538 Riddler: Can You Solve This Napoleonic Puzzle?

This week's Riddler is "short but deadly":

Complete this series:

10, 11, 12, 13, 14, 15, 16, 17, 21, 23, 30, 33, …

(Yep, that’s it.)

I initially assumed the title of the column was related in some way to the solution (and wasted a bit of time chasing down that dead end), but as far as I can tell the only thing "Napoleonic" about this puzzle is that it is short (but probably not as deadly).

I also don't really see how it's related to this tweet from the Riddler editor earlier this week:

So if it's not related to Napoleon or Wisconsin delegate counts or anything like that, what is it?  In theory, we could complete the sequence with any number by making up arbitrary and convoluted rules that govern it, but that's no fun and I was sure that wasn't what the author intended.

After some tinkering, I noticed that none of the numbers include the digits 8 or 9, so I thought maybe that instead of decimal representations, these were base-8 representations of numbers instead.  Converting them all to base 8 gives us:

1 * 8^1 + 0 * 8^0 = 8

1 * 8^1 + 1 * 8^0 = 9

1 * 8^1 + 2 * 8^0 = 10

1 * 8^1 + 3 * 8^0 = 11

1 * 8^1 + 4 * 8^0 = 12

1 * 8^1 + 5 * 8^0 = 13

1 * 8^1 + 6 * 8^0 = 14

1 * 8^1 + 7 * 8^0 = 15

2 * 8^1 + 1 * 8^0 = 17

2 * 8^1 + 3 * 8^0 = 19

3 * 8^1 + 0 * 8^0 = 24

3 * 8^1 + 3 * 8^0 = 27

That doesn't look any more solvable than the original, unfortunately.  But with no better leads to follow I pursued the idea that these were, in fact, representations in different bases.  Perhaps, instead of each term being a different number in the same base, they were all the same number in different bases.  The sudden jump from 17 to 21 in the original sequence still indicated to me that 17 was in base 8, which is equal to 15 in decimal.

And sure enough, upon inspection we can see that the sequence is the decimal number 15 represented in different bases, starting at 15 and decreasing by 1 in each step:

1 * 15^1 + 0 * 15^0 = 15

1 * 14^1 + 1 * 14^0 = 15

1 * 13^1 + 2 * 13^0 = 15

1 * 12^1 + 3 * 12^0 = 15

1 * 11^1 + 4 * 11^0 = 15

1 * 10^1 + 5 * 10^0 = 15

1 * 9^1 + 6 * 9^0 = 15

1 * 8^1 + 7 * 8^0 = 15

2 * 7^1 + 1 * 7^0 = 15

2 * 6^1 + 3 * 6^0 = 15

3 * 5^1 + 0 * 5^0 = 15

3 * 4^1 + 3 * 4^0 = 15

So, the remaining terms in the sequence would be 15 represented in bases 3, 2 and 1, which are 120, 1111, and 111111111111111, respectively.

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