Difference between revisions of "2020 AMC 10A Problems/Problem 9"
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| − | This is similar to Solution 1, with the same basic idea, but a little different. Since both <math>7</math> and <math>11</math> are prime, their LCM must be their product. However, we don't want to find the LCM, but rather the number of seats required to fit the children and adults. So the answer would be <math>7 + 11 = \boxed{\text{ (B) } 18}</math>. | + | This is similar to Solution 1, with the same basic idea, but a little different. Since both <math>7</math> and <math>11</math> are prime, their LCM must be their product. However, we don't want to find the LCM, but rather the number of seats required to fit the children and adults. So the answer would be <math>7 + 11 = \boxed{\text{(B)} 18}</math>. |
Revision as of 00:08, 2 February 2020
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
The least common multiple of and is 
. Therefore, there must be adults and children. The total number of benches is 
.
Solution 2
This is similar to Solution 1, with the same basic idea, but a little different. Since both and are prime, their LCM must be their product. However, we don't want to find the LCM, but rather the number of seats required to fit the children and adults. So the answer would be 
.
~Baolan
Video Solution
~IceMatrix
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. 
