 O 2 , 5 8 O , 4 O 0,

O , 2 O 6 , 5 4 O , 8 O 0.

O , 8 O 6 , 5 4 O , 2 O 0.

where O is one of 1, 3, 7, and 9. (The last digit is 0 and not an odd number.) There are **96 possible answers left**.

**Step 8**. The number formed by the first eight digits must be a multiple of 8. By the divisibility rules, this means that the number formed by the sixth, seventh, and eighth digits must be a multiple of 8.

For the first form, that means that must be a multiple of 8. We can directly test this:

**: a multiple of 8.**

: not a multiple of 8.

: not a multiple of 8.

**: a multiple of 8.**

For the second form, that means that must be a multiple of 8. This is impossible. Let . Then

.

We see that is odd, and therefore is not a multiple of 2. Therefore, is not a multiple of 4 (let alone 8).

For the third form, that means that must be a multiple of 8. This is also impossible. Let . Then

.

We see that is odd, and therefore is not a multiple of 2. Therefore, is not a multiple of 4 (let alone 8).

For the fourth form, that means that must be a multiple of 8. We can directly test this:

: not a multiple of 8.

**: a multiple of 8.**

**: a multiple of 8.**

: not a multiple of 8.

In other words, we’re down to

O , 4 O 2 , 5 8 1 , 6 O 0,

O , 4 O 2 , 5 8 9 , 6 O 0,

O , 8 O 6 , 5 4 3 , 2 O 0,

O , 8 O 6 , 5 4 7 , 2 O 0.

For each of these, there are 3! = 6 ways of choosing the remaining odd digits. Since there are four forms, there are 4 x 6= **24 possible answers left**.

In tomorrow’s post, I’ll cut this number down to 10.