Question

A number divided by 296 leaves a remainder of 75, if the same number is divided by 37 then remainder obtained is
(a) 2
(b) 1
(c) 11
(d) 8

Answer 1

- Hint: Write the given information in equation form by assuming the number to be x. A number leaves a remainder of 75 on dividing by 296 means that number is 75 more than the multiple of 296.
x = 296q + 75 where q is an integer known as quotient tells how many times 296 is contained in x.

Complete step-by-step solution -
Let the remainder be r when x is divided by 37. Then we get
x = 37s + r
Now eliminate x from both the equation and try to write the expression in the multiple of 37 with some remainder.

When a number $a$ is divided by $b$ and leaves a remainder $r$ with $b$ is contained maximum $q$ times in $a$. Then this information is written in equation form as
$a=bq+r$ with $0\le r < b\,\,,\,\,r\in \mathbb{Z}\,\,\,\,\,\,\,\,\cdot \cdot \cdot (\text{i)}$
In the question we are given that a number when divided by 296 leaves a remainder 75. Let that number be x. Then writing this information in the form of equation (i) we get
$x=296q+75\,\,\,\text{for some }q\in \mathbb{Z}\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot (\text{ii})$
Let when x is divided by 37 leaves a remainder r. Then writing again as equation (i) we get
$x=37s+r\,\,\,\,\text{for some }s\in \mathbb{Z}\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot (\text{iii})$
Now eliminating x from equation (ii) and equation (iii) we get
$\begin{align}
  & 296q+75=37s+r \\
 & \Rightarrow r=296q-37s+75 \\
 & \Rightarrow r=37\times 8q-37s+37\times 2+1 \\
 & \Rightarrow r=37\left( 8q-s+2 \right)+1 \\
\end{align}$
Putting this r in equation (iii) we get
$x=37s+37\left( 8q-s+2 \right)+1$
We know that r must be less than 37, so combining both the multiple of 37 we get,
$x=37\left( 8q+2 \right)+1$
Since q is an integer so 8q+2 is also an integer and 0 ≤ 1 < 37. So we get the remainder as 1.
Hence, when that number is divided by 37, it leaves the remainder of 1.

Note: You can directly use the equation (i) to simplify as given below
$\begin{align}
  & x=296q+75 \\
 & \Rightarrow x=37\times 8q+37\times 2+1 \\
 & \Rightarrow x=37\left( 8q+2 \right)+1 \\
\end{align}$
So the remainder will be 1.