# 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 $$b$$?

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

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

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

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

• Although it doesn't affect the solution, $b > 8$, as base $8$ can't have $8$ appear as a digit. – eyeballfrog Mar 13 at 17:32

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