If $85_b = 58_c$, what is the smallest possible value of $b$?

Question

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 $b$?

Answer 1

$85_b = 58_c$ 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 $85_{11} = 58_{17} = 93_{10}$

Answer 2

$$8b+5=5c+8\implies 5c =8b-3\implies 5\mid 8b-3$$

So $$5\mid 3b-3\implies 5\mid b-1\implies b=5k+1$$

Since $b\geq 9$ we have $b_{\min}=11$

Answer 3

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 $85_{11} = 58_{17}=93$.