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

I KNOW THIS IS SOLUTION BUT I DON'T KNOW WHY?

We first find the difference of the numbers and then find the HCF of the got numbers.

183−91=92

183−43=140

91−43=48

Now find HCF of 92, 140 and 48, we get

92=2×2×23

140=2×2×5×7

48=2×2×2×2×3

HCF(92, 140, 48) = 4

Therefore, 4 is the required number.

Can you explain how we get correct answer using this method.

Call the greatest number $$n$$ and the common remainder $$x$$, so our problem is
\begin{align} x&\equiv 43\pmod{n}\\ x&\equiv 91\pmod{n}\\ x&\equiv 183\!\!\!\pmod{n} \end{align}\qquad
\begin{align} &n\mid 91\!-\!43,\, 183\!-\!91,\, 184\!-\!43\\ \iff \ \ &n\mid 48,92,140\\ \iff\ \ &n\mid \gcd(48,92,140) = 4\end{align}\ \