Question

Jerry likes to play with numbers. One day, he wrote all the integers from $1$ to $2024$ 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 $1, 2, 3,$ and $5,$ and then write either $11,$ their sum, or $30,$ 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?

$1010$

$1011$

$1012$

$1013$

$1014$

Answer: A

Concepts:parity invariant optimization

Difficulty rating: 1800

Solution:

Among $1, \ldots, 2024$ there are $1012$ even numbers and $1012$ odd numbers. Each operation replaces $4$ entries by $1,$ so the total count falls by $3.$ If a move consumes $e$ even entries, its output is either odd, reducing the even count by $e,$ or even, reducing it by $e-1.$ In either case the even count falls by at most $3.$ Therefore eliminating all $1012$ even entries takes at least $\lceil1012/3\rceil=338$ moves. This is achievable: use $337$ 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 $2024-3\cdot338=1010.$ Therefore, the answer is A.

2024 AMC 10B Problem 16

Answer: A

Difficulty rating: 1800

Solution:

Problem 16 in Other Years