Difference between revisions of "2021 AMC 10B Problems/Problem 5"
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We are required to find the sum of the ages, which is <cmath>3 + 8 + 5 + 6 = \boxed{(B) \text{ } 22}.</cmath> | We are required to find the sum of the ages, which is <cmath>3 + 8 + 5 + 6 = \boxed{(B) \text{ } 22}.</cmath> | ||
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== Video Solution by OmegaLearn (Using Factors) == | == Video Solution by OmegaLearn (Using Factors) == | ||
Revision as of 12:08, 21 February 2021
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 ![\[3 + 8 + 5 + 6 = \boxed{(B) \text{ } 22}.\]](http://latex.artofproblemsolving.com/5/3/3/533b83ff6a78f1348e3e6954bcd78d80f5290f18.png)
Video Solution by OmegaLearn (Using Factors)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/gLahuINjRzU?t=857
~IceMatrix
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. 
