MonkaS's blog

How do you solve the New Year of Puzzles problem???
By MonkaS, history, 3 hours ago, In English

Here is a link to the problem I'm talking about. Does anybody know how to solve this? https://www.newyearofpuzzles.com/teaser

Decided to post on CF because there are lots of smart people here. This is my idea right now:

Solution idea

The hint says you don't need a calculator but the only solution I can come up with would need a calculator. Unless there is some way to do these combinatorics in your head.

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3 hours ago, # |
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have you tried FFT?

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2 hours ago, # |
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We have initial power of 4, and intended final power of 12,848. So, the net increase in power needs to be 12848 - 4 = 12844.

Since power 4 is 1st shot and 12,848 is the 100,000th shot, we have 99,999 moves to increase the power by a net difference of 12844. Since we can change power by only exactly 2 in each move, out of the 99,999 moves, we will have 12844/2 = 6422 extra +2 power moves than -2 moves. You have solved till this point.

That means 99,999 - 6422 = some_odd_number. Now, this some_odd_number must be evenly divided into +2 and -2 (so that they cancel out), but it is impossible, since it is an odd number. So the answer is 0.

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