Difference between revisions of "2024 AMC 10B Problems/Problem 16"
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First, we observe that each time Joey replaces four number with one (let's call this an "<math>operation</math>"), we lose exactly three numbers regardless of the type of <math>operation</math> (<math>Additive</math> or <math>Multiplicative</math>). | First, we observe that each time Joey replaces four number with one (let's call this an "<math>operation</math>"), we lose exactly three numbers regardless of the type of <math>operation</math> (<math>Additive</math> or <math>Multiplicative</math>). | ||
| − | <math>\Rightarrow x \equiv 2024 \equiv 2</math> | + | <math>\Rightarrow x \equiv 2024 \equiv 2\pmod3</math> |
Already, we see that the only possible options are <math>\textbf{(A) } 1010</math> or <math>\textbf{(D) } 1013</math> | Already, we see that the only possible options are <math>\textbf{(A) } 1010</math> or <math>\textbf{(D) } 1013</math> | ||
Revision as of 11:32, 14 November 2024
Contents
Problem
Jerry likes to play with numbers. One day, he wrote all the integers from 
to 
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 , 
, 
, and 
, and then write either 
, their sum, or 
, 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?
Solution 1
Consider the numbers as 
. Note that the number of odd integers is monotonously decreasing.
We need to get rid of 
's, so we either add or multiply the s together to get 
To get rid of the final , we need to consume three other 's to result in one . Thus the answer is 
,
~Mintylemon66
Solution 2
(I will denote even numbers as 
and odd numbers as 
for the purpose of this question, and 
as the answer we need)
First, we observe that each time Joey replaces four number with one (let's call this an "
"), we lose exactly three numbers regardless of the type of (
or 
).
Already, we see that the only possible options are 
or 
We need there to be no 's on the board at the end, so we attempt to convert them into while retaining as many existing 's as possible.
Notice that converting 4 's into an is not possible. We can apply an 
on 3 's and 1 to get back 1 (
) By doing so, we retain all the 's we use in an (and don't generate any new ones). In the first numbers, there are 's. This means that 
, and the only option left is 
Note: For the exact steps to arrive at 1010 odd numbers, perform
(337 times)
(1 time)
~laythe_enjoyer211
Solution 3 (guessing strategy)
Notice that with every step, Jerry removes 
numbers and adds number to the board. Thus, numbers are removed from the board in every step. Since 
, the answer choice must also be congruent to mod . Only 
and 
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 even numbers and odd numbers, it is is not possible to get more odds than there were initially. Thus, we choose 
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
