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2024 AMC 10B Problem 16

2024 AMC 10B Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Games & Processes · difficulty 3 of 5 · about 5 min · position P16-20

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?

A)
$1010$
B)
$1011$
C)
$1012$
D)
$1013$
E)
$1014$

Answer

Key insight

Each move removes three numbers, so the count stays 2 mod 3, and odd numbers never exceed the initial 1012; both limits leave 1010.

Solution

Two observations bound the answer.

Count mod 3. Every operation erases four numbers and writes one, a net loss of three. Starting from $2024 \equiv 2 \pmod 3$ , the count is always $2 \pmod 3$ . Among the choices only $1010$ and $1013$ qualify.

Odd numbers never increase. An operation produces at most one odd number, and to produce an odd result it must consume at least one odd input (a sum is odd only if an odd number of its terms are odd; a product is odd only if all four are). So the number of odd numbers on the board never goes up. It starts at $1012$ , so the final count is at most $1012$ , ruling out $1013$ .

Construction for $1010$ . Repeatedly replace three evens and one odd by their sum, which is odd; this keeps the odd count at $1012$ and removes three evens each time. After $337$ such moves, $1011$ evens are gone and one even remains. Absorb it with the sum of that even and three odds, which is odd. The board now holds $1012 - 2 = 1010$ numbers, all odd.

The answer is $\boxed{\textbf{(A)}\ 1010}$ .

Why this works

Process problems that ask for a maximum are solved by finding invariants or monovariants that cap the answer, then exhibiting a sequence that hits the cap. Here two independent facts (the count mod $3$ , and the non-increasing number of odds) squeeze the answer to a single value. Always write out the parity table for sum and product before reasoning.

The trap

Assuming the 1012 odd numbers can all survive; one operation is needed to absorb the last even number, and it must consume at least one odd, which drops the count to 1010 by the mod-3 constraint.

Common mistakes

Assuming the 1012 odd numbers can all survive; one operation is needed to absorb the last even number, and it must consume at least one odd, which drops the count to 1010 by the mod-3 constraint.
Trying to eliminate evens four at a time by multiplying ( $E \cdot E \cdot E \cdot E$ is still even) and concluding the last even can never be removed.

Techniques

Use the answer choices (mod checks, size, form) to eliminate or select · Use an invariant, parity, or coloring argument