Pizza Hut Pi Day Challenge (Part 7)

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 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.

where O represents the remaining three odd digts. (The last digit is 0 and not an odd number.) There are 24 possible answers left.

Step 9. The number formed by the first three digits must be a multiple of 3. By the divisibility rules, this means that the sum of the first three digits must be a multiple of 3.

For the first form, that means that O + 4 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 3, 7, and 9. We can directly test this to see that it’s impossible:

3 + 4 + 7 = 14, not a multiple of 3.

3 + 4 + 9 = 16, not a multiple of 3.

7 + 4 + 9 = 20, not a multiple of 3.

For the second form, that means that O + 4 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 1, 3, and 7. We can directly test this:

1 + 4 + 3 = 8, not a multiple of 3.

1 + 4 + 7 = 12, a multiple of 3.

3 + 4 + 7 = 14, not a multiple of 3.

Therefore, for the second form, the first three digits could be either 147 or 714.

For the third form, that means that O + 8 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 1, 7, and 9. We can directly test this to see that it’s impossible:

1 + 8 + 7 = 16, not a multiple of 3.

1 + 8 + 9 = 18, a multiple of 3.

7 + 8 + 9 = 24, a multiple of 3.

Therefore, for the third form, the first three digits could be either 189, 981, 789, or 987.

For the fourth form, that means that O + 8 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 1, 3, and 9. We can directly test this to see that it’s impossible:

1 + 8 + 3 = 12, a multiple of 3.

1 + 8 + 9 = 18, a multiple of 3.

3 + 8 + 9 = 20, not a multiple of 3.

Therefore, for the fourth form, the first three digits could be 183, 381, 189, or 981.

1 , 4 7 2 , 5 8 9 , 6 O 0,

7 , 4 1 2 , 5 8 9 , 6 O 0,

1 , 8 9 6 , 5 4 3 , 2 O 0,

9 , 8 1 6 , 5 4 3 , 2 O 0,

7 , 8 9 6 , 5 4 3 , 2 O 0,

9 , 8 7 6 , 5 4 3 , 2 O 0,

1 , 8 3 6 , 5 4 7 , 2 O 0,

3 , 8 1 6 , 5 4 7 , 2 O 0,

1 , 8 9 6 , 5 4 7 , 2 O 0,

9 , 8 1 6 , 5 4 7 , 2 O 0.

Indeed, for each of these, there is only one odd digit left, which means we automatically know what it has to be for each of these 10 answers by process of elimination:

1 , 4 7 2 , 5 8 9 , 6 3 0,

7 , 4 1 2 , 5 8 9 , 6 3 0,

1 , 8 9 6 , 5 4 3 , 2 7 0,

9 , 8 1 6 , 5 4 3 , 2 7 0,

7 , 8 9 6 , 5 4 3 , 2 1 0,

9 , 8 7 6 , 5 4 3 , 2 1 0,

1 , 8 3 6 , 5 4 7 , 2 9 0,

3 , 8 1 6 , 5 4 7 , 2 9 0,

1 , 8 9 6 , 5 4 7 , 2 3 0,

9 , 8 1 6 , 5 4 7 , 2 3 0.

So we’re down to 10 possible answers left.

In tomorrow’s post, I’ll finally find the answer.