Find the triples $(a,b,c)$ of positive integers that satisfy $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3. $$
I found this on a local question paper, and I am unable to solve it.
Any help will be appreciated.
Find the triples $(a,b,c)$ of positive integers that satisfy $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3. $$
I found this on a local question paper, and I am unable to solve it.
Any help will be appreciated.
On
We have $(1+\frac{1}{3})(1+\frac{1}{2})^2=3$, so that is one solution. It also shows that at least one of $a,b,c$ must be $<3$. wlog we may take $a\le b\le c$. So $a=1$ or $2$.
Suppose $a=1$. Then $(1+\frac{1}{b})(1+\frac{1}{c})=\frac{3}{2}$. Since $(1+\frac{1}{5})^2<\frac{3}{2}$ we must have $b<5$. Obviously we need $b>2$. We find $b=3$ gives the solution $(a,b,c)=(1,3,8)$ and $b=4$ gives the solution $(1,4,5)$.
Suppose $a=2$. Then we have $(1+\frac{1}{b})(1+\frac{1}{c})=2$. Since $(1+\frac{1}{3})^2<2$ we must have $b<3$. That gives the solution already noted of $(2,2,3)$.
Let $a=b=2$ and $c=3$. So you have $(3/2)(3/2)(4/3)=3$.