2024 AMC 10B Problems/Problem 16

Revision as of 08:30, 14 November 2024 by Mintylemon66 (talk | contribs) (Solution 1)

Problem

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?

$\textbf{(A) } 1010 \qquad \textbf{(B) } 1011 \qquad \textbf{(C) } 1012 \qquad \textbf{(D) } 1013 \qquad \textbf{(E) } 1014$

Solution 1

Consider the numbers as $1,0,1,0,...,1,0$. Note that the number of odd integers is monotonously decreasing.

We need to get rid of $1012$ $0$'s, so we either add or multiply the $0$s together to get $1012\rightarrow 253 \rightarrow 64+1=64 \rightarrow 16 \rightarrow 4 \rightarrow 1.$

To get rid of the final $0$, we need to consume three other $1$'s to result in one $1$. Thus the answer is $1012-2=\boxed{\textbf{(A) } 1010 }$,

~Mintylemon66

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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