Difference between revisions of "2020 AMC 10A Problems/Problem 9"
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== Solution 2 == | == Solution 2 == | ||
− | This is similar to Solution 1, with the same basic idea, but we don't need to calculate the LCM. Since both <math>7</math> and <math>11</math> are prime, their LCM must be their product. So the answer would be <math>7 + 11 = \boxed{\text{(B) } 18}</math>. ~Baolan | + | This is similar to Solution 1, with the same basic idea, but we don't need to calculate the LCM. Since both <math>7</math> and <math>11</math> are relatively prime, their LCM must be their product. So the answer would be <math>7 + 11 = \boxed{\text{(B) } 18}</math>. ~Baolan |
==Video Solution== | ==Video Solution== |
Revision as of 10:18, 15 March 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 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
~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.