1. Find the number of triples $(a, b, c)$ of positive integers such that $a+a b+a b c=11$.
Answer:
3
We can write $a+a b+a b c=a(1+b+b c)$. Since 11 is prime, $a=11$ or $a=1$. But since $b, c$ are both positive integers, we cannot have $a=11$, and so $a=1$. Then $1+b+b c=11 \Longrightarrow b+b c=10 \Longrightarrow$ $b(c+1)=10$, and since $c$ is a positive integer, only $b=1,2,5$ are possible. This gives the 3 triples $(a, b, c)=(1,1,9),(1,2,4),(1,5,1)$.