Difference between revisions of "1993 AHSME Problems/Problem 18"
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== Solution == | == Solution == | ||
− | <math>\fbox{E}</math> | + | We note that Al and Barb's work schedules clearly form a cycle, with Al's work cycle being 4 days long and Barb's work schedule being 10 days long. Since <math>\text{LCM}\left(4,10\right)=20</math>, we know that Al's and Barb's work cycles coincide, and thus collectively repeat, every 20 days. As a result, we only need to figure out how many coinciding rest days they have in the first 20 days, and multiply that number by <math>\frac{1000}{20}=50</math> to determine the total number of coinciding rest days. These first 20 days are written out below: |
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+ | Day 1: Al works; Barb works | ||
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+ | Day 2: Al works; Barb works | ||
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+ | Day 3: Al works; Barb works | ||
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+ | Day 4: Al rests; Barb works | ||
+ | |||
+ | Day 5: Al works; Barb works | ||
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+ | Day 6: Al works; Barb works | ||
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+ | Day 7: Al works; Barb works | ||
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+ | Day 8: Al rests; Barb rests | ||
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+ | Day 9: Al works; Barb rests | ||
+ | |||
+ | Day 10: Al works; Barb rests | ||
+ | |||
+ | Day 11: Al works; Barb works | ||
+ | |||
+ | Day 12: Al rests; Barb works | ||
+ | |||
+ | Day 13: Al works; Barb works | ||
+ | |||
+ | Day 14: Al works; Barb works | ||
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+ | Day 15: Al works; Barb works | ||
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+ | Day 16: Al rests; Barb works | ||
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+ | Day 17: Al works; Barb works | ||
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+ | Day 18: Al works; Barb rests | ||
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+ | Day 19: Al works; Barb rests | ||
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+ | Day 20: Al rests; Barb rests | ||
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+ | In this 20-day cycle, Al and Barb's rest days coincide twice: on day 8 and day 20. As a result (as discussed above), the answer is <math>2\cdot50=\fbox{100\text{ (E)}}</math>. ~cw357 | ||
== See also == | == See also == |
Latest revision as of 15:49, 9 September 2020
Problem
Al and Barb start their new jobs on the same day. Al's schedule is 3 work-days followed by 1 rest-day. Barb's schedule is 7 work-days followed by 3 rest-days. On how many of their first 1000 days do both have rest-days on the same day?
Solution
We note that Al and Barb's work schedules clearly form a cycle, with Al's work cycle being 4 days long and Barb's work schedule being 10 days long. Since , we know that Al's and Barb's work cycles coincide, and thus collectively repeat, every 20 days. As a result, we only need to figure out how many coinciding rest days they have in the first 20 days, and multiply that number by to determine the total number of coinciding rest days. These first 20 days are written out below:
Day 1: Al works; Barb works
Day 2: Al works; Barb works
Day 3: Al works; Barb works
Day 4: Al rests; Barb works
Day 5: Al works; Barb works
Day 6: Al works; Barb works
Day 7: Al works; Barb works
Day 8: Al rests; Barb rests
Day 9: Al works; Barb rests
Day 10: Al works; Barb rests
Day 11: Al works; Barb works
Day 12: Al rests; Barb works
Day 13: Al works; Barb works
Day 14: Al works; Barb works
Day 15: Al works; Barb works
Day 16: Al rests; Barb works
Day 17: Al works; Barb works
Day 18: Al works; Barb rests
Day 19: Al works; Barb rests
Day 20: Al rests; Barb rests
In this 20-day cycle, Al and Barb's rest days coincide twice: on day 8 and day 20. As a result (as discussed above), the answer is . ~cw357
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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