A problem by أحمد الحلاق

We have that $\log_{6}a + \log_{6}b + \log_{6}c = \log_{6}(abc) = 6 \Longrightarrow abc = 6^{6}$ .

Now with $r \gt 1$ we have that $b = ar, c = ar^{2}$ and $abc = a^{3}r^{3} = (ar)^{3} = 6^{6} \Longrightarrow ar = 6^{2} = 36$ .

This in turn implies that $ac = 6^{4} = 2^{4}3^{4}$ .

Next, we require that $b - a = ar - a = 36 - a = n^{2}$ for some (positive) integer $n$ . Since $a \gt 0$ we see that $n$ is limited to $1,2,3,4,5$ , yielding respective values for $a$ of $35, 32, 27, 20, 11$ . Now as $ac = 2^{4}3^{4}$ we can only have primes $2$ and $3$ in the prime factorization of $a$ , and in each case only up to a power of $4$ . The only possible value for $a$ among the ones listed above is $27 = 3^{3}$ , which yields a value for $c$ of $2^{4}*3 = 48$ .

Thus $a + b + c = 27 + 36 + 48 = \boxed{111}$ .