Difference between revisions of "2025 AMC 8 Problems/Problem 13"
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| − | ==Solution 1== | + | == Problem == |
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| + | Each of the even numbers <math>2, 4, 6, \ldots, 50</math> is divided by <math>7</math>. The remainders are recorded. Which histogram displays the number of times each remainder occurs? | ||
| + | |||
| + | == Solution 1 == | ||
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 <math>\boxed{\textbf{(A)}}</math>. | 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 <math>\boxed{\textbf{(A)}}</math>. | ||
| − | ~ | + | ~Sigmacuber |
| − | ==Solution 2== | + | == Solution 2 == |
Writing down all the remainders gives us | Writing down all the remainders gives us | ||
| − | <cmath> | + | <cmath>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.</cmath> |
| − | 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. | ||
| − | </cmath> | ||
In this list, there are <math>3</math> numbers with remainder <math>0</math>, <math>4</math> numbers with remainder <math>1</math>, <math>4</math> numbers with remainder <math>2</math>, <math>3</math> numbers with remainder <math>3</math>, <math>4</math> numbers with remainder <math>4</math>, <math>3</math> numbers with remainder <math>5</math>, and <math>4</math> numbers with remainder <math>6</math>. Manually computation of every single term can be avoided by recognizing the pattern alternates from <math>0, 2, 4, 6</math> to <math>1, 3, 5</math> and there are <math>25</math> terms. The only histogram that matches this is <math>\boxed{\textbf{(A)}}</math>. | In this list, there are <math>3</math> numbers with remainder <math>0</math>, <math>4</math> numbers with remainder <math>1</math>, <math>4</math> numbers with remainder <math>2</math>, <math>3</math> numbers with remainder <math>3</math>, <math>4</math> numbers with remainder <math>4</math>, <math>3</math> numbers with remainder <math>5</math>, and <math>4</math> numbers with remainder <math>6</math>. Manually computation of every single term can be avoided by recognizing the pattern alternates from <math>0, 2, 4, 6</math> to <math>1, 3, 5</math> and there are <math>25</math> terms. The only histogram that matches this is <math>\boxed{\textbf{(A)}}</math>. | ||
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~alwaysgonnagiveyouup | ~alwaysgonnagiveyouup | ||
| − | ==Video Solution | + | == Video Solution 1 == |
https://youtu.be/VP7g-s8akMY?si=JkH2XHbYOhk5-PSc&t=1269 | https://youtu.be/VP7g-s8akMY?si=JkH2XHbYOhk5-PSc&t=1269 | ||
~hsnacademy | ~hsnacademy | ||
| − | ==Video Solution by Thinking Feet== | + | == Video Solution 2 by Thinking Feet == |
https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
| − | ==See Also== | + | == See Also == |
| + | |||
{{AMC8 box|year=2025|num-b=12|num-a=14}} | {{AMC8 box|year=2025|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 12:51, 3 February 2025
Contents
Problem
Each of the even numbers 
is divided by 
. The remainders are recorded. Which histogram displays the number of times each remainder occurs?
Solution 1
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 
.
~Sigmacuber
Solution 2
Writing down all the remainders gives us
![\[2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1.\]](http://latex.artofproblemsolving.com/e/a/d/ead2ad7bd4fcb3c595a09558d722006e01c9b5ee.png)
In this list, there are 
numbers with remainder 
, 
numbers with remainder 
, numbers with remainder 
, numbers with remainder , numbers with remainder , numbers with remainder 
, and numbers with remainder 
. Manually computation of every single term can be avoided by recognizing the pattern alternates from 
to 
and there are 
terms. The only histogram that matches this is .
~alwaysgonnagiveyouup
Video Solution 1
https://youtu.be/VP7g-s8akMY?si=JkH2XHbYOhk5-PSc&t=1269 ~hsnacademy
Video Solution 2 by Thinking Feet
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
