Impossible to Solve?

(The statement of the problem is updated. If you come here from Facebook with the original problem, it was unclear because $b$ is not quantified; people may (rightly) assume that the statement holds for all $b$ instead of for some $b$ , which will give no possible value for $a$ .)

Since $a$ divides both $4b-1$ and $5b+2$ , it must also divide any linear combination of it. In particular, $a$ divides $4(5b+2) + (-5)(4b-1) = 13$ (both $4(5b+2)$ and $(-5)(4b-1)$ are multiples of $a$ , so their sum is also a multiple of $a$ ). But the only possible divisor of $13$ that is greater than $2$ is $13$ . It remains to show that such $b$ exists. But this is straightforward with $b = 10$ ( $4b-1 = 39, 5b+2 = 52$ are both multiples of $13$ ). So the statement is verified, and thus $a = \boxed{13}$ .

Assume 4b-1=a k1 and 5b+2=a k2 Multiply first equation by 5 and second one by 4 and subtract you get 13=a(4 k2-5 k1) Then from here one possibility is that a=13