If $85_b = 58_c$, what is the smallest possible value of $b$? I have searched and watched videos online and can't find a method to solve this problem:

If $85_b=58_c$ for some positive integer bases $b$ and $c$, what is the smallest possible value for $b$?

 A: $85_b = 58_c$ means $8b + 5 = 5 c + 8$. Thus, $b = (3+5c)/8$. This is an integer whenever $c = 8n + 1$ for some integer $n$. In this case, $b = 5n+1$. Now, since $8$ appears as a digit, we must have $b > 8$. The smallest integer of the form $5n+1$ that is larger than $8$ is $11$.
So the smallest solution is $b = 11$, $c = 17$, as $85_{11} = 58_{17} = 93_{10}$
A: $$8b+5=5c+8\implies 5c =8b-3\implies 5\mid 8b-3$$
So $$5\mid 3b-3\implies 5\mid b-1\implies b=5k+1$$
Since $b\geq 9$ we have $b_{\min}=11$
A: well think it out: 
$8b + 5 = 5c + 8$.
So $8b - 5c = 3$. 
$b = 1; c= 1$ is a solution but it is too small as we must have $b,c > 8$.  But there are an infinite number of solutions:
If $8b - 5c = 3$ then $8(b+ 5k)-(c+8k)= (8b-5c) + 40k - 40k$ will also be a solution.
$8$ and $5$ are relatively prime so all solutions are of the form:
$8(b + 5k) - (c + 8k) = 3$. And as $b=c = 1$ is one solution all solutions are of form:
$8(1 + 5k) - (1 + 8k)$.
So we just need to find the smallest $k$ where $1 + 5k, 1+8k > 8$.  That number is clearly $k = 2$ and $b = 11> 8$ and $c = 17> 8$. 
And $85_{11} = 58_{17}=93$.
