Question

A number when divided by 342 gives a remainder 47. When the same number is divided by 19, what would be the remainder?

Answer 1

Hint: We will use the division rule in this question that is $n=mq+r$, where n is dividend, m is divisor, q is quotient and r is remainder. Here we will assume that number is n and the quotient is x. So, we will be able to write \[n=342x+47\]. Now, we write this number n in terms of $19y+r$.

Complete step-by-step answer:
The division rule says that if n is dividend, m is divisor, q is quotient and r is remainder then the number n can be written as,
$n=mq+r.............\left( 1 \right)$
Let us assume that the number is n and we are given that the divisor is 342 and remainder is 47.
So, assume that the quotient is y.
Using equation (1), we will have,
\[n=342y+47..........\left( 2 \right)\]
Where y is an integer.
Now, we can write $342=19\times 18$
The equation (2) will become,
$\begin{align}
  & \Rightarrow n=19\times 18y+47 \\
 & \Rightarrow n=19\times 18y+19\times 2+9 \\
 & \Rightarrow n=19\times \left( 18y+2 \right)+9 \\
\end{align}$
As y is an integer. So, $18y+2$ will be integer too.
Now, we will assume $18y+2=k$.
$\Rightarrow n=19k+9...........\left( 3 \right)$
Now, applying the division rule, we can see from equation (3) that the quotient is k and remainder is 9.
So, the remainder when the number divided by 19 is 9.

Note: Students can make mistakes in the step where we write 47 as $19\times 2+9$. Instead of this they can write as \[47=19\times 3-9\]. But students will have to understand that the remainder should always be positive in fact 0$\leq$ r$<$m (m = divisor).