# Problem of the Day

John and Deborah work together at their college library. If John works on a given day, he won’t work again until the day after that. If Deborah works on a given day, she won’t work again until the day after that. In days, the semester will end and John and Deborah will no longer work at the library. If they both work Â today, will they both work on the same day again before the semester ends?

(1) is even, and is odd

(2) Â None of the prime factors that divide evenly divide evenly.

B. Statement 2 alone is sufficient, but statement 1 alone is not sufficient.

See the Solution

### Solution

[latexpage]

If John works the $j^{th}$ day after every day that he works, and he works today, then he will work on the day that is $j$, $2j$, $3j$… days from today. That is, the number of days until some future work day is always a Â multiple of $j$ if he works today. Similarly, Deborah’s work days will be some multiple of $d$ days from today. Thus, if they both work today, they won’t work together again until some number of days has passed that is a multiple of BOTH $j$ and $d$. The first such day will be the least common multiple of $j$ and $d$. If $j$ and $d$ have no factors in common, then their least common multiple will be $jd$. Otherwise, it will be smaller. For example, if $j=3$ and $d=8$, LCM$(3,8)=3 \cdot 8=24$, because 3 and 8 do not share any factors. However, if $j=2$ and $d=4$, then LCM$(2,4)=4$, which is not that same as $2 \cdot 4$.

Statement 1: $j$ is even, and $d$ is odd.Â If $j=3$ and $d=8$, then the semester will end in 17 days, but they won’t work together for another 24 days. However, if $j=3$ and $d=6$, then the semester will end in 11 days, but they will work together in 6 days. INSUFFICIENT.

Statement 2: Â Â Â None of the prime factors that divide $j$ evenly divide $d$ evenly. This tells us that $j$ and $d$ have no common factors, which means LCM$(j,d)=jd$. The semester ends in $jd-7$ days, but they won’t work together again for $jd$ more days. Thus, they definitely won’t work together again. SUFFICIENT.