Problem:
Find the number of triples $(a, b, c)$ of positive integers such that $a + ab + abc = 11$.
Solution:
Note that the given equation can be factored as $$a(1+b(1+c))=11$$
If $a=11$, then we must have $$1+b(1+c) = 1 \implies b(1+c)=0,$$ which is not possible for positive integers $b, c$.
Hence, $a$ must be $1$. Then, we must have $$1+b(1+c)=11 \implies b(1+c)=10$$
For this, the factors and valid solutions are
$$b=1, (1+c)=10 \implies (b, c) = (1, 9)$$
$$b=2, (1+c)=5 \implies (b, c) = (2, 4)$$
$$b=5, (1+c)=2 \implies (b, c) = (5, 1)$$
which gives us $\boxed{3}$ solutions.
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