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.

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