Question

A number when divided by 3 leaves a remainder 1. When the quotient is divided by 2 it leaves a remainder 1. What will be the remainder when the number is divided by 6?
(a)3
(b)4
(c)5
(d)2

Answer 1

Hint: It is given that a number when divided by 3 leaves a remainder 1 so we can write this statement in the form of dividend, divisor, quotient and remainder. We know that their relation is equal to: $Dividend=Divisor\times quotient+remainder$. Here, the divisor is 3 and remainder is 1. Let us assume that quotient is “q” and dividend is “p” so substituting these values in the equation that we have just described we get, $p=3q+1$ and the equation when the quotient is divided by 2 it leaves a remainder 1 is $q=2x+1$. As you can see that q is an odd number so let us take different values of q as 3, 5, 7, 9 etc. Substituting $q=3$ in $p=3q+1$ we get, $p=3\left( 3 \right)+1=10$. Dividing p by 6 will give the remainder. Similarly, you can check for other odd values of q also.

Complete step-by-step answer:
It is given that a number when divided by 3 leaves a remainder 1 so let us assume the number is p.
Now, we are going to write this statement using remainder theorem that:
$Dividend=Divisor\times quotient+remainder$……….. Eq. (1)
The dividend is p, divisor is 3 and remainder is 1 so let us assume that the quotient is q. Now, substituting these values in the above equation we get,
$p=3q+1$……….. Eq. (2)
It is also given that when the quotient is divided by 2 it leaves a remainder 1 so here dividend is q, divisor is 2 and the remainder is 1 and let us assume that quotient when divided by 2 is “x”. Substituting these values in eq. (1) we get,
$q=2x+1$
In the above equation q is equal to $2x+1$ which is the form of an odd number so we can say that q is an odd number. Hence, q can be any odd number which is 3, 5, 7, 9 etc.
So, let us put q equal to 3 in eq. (2) we get,
$\begin{align}
  & p=3q+1 \\
 & \Rightarrow p=3\left( 3 \right)+1 \\
 & \Rightarrow p=9+1=10 \\
\end{align}$
The given number in the question is 10 so on dividing this number by 6 we get,
$6\overset{1}{\overline{\left){\begin{align}
  & 10 \\
 & \dfrac{-6}{4} \\
\end{align}}\right.}}$
 From the above division you can see that the remainder is 4.
The remainder when divided the number by 6 is equal to 4.
Hence, the correct option is (b).

Note: Let us check for other odd values of q whether we are getting the remainder 4 or not.
Let us equate q to 5 and then substituting this value in eq. (2) we get,
$\begin{align}
  & p=3q+1 \\
 & \Rightarrow p=3\left( 5 \right)+1 \\
 & \Rightarrow p=16 \\
\end{align}$
On dividing the above value of p by 6 we get,
$6\overset{2}{\overline{\left){\begin{align}
  & 16 \\
 & \dfrac{12}{4} \\
\end{align}}\right.}}$
From the above division, you can see that the remainder is 4.
Hence, we have proved that by plugging any odd value of q in $p=3q+1$ and then dividing this number by 6 we get remainder 4.