Difference between revisions of "2021 AMC 10B Problems/Problem 5"
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| + | ==Problem== | ||
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The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give <math>24</math>, while the other two multiply to <math>30</math>. What is the sum of the ages of Jonie's four cousins? | The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give <math>24</math>, while the other two multiply to <math>30</math>. What is the sum of the ages of Jonie's four cousins? | ||
<math>\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25</math> | <math>\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25</math> | ||
| − | ==Solution | + | ==Solution== |
First look at the two cousins' ages that multiply to <math>24</math>. Since the ages must be single-digit, the ages must either be <math>3 \text{ and } 8</math> or <math>4 \text{ and } 6.</math> | First look at the two cousins' ages that multiply to <math>24</math>. Since the ages must be single-digit, the ages must either be <math>3 \text{ and } 8</math> or <math>4 \text{ and } 6.</math> | ||
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Next, look at the two cousins' ages that multiply to <math>30</math>. Since the ages must be single-digit, the only ages that work are <math>5 \text{ and } 6.</math> Remembering that all the ages must all be distinct, the only solution that works is when the ages are <math>3, 8</math> and <math>5, 6</math>. | Next, look at the two cousins' ages that multiply to <math>30</math>. Since the ages must be single-digit, the only ages that work are <math>5 \text{ and } 6.</math> Remembering that all the ages must all be distinct, the only solution that works is when the ages are <math>3, 8</math> and <math>5, 6</math>. | ||
| − | We are required to find the sum of the ages, which is <cmath>3 + 8 + 5 + 6 = \boxed{(B) | + | We are required to find the sum of the ages, which is <cmath>3 + 8 + 5 + 6 = \boxed{\textbf{(B)} ~22}.</cmath> |
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| + | -[[User:PureSwag|PureSwag]] | ||
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| + | == Video Solution by OmegaLearn (Using Factors) == | ||
| + | https://youtu.be/oe7lnbDO8bo | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution by TheBeautyofMath== | ||
| + | https://youtu.be/gLahuINjRzU?t=857 | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | ==Video Solution by Interstigation== | ||
| + | https://youtu.be/DvpN56Ob6Zw?t=358 | ||
| + | |||
| + | ~Interstigation | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/cQC4YORt3OU | ||
| − | + | ~Education, the Study of Everything | |
== See Also == | == See Also == | ||
Latest revision as of 17:18, 16 August 2022
Contents
Problem
The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give 
, while the other two multiply to 
. What is the sum of the ages of Jonie's four cousins?
Solution
First look at the two cousins' ages that multiply to . Since the ages must be single-digit, the ages must either be 
or 
Next, look at the two cousins' ages that multiply to . Since the ages must be single-digit, the only ages that work are 
Remembering that all the ages must all be distinct, the only solution that works is when the ages are 
and 
.
We are required to find the sum of the ages, which is ![\[3 + 8 + 5 + 6 = \boxed{\textbf{(B)} ~22}.\]](http://latex.artofproblemsolving.com/0/8/a/08a7d7abac9e4a0a4564b8008acc3e1aded4f510.png)
Video Solution by OmegaLearn (Using Factors)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/gLahuINjRzU?t=857
~IceMatrix
Video Solution by Interstigation
https://youtu.be/DvpN56Ob6Zw?t=358
~Interstigation
Video Solution
~Education, the Study of Everything
See Also
| 2021 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
