Question

find the greatest number that will divide 43 91 and 183 so as to leave the same remainder in each case​

Answer 1

$\sf{\underline{To\:find:}}$ The greatest number.

$\sf{\underline{Now:}}$

$\implies$ $\sf{91-43 = 48}$

$\implies$ $\sf{183-91 = 92}$

$\implies$ $\sf{183-43 = 140}$

So, the required numbers are: 48, 92, 140

$\sf{\underline{Now:}}$

We have to find the HCF of: $\boxed{\sf{48, 92, 140}}$

$\sf{\underline{Their\:HCF\:is:}}$ 4

$\sf{\underline{Note:}}$ Thus, 4 is the greatest number, which will divide 43, 91 and 183 so as to leave the same remainder in each case.

$\sf{\underline{HCF\:can\:be\: calculated\:as:}}$

$\sf{48 = 2 \times 2 \times 2 \times 2 \times 3}$

$\sf{92 = 2 \times 2 \times 23}$

$\sf{140 = 2 \times 2 \times 5 \times 7}$

$\sf{\underline{Here:}}$

Only 2 is twice common in factor of all three numbers.

$\sf{\underline{So:}}$

HCF $= 2 \times 2 = 4$

$\sf{\underline{Therefore:}}$

4 is the required number.