Question

When a number is successively divided by 7, 5 and 4, it leaves remainder of 4,2 and 3 respectively. What will be the respective remainders when the smallest such number is successively divided by 8, 5 and 6? \[\]
A. $5,0,3$\[\]
B.$2,2,4$\[\]
C.$3,0,3$\[\]
D. $2,4,2$\[\]

Answer 1

Hint: We assume the smallest number as $n$. Let us denote the quotient as $x$ when $n$ is divided by 7 leaving remainder 4 , as $y$ when $x$ is divided by 5 leaving remainder 2 and as $z$ when $y$ is divided by 4 leaving remainder 3. We use Euclid’s division lemma$\left( n=dq+r \right)$ to express $n$ in terms of $z$. We find the smallest value form the expression and verify by dividing by 7,5 and 4 8, 5 and 6?. We divide the smallest value 8, 5 and 6 successively and collect the successive remainders. \[\]

Complete step-by-step answer:
We know that in arithmetic operation of division the number we are going to divide is called dividend, the number by which divides the dividend is called divisor. We get a quotient which is the number of times the divisor is of dividend and also remainder obtained at the end of division. If the number is $n$, the divisor is $d$, the quotient is $q$ and the remainder is $r$, they are related by the following equation,
\[n=dq+r\]
Here the divisor can never be zero. The above relation is called Euclid’s division Lemma. \[\]
We first need to find the smallest number such that when it will be successively divided by 7, 5 and 4, it leaves remainders of 4,2 and 3 respectively. Let the smallest number be $n$. Let us denote the quotient as $x$ when $n$ is divided by 7 leaving remainder 4 , denote the quotient as $y$ when $x$ is divided by 5 leaving remainder 2 and denote the quotient as $z$ when $y$ is divided by 4 leaving remainder 3. We use Euclid’s division lemma and have,
\[\begin{align}
  & n=7x+4....\left( 1 \right) \\
 & x=5y+2....\left( 2 \right) \\
 & y=4z+3.....\left( 3 \right) \\
\end{align}\]
We put $x$ from equation (2) in equation (1) to have,
\[\begin{align}
  & n=7x+4 \\
 & \Rightarrow n=7\left( 5y+2 \right)+4 \\
 & \Rightarrow n=35y+18 \\
\end{align}\]
We put the value of $y$ from (3) in above step to have,
\[\begin{align}
  & \Rightarrow n=35\left( 4z+3 \right)+18 \\
 & \Rightarrow n=140z+105+18 \\
 & \Rightarrow n=140z+123 \\
\end{align}\]
We see from the above equation that the smallest positive integral value of $n$ occurs when $z=0$ which means $n=123$. We verify by dividing 123 by 7, 5 and 4 successively to get remainders 4, 2 and 3 respectively with successive quotients 17, 3 and 3.\[\]
We are asked in the question to find the remainders when divided by 8, 5 and 6 successively. So we divide 123 by 8 to get the remainder 3 and first quotient 15. We divide the first quotient 15 by 5 to get second quotient 3 and remainder 0. We divide the second quotient by 6 to get the third quotient 30 and remainder 3. So the remainders are $3,0,3$ and hence the correct option is C.\[\]

So, the correct answer is “Option C”.

Note: We can alternatively solve by guessing by trial and error for $x$ in $7x+4$ such that $7x+4$ satisfies the given condition. We assumed $n$ to be a positive integer since the divisors and remainders are positive. If 123 would not have satisfied we would have used trial and error for $z$ and $n$.