2015 HMMT November General #1

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 $$1+b(1+c)=1⟹b(1+c)=0,$$ which is not possible for positive integers $b,c$.

Hence, $a$ must be $1$. Then, we must have $$1+b(1+c)=11⟹b(1+c)=10$$ 

For this, the factors and valid solutions are 

$$b=1,(1+c)=10⟹(b,c)=(1,9)$$

$$b=2,(1+c)=5⟹(b,c)=(2,4)$$

$$b=5,(1+c)=2⟹(b,c)=(5,1)$$

which gives us $\boxed{{\displaystyle 3}}$ solutions.