Difference between revisions of "2013 AMC 10A Problems/Problem 17"
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Remember to subtract 1, because you do not wish to count the 60th day, when all three visit. | Remember to subtract 1, because you do not wish to count the 60th day, when all three visit. | ||
− | A and B visit <math>\frac{60}{3 | + | A and B visit <math>\frac{60}{3 \cdot 4}-1 = 4</math> times. |
− | B and C visit <math>\frac{60}{4 | + | B and C visit <math>\frac{60}{4 \cdot 5}-1 = 2</math> times. |
− | C and A visit <math>\frac{60}{3 | + | C and A visit <math>\frac{60}{3 \cdot 5}-1 = 3</math> times. |
This is a total of <math>9</math> visits per <math>60</math> day period. | This is a total of <math>9</math> visits per <math>60</math> day period. | ||
− | Therefore, the total number of 2-person visits is <math>9 | + | Therefore, the total number of 2-person visits is <math>9 \cdot 6 = \boxed{\textbf{(B) }54}</math>. |
==See Also== | ==See Also== |
Revision as of 01:40, 2 January 2016
Problem
Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?
Solution
The 365-day time period can be split up into 60-day time periods, because after days, all three of them visit again (Least common multiple of , , and ). You can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by 60. Remember to subtract 1, because you do not wish to count the 60th day, when all three visit.
A and B visit times. B and C visit times. C and A visit times.
This is a total of visits per day period. Therefore, the total number of 2-person visits is .
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
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