Question 1096548
<br>There are an infinite number of solutions if there are no restrictions on the values of a, b, and c.  So I will assume the problem requires the values to be positive integers.  In that case....<br>
{{{a + 1/(b+1/c) = a + 1/((bc+1)/c) = a + c/(bc+1) = 37/16 = 2 + 5/16}}}<br>
Since a, b, and c have to be positive integers, there are only two possibilities:
(1) {{{a = 1}}} and {{{c/(bc+1) = 21/16}}}; or
(2) {{{a = 2}}} and {{{c/(bc+1) = 5/16}}}<br>
Looking for a solution for case (1)...<br>
{{{c/(bc+1) = 21/16}}}
{{{16c = 21bc+21}}}
{{{16c-21bc = 21}}}
{{{c(16-21b) = 21}}}
{{{c = 21/(16-21b)}}}<br>
Clearly there are no positive integer values for b that produce positive integer values for c.  So there are no solutions for this case.<br>
Looking for a solution for case (2)...<br>
{{{c/(bc+1) = 5/16}}}
{{{16c = 5bc+5}}}
{{{16c-5bc = 5}}}
{{{c(16-5b) = 5}}}
{{{c = 5/(16-5b)}}}<br>
There is exactly one positive integer value for b that produces a positive integer value for c: b=3 gives c=5.<br>
So there is a unique solution to the problem in positive integers:
a=2;
b=3;
c=5<br>
CHECK:<br>
{{{2 + 1/(3+1/5) = 2 + 1/(16/5) = 2 + 5/16 = 37/16}}}<br>
The answer to the problem is then a+b+c = 10.