Question

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
(A) 4
(B) 7
(c) 9
(D) 13

Answer 1

Hint: Represent the number as $\text{Divisor}\times \text{Quotient}+\text{Remainder}$, here divisor and remainder take it as d and r. As it will be same for all and quotient as ${{q}_{1,}}{{q}_{2,}}{{q}_{3}}$. Then find the expression of $d\left( {{q}_{2}}-{{q}_{1}} \right),d\left( {{q}_{3}}-{{q}_{2}} \right)\text{ and }d\left( {{q}_{3}}-{{q}_{1}} \right)$. Then finally find the values of three numbers to get the solution.

Complete step-by-step solution:
As we know the rule of division we can represent it in the form,
$\left( \text{Dividend} \right)=\left( \text{Quotient} \right)\times \left( \text{Divisor} \right)+\left( \text{Remainder} \right)$
Here in the dividend we will put 43,91,183 to find three equations where divisors and quotients will remain the same and remainder will only differ.
So for 43,let the divisor be d, quotient be q and remainder be r, so we can write it as
${{q}_{2}}>\text{ }{{\text{q}}_{1}}$
So, r can be written as $43-d\times {{q}_{1}}$
Now for 91, the divisor and remainder would be d and r respectively and be ${{q}_{2}}$ so we can write it as,
$\text{Divisor}\times \text{Quotient}+\text{Remainder}$
So, r can be written as $91-d{{q}_{2}}$
Now for 183, the divisor and remainder would be d and r respectively and quotient be ${{q}_{3}}$ and so we can write it as,
$183=d\times {{q}_{3}}+r$
So, r can be written as $183-d\times {{q}_{3}}$
Now for r there are three different form of it which are,
$r=43-d{{q}_{1}},r=91-d{{q}_{2}},r=183-d{{q}_{3}}$
So write it as
\[\begin{align}
  & 43-d{{q}_{1}}=91-d{{q}_{2}} \\
 & \Rightarrow d\left( {{q}_{2}}-{{q}_{1}} \right)=48 \\
 & \Rightarrow d\left( {{q}_{2}}-{{q}_{1}} \right)=48 \\
\end{align}\]
Now we can write it as,
$\begin{align}
  & 91-d{{q}_{2}}=183-d{{q}_{3}} \\
 & \Rightarrow d\left( {{q}_{3}}-{{q}_{2}} \right)=92 \\
 & \Rightarrow d\left( {{q}_{3}}-{{q}_{2}} \right)=92 \\
\end{align}$
Now at last we can write it as,
$\begin{align}
  & 183-d{{q}_{3}}=43-d{{q}_{2}} \\
 & \Rightarrow d\left( {{q}_{3}}-{{q}_{1}} \right)=140 \\
 & \Rightarrow d\left( {{q}_{3}}-{{q}_{1}} \right)=140 \\
\end{align}$
By the expression $d\left( {{q}_{2}}-{{q}_{1}} \right)=48$
We can conclude that ${{q}_{2}}>\text{ }{{\text{q}}_{1}}$ ,
$d\left( {{q}_{3}}-{{q}_{2}} \right)=92$ we can say that ${{\text{q}}_{3}}>{{\text{q}}_{2}}$ and lastly from \[\text{d}\left( {{q}_{3}}-{{q}_{1}} \right)=140\], we can say that ${{\text{q}}_{3}}>{{\text{q}}_{1.}}$.
So, we can say that ${{q}_{3}}>\text{ }{{\text{q}}_{2}}>\text{ }{{\text{q}}_{1}}$ .
Now the d will be the HCF of numbers 48, 92, 140 and d will be the greatest common divisor.
So let’s factorize 48, 92,140 in the product of prime factors and find the greatest common divisor between them.
So we can write,
$48={{2}^{4}}\times 3,\text{ }92={{2}^{2}}\times 23,140={{2}^{2}}\times 5\times 7$
So on comparing the factors we can say that ${{2}^{2}}$ or $4$ is the highest common factor between 48, 92, 140.
So the value of d or divisor is $4$ .
The correct option is “A”.

Note: In such types of problems we need to find the factors of given numbers. Here the remainder is the same and we have 3 equations and 3 variables. After solving these 3 equations we get the value of the divisor which is represented by d.