# 1993 AHSME Problems/Problem 18

## 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? $\text{(A) } 48\quad \text{(B) } 50\quad \text{(C) } 72\quad \text{(D) } 75\quad \text{(E) } 100$

## 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 $\text{LCM}\left(4,10\right)=20$, 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 $\frac{1000}{20}=50$ 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 $2\cdot50=\fbox{100\text{ (E)}}$. ~cw357

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