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# 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. $2,4,2$

Last updated date: 17th Sep 2024
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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. 

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 $n$.