Difference between revisions of "2020 AMC 10A Problems/Problem 9"
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The least common multiple of <math>7</math> and <math>11</math> is <math>77</math>. Therefore, there must be <math>77</math> adults and <math>77</math> children. The total number of benches is <math>\frac{77}{7}+\frac{77}{11}=11+7=\boxed{\text{(B) }18}</math>. | The least common multiple of <math>7</math> and <math>11</math> is <math>77</math>. Therefore, there must be <math>77</math> adults and <math>77</math> children. The total number of benches is <math>\frac{77}{7}+\frac{77}{11}=11+7=\boxed{\text{(B) }18}</math>. | ||
| − | + | -taarunganesh | |
== Solution 2 == | == Solution 2 == | ||
Revision as of 14:03, 27 December 2020
Contents
Problem
A single bench section at a school event can hold either 
adults or 
children. When 
bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of 
Solution 1
The least common multiple of and is 
. Therefore, there must be adults and children. The total number of benches is 
.
-taarunganesh
Solution 2
This is similar to Solution 1, with the same basic idea, but we don't need to calculate the LCM. Since both and are relatively prime, their LCM must be their product. So the answer would be 
. ~Baolan
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
~savannahsolver
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
