you just plug in h=1,10, c=1,10
and sort them
I guess what I'd do is solve 13h+5c = (41,47,59,61,66)
and remember that h and c have to be non-negative integers
\(13h+5c = 41\\ c(h) = \dfrac{41-13h}{5}\\ c(2) = 3\\ \text{so 41 is a legit choice}\)
\(13h+5c=47\\ c(h)=\dfrac{47-13h}{5}\\ \text{there are no integer solutions to this equation}\\ \text{so this must the the non legit total}\)
checking for thoroughness
\(13\cdot 3 + 5\cdot 4 = 59\)
\(13\cdot 2 + 5 \cdot 7 = 61\)
\(13 \cdot 2 + 5 \cdot 8 = 66\)
.