Difference between revisions of "2021 AMC 10B Problems/Problem 5"
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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> |
Revision as of 01:46, 12 February 2021
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
-PureSwag
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.