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Difference between revisions of "2024 AMC 10B Problems/Problem 16"
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==Solution 3 (guessing strategy)==
 
==Solution 3 (guessing strategy)==
−
Notice that with every step, Jerry removes <math>4</math> numbers and adds <math>1</math> number to the board. Thus, <math>3</math> numbers are removed from the board in every step. Since <math>2024\equiv2\pmod{3}</math>, the answer choice must also be congruent to <math>2</math> mod <math>3</math>. Only <math>1010</math> and <math>1013</math> among the answer choices satisfy this condition. This alone eliminates three choices, and one can guess from here for a 3 point expected value. However, since we start out with <math>1012</math> even numbers and <math>1012</math> odd numbers, it seems unlikely that it is possible to get more odds than it started out with. Thus, we choose <math>\boxed{\textbf{(A) } 1010.}</math>
+
Notice that with every step, Jerry removes <math>4</math> numbers and adds <math>1</math> number to the board. Thus, <math>3</math> numbers are removed from the board in every step. Since <math>2024\equiv2\pmod{3}</math>, the answer choice must also be congruent to <math>2</math> mod <math>3</math>. Only <math>1010</math> and <math>1013</math> among the answer choices satisfy this condition. This alone eliminates three choices, and one can guess from here for a 3 point expected value. However, since we start out with <math>1012</math> even numbers and <math>1012</math> odd numbers, it is is not possible to get more odds than there were initially. Thus, we choose <math>\boxed{\textbf{(A) } 1010.}</math>
  
 
~Elephant200
 
~Elephant200

Revision as of 11:31, 14 November 2024

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


Solution 2

(I will denote even numbers as $e$ and odd numbers as $\normalsize o$ for the purpose of this question, and $x$ as the answer we need)

First, we observe that each time Joey replaces four number with one (let's call this an "$operation$"), we lose exactly three numbers regardless of the type of $operation$ ($Additive$ or $Multiplicative$).

$\Rightarrow x \equiv 2024 \equiv 2$ ($mod 3$)

Already, we see that the only possible options are $\textbf{(A) } 1010$ or $\textbf{(D) } 1013$

We need there to be no $e$'s on the board at the end, so we attempt to convert them into $\normalsize o$ while retaining as many existing $\normalsize o$'s as possible.

Notice that converting 4 $e$'s into an $\normalsize o$ is not possible. We can apply an $Additive \, operation$ on 3 $e$'s and 1 $\normalsize o$ to get back 1 $\normalsize o$ ($e+e+e+\normalsize o=\normalsize o$) By doing so, we retain all the $\normalsize o$'s we use in an $operation$ (and don't generate any new ones). In the first $2024$ numbers, there are $1012$ $\normalsize o$'s. This means that $x \le 1012$, and the only option left is $\boxed{\textbf{(A) } 1010}$


Note: For the exact steps to arrive at 1010 odd numbers, perform

$e+e+e+o=o$ (337 times)

$e+o+o+o=o$ (1 time)

~laythe_enjoyer211

Solution 3 (guessing strategy)

Notice that with every step, Jerry removes $4$ numbers and adds $1$ number to the board. Thus, $3$ numbers are removed from the board in every step. Since $2024\equiv2\pmod{3}$, the answer choice must also be congruent to $2$ mod $3$. Only $1010$ and $1013$ among the answer choices satisfy this condition. This alone eliminates three choices, and one can guess from here for a 3 point expected value. However, since we start out with $1012$ even numbers and $1012$ odd numbers, it is is not possible to get more odds than there were initially. Thus, we choose $\boxed{\textbf{(A) } 1010.}$

~Elephant200

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/c6nhclB5V1w?feature=shared

~ Pi Academy

See also

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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