# Pizza Hut Pi Day Challenge (Part 1)

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?

To illustrate this idea, let me give an example that almost works: 9,632,581,470, a ten-digit number that uses each digit exactly once.

1. The first digit is 9, which is clearly a multiple of 1.
2. The first two digits form the number 96, which is clearly a multiple of 2: $96 = 2 \times 48$.
3. The first three digits form the number 963, which is a multiple of 3: $963 = 3 \times 321$.
4. The first four digits form the number 9632, which is a multiple of 4: $9632 = 4 \times 2408$.
5. The first five digits form the number 96,325, which is a multiple of 5: $96,325 = 5 \times 19,265$.
6. The first six digits form the number 963,258, which is a multiple of 6: $963,258 = 6 \times 160,523$.
7. The first seven digits form the number 9,632,581, which is a multiple of 7: $9,632,581 = 7 \times 1,376,083$.
8. Doesn’t work: $96,325,814 = 8 \times 12,040,726 + 6$.
9. The first nine digits for the number 963,258,147, which is a multiple of 9: $963,258,147 = 9 \times 107,028,683$
10. The ten-digit number is a multiple of 10: $9,632,581,470 = 10 \times 963,258,147$.

So, this number works for 9 of the 10 cases… but we need the number that works for all 10 cases.

In this series, I plan to discuss my solution of this problem using the rules of divisibility. I’ll start my solution with tomorrow’s post.

In the meantime, if you’d like to think about the solution on your own, I offer this green thought bubble to give you some time to think about it on your own.