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 with 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 board were odd. What is the maximum possible number of integers on the board at that time?

$1010$

$1011$

$1012$

$1013$

$1014$

Solution:

Among $1, \ldots, 2024$ there are $1012$ even numbers and $1012$ odd. Each operation eats $4$ numbers and writes back $1,$ so the count falls by $3;$ to keep it high we want as few operations as possible. Every even number has to go, and a move that produces an odd result can clear at most $3$ evens at once (one odd plus three evens sums to odd). Clearing all $1012$ evens therefore takes at least $\lceil 1012/3 \rceil = 338$ moves. And that's achievable: $337$ moves of "one odd $+$ three evens $\to$ odd sum" wipe out $1011$ evens, then one move of "three odds $+$ one even $\to$ odd sum" gets the last. That leaves $2024 - 3 \cdot 338 = 1010$ numbers. Therefore, the answer is A.

2024 AMC 10B Problem 16

Solution:

Problem 16 in Other Years