# Pizza Hut Pi Day Challenge (Part 5)

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 , 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.

where E is one of 2, 4, 6, and 8 (not already included) and O is one of 1, 3, 7, and 9. (The last digit is 0 and not an odd number.) There are 192 possible answers left.

Step 7. The number formed by the first four digits must be a multiple of 4. By the divisibility rules, this means that the number formed by the third and fourth digits must be a multiple of 4.

In other words, the two digit number $OE$ has to be a multiple of 4, where $O$ is odd and $E$ is even. Let $O = 2n+1$ and $E = 2k$. Then

$OE = 10 \times O + E$

$= 10(2n+1) + 2k$

$= 20n+2k+10$

$= 2(10n+k+5)$

also has to be a multiple of 4, which means that $10n+k+5$ has to be a multiple of 2. Since $10n$ is already of a multiple of 2, that means that $k + 5$ must be a multiple of 2, or that $k$ must be odd.

In other words, $E = 2k$, where $k$ is an odd number. Therefore, the fourth digit must be either 2 or 6. This eliminates two of the above forms, so we’re down to

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

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

At this point, I can include the remaining ways of choosing the even digits:

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.

For each of these, there are 4! = 24 ways of choosing the remaining odd digits. Since there are four forms, there are 4 x x 24 = 96 possible answers left.

In tomorrow’s post, I’ll cut this number down to 24.

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