2025 AMC 8 Problems/Problem 13: Difference between revisions

Revision as of 22:25, 31 January 2025

Problem

Each of the even numbers $2, 4, 6, \ldots, 50$ is divided by $7$ . The remainders are recorded. Which histogram displays the number of times each remainder occurs?

Solution

Let's take the numbers 2 through 14 (evens). The remainders will be 2, 4, 6, 1, 3, 5, and 0. This sequence keeps repeating itself over and over. We can take floor(50/14) = 3, so after the number 42, every remainder has been achieved 3 times. However, since 44, 46, 48, and 50 are left, the remainders of those will be 2, 4, 6, and 1 respectively. The only histogram in which those 4 numbers are set at 4 is histogram $\boxed{\text{(A)}}$ .

~Sigmacuber

Solution 2

Writing down all the remainders gives us

$2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1.$

In this list, there are $3$ numbers with remainder $0$ , $4$ numbers with remainder $1$ , $4$ numbers with remainder $2$ , $3$ numbers with remainder $3$ , $4$ numbers with remainder $4$ , $3$ numbers with remainder $5$ , and $4$ numbers with remainder $6$ . Manually computation of every single term can be avoided by recognizing the pattern alternates from $0, 2, 4, 6$ to $1, 3, 5$ and there are $25$ terms. The only histogram that matches this is $\boxed{\textbf{(A)}}$ .

~alwaysgonnagiveyouup

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=JkH2XHbYOhk5-PSc&t=1269 ~hsnacademy

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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 AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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