Find the greatest number that will divide $43,91$ and $183$ so as to leave the same remainder in each case.

Here, we need to find differences between the given numbers. If two numbers give the same remainder when divided by some other number, then their difference must give a remainder of zero when divided by that number.
Our numbers here are $91-43=48,183-91=92,183-43=140$ 
So we have the set of numbers $\{48,92,140\}$ and we want to know the biggest number that divides all these numbers. 

So,

$48=2\times 2\times 2\times 3$

$92=2\times 2\times 23$

$140=2\times 2\times 5\times 7$

The greatest common divisor of $\{48,92,140\}$ is $4.$ 

So, answer is $4.$