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2024 AMC 10B Problem 16

Problem 16 of 25IntermediateAlgebraNumber Theory

Jerry likes to play with numbers. One day, he wrote all the integers from 11 to 20242024 on the whiteboard. Then he repeatedly chose four numbers on the whiteboard, erased them, and replaced them by either their sum or their product. (For example, Jerry’s first step might have been to erase 1,1, 2,2, 3,3, and 5,5, and then write either 11,11, their sum, or 30,30, their product, on the whiteboard.) After repeatedly performing this operation, Jerry noticed that all the remaining numbers on the whiteboard were odd. What is the maximum possible number of integers on the whiteboard at that time?

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Solution

Among 1,,20241, \ldots, 2024 there are 10121012 even numbers and 10121012 odd numbers. Each operation replaces 44 entries by 1,1, so the total count falls by 3.3. If a move consumes ee even entries, its output is either odd, reducing the even count by e,e, or even, reducing it by e1.e-1. In either case the even count falls by at most 3.3. Therefore eliminating all 10121012 even entries takes at least 10123=338\lceil\frac{1012}{3}\rceil=338 moves. This is achievable: use 337337 sums containing one odd and three evens, then one sum containing three odds and the final even. Every output is odd. Thus the maximum remaining count is 20243338=1010.2024-3\cdot338=1010. Therefore, the answer is A.

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Concepts: parity · invariant · optimization

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.