Can you find $a+b+c$ ???

This is simple if we can express $1.5=\frac{3}{2}$ in the form of a continued fraction: $\frac{3}{2}=1+\frac{1}{2}=1+\cfrac{1}{1+\frac{1}{1}}\rightarrow a=b=c=1\rightarrow a+b+c=1+1+1=\boxed{3}$

If you don't know much about continued fractions,read the Wikipedia article on it.

Here is how I solved it:

$a + \frac{1}{b + \frac{1}{c}} = a + \frac{1}{\frac{bc+1}{c}} = a + \frac{c}{bc+1} = \frac{abc + a + c}{bc + 1} = \frac{3}{2}$

where the last equality is given by the problem statement.

Now we can say $bc + 1 = 2 \Rightarrow bc = 1$ and since $a, b, c$ are positive integers we have $b = 1, c =1$ .

Similarly we can say $abc + a + c = 3$ , and since $b = 1, c = 1$ we have $2a + 1 = 3 \Rightarrow a = 1$

Therefore $a + b + c = 1 + 1 + 1 = 3$

Note that $a=1$

$\frac{1}{b+ \frac{1}{c}} = \frac{1}{2}$ $\ b + \frac{1}{c} = 2$ $=> b = c = 1$ Thus, $a+b+c = 1+1+1 = 3$

There is a grammar error in the problem as stated. It should read:

Find a+b+c. If a, b, and c are positive integers.

Otherwise you could technically assume that a and b are not restricted to positive integers which would give you an infinite number of solutions.