36-16 Art and Bob start their new jobs on the same day. Art's work schedule is 3 days on and 1 day off, while Bob's work schedule is 7 days on and 3 days off. In the first 1000 days, the number of days they both have off is 
(A) 48.
(B) 50.
(C) 72.
(D) 75.
(E) 100.
(44th American High School Mathematics Examination, 1993)
[Solution 1] El has a cycle of 4 days, and Bob has a cycle of 10 days. The least common multiple of 4 and 10 is 20, so they have a 20-day cycle together.
There are 50 such cycles in 1000 days. We mark each day of a cycle as 1, $2, \cdots, 20$.

El rests on days $4, 8, 12, 16, 20$, and Bob rests on days $8, 9, 10, 18, 19, 20$, so in each cycle, they rest together on 2 days.
Therefore, the number of days they both rest in the first 1000 days is $2 \cdot 50 = 100$.
Thus, the answer is $(E)$.
[Solution 2] From the problem, we know that El's rest days are the $4k$-th days, and Bob's rest days are the $10t+8, \quad 10t+9, \quad 10t+10 \quad(1 \leqslant t \leqslant 100$, and $t$ is an integer $)$.
From $4k = 10t + 8$, i.e., $2k = 5t + 4$, we know that $t$ is even, so there are 50 solutions;
While $4k = 10t + 9$ has no solution (the left side is even, the right side is odd);
And $4k = 10t + 10$, i.e., $2k = 5t + 5$, we know that $t$ is odd, so there are also 50 solutions.
In summary, they have 100 common rest days.
Thus, the answer is $(E)$.