2024 AMC 10B Problem 16

This is a "maximum" question, so there are two halves: find limits that no strategy can beat, then show a strategy that reaches the limit.

Limit 1: how the count changes.

1. Each move erases $4$ numbers and writes $1$, so the board loses exactly $3$ numbers every time.
2. The board starts with $2024$ numbers, and $2024 = 3 \cdot 674 + 2$. Taking away $3$ at a time, the count always leaves remainder $2$ when divided by $3$: $2024, 2021, 2018, \ldots$
3. Check the choices: $1010 = 3 \cdot 336 + 2$ and $1013 = 3 \cdot 337 + 2$ fit. $1011$, $1012$ and $1014$ do not.

Limit 2: the number of odd numbers never goes up.

1. Recall the rules: a sum is odd only when it has an odd number of odd terms, and a product is odd only when every factor is odd.
2. So a move can create an odd number only if at least one of the four erased numbers was odd. A move erases at least one odd and writes at most one odd, so the number of odds on the board never goes up.
3. From $1$ to $2024$ there are $1012$ odd numbers. At the end everything is odd, so there are at most $1012$ numbers. That rules out $1013$.

So the answer is at most $1010$.

Reaching $1010$.

1. Take three even numbers and one odd number and replace them with their sum. Even $+$ even $+$ even $+$ odd is odd. This removes three evens and keeps the number of odds at $1012$.
2. There are $1012$ evens. Doing step 7 $337$ times removes $3 \cdot 337 = 1011$ evens, leaving $1$ even.
3. Now take that last even number with three odd numbers and replace them with their sum: even $+$ odd $+$ odd $+$ odd is odd. This uses up three odds and writes one, so $1012 - 3 + 1 = 1010$ odds remain.
4. The board now holds $1010$ numbers, all odd.

The limit is $1010$ and it can be reached.

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

For a process that repeats a move, look for things the move cannot change (the remainder of the count when divided by $3$) or can only change in one direction (the number of odds, which never grows). Each gives a limit, and together they leave a single possible answer. Then build an actual sequence of moves that reaches it.

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 $\boxed{\textbf{(A)}\ 1010}$.