# Pizza Hut Pi Day Challenge (Part 6)

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 $8O6$ must be a multiple of 8. We can directly test this:

$816/8 = 102$: a multiple of 8.

$836/8 = 104.5$: not a multiple of 8.

$876/8 = 109.5$: not a multiple of 8.

$896/8 = 112$: a multiple of 8.

For the second form, that means that $8O4$ must be a multiple of 8. This is impossible. Let $O = 2n+1$. Then

$8O4 = 800 + 10 \times O + 4$

$= 800 + 10(2n+1) + 4$

$= 800 + 20n + 14$

$= 2(400 + 10n + 7)$.

We see that $7$ is odd, and therefore $400 + 10n + 7$ is not a multiple of 2. Therefore, $2(400 + 10n + 7)$ is not a multiple of 4 (let alone 8).

For the third form, that means that $4O8$ must be a multiple of 8. This is also impossible. Let $O = 2n+1$. Then

$4O8 = 400 + 10 \times O + 8$

$= 400 + 10(2n+1) + 8$

$= 400 + 20n + 18$

$= 2(200 + 10n + 9)$.

We see that $9$ is odd, and therefore $200 + 10n + 9$ is not a multiple of 2. Therefore, $2(200 + 10n + 9)$ is not a multiple of 4 (let alone 8).

For the fourth form, that means that $4O2$ must be a multiple of 8. We can directly test this:

$412/8 = 51.5$: not a multiple of 8.

$432/8 = 54$: a multiple of 8.

$472/8 = 59$: a multiple of 8.

$492/8 = 61.5$: 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.