# Pizza Hut Pi Day Challenge (Part 4)

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 the form

O , E O E , 5 E O , E O 0,

where E is one of 2, 4, 6, and 8 and O is one of 1, 3, 7, and 9. (The last digit is 0 and not an odd number.) There are 576 possible answers left.

Step 6. The number formed by the first three digits must be a multiple of 3. Also, the number formed by the first six digits must be a multiple of 6, which means it’s also a multiple of 3. By the divisibility rules, the sum of the first three digits must be a multiple of 3, and the sum of the first six digits must be a multiple of 3. Therefore, the sum of the fourth, fifth, and sixth digits must be 3.

Stated another way, the fourth and sixth digits have to be different even numbers so that 5 plus the sum of these two even numbers is a multiple of 3. Some quick testing reveals that these two even numbers have to be either 4 and 6 or else 2 and 8 in some order:

5 + 2 + 4 = 11, not a multiple of 3.

5 + 2 + 6 = 13, not a multiple of 3.

5 + 2 + 8 = 15, a multiple of 3.

5 + 4 + 6 = 15, a multiple of 3.

5 + 4 + 8 = 17, a multiple of 3.

5 + 6 + 8 = 19, not a multiple of 3.

Therefore, the solution must have one of the following forms:

O , E O 2 , 5 8 O , E O 0,

O , E O 8 , 5 2 O , E O 0,

O , E O 4, 5 6 O , E O 0,

O , E O 6 , 5 4 O , E O 0.

For each of these there are 2! = 2 ways of choosing the remaining even digits and 4! = 24 ways of choosing the remaining odd digits. Since there are four forms, there are 4 x 2 x 24 = 192 possible answers left.

In tomorrow’s post, I’ll cut this number in half.