Question

When 201 is divided by 2, then the remainder is ……..
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: we use the help of long division and remainder theorem to find the remainder. For the dividend 201 and the divisor 2, we apply the theorem $q=pk+r,0\le r < p$. We get the value of k and r to solve the problem.

Complete step by step answer:
We apply the concept of long division and remainder theorem to find the remainder When 201 is divided by 2.
\[2\overset{100}{\overline{\left){\begin{align}
  & 201 \\
 & \underline{2} \\
 & 00 \\
 & \underline{00} \\
 & 01 \\
\end{align}}\right.}}\]
We first tried to equate the first digit of the dividend with 2 the divisor 2 and that’s why we multiplied with 1. We get 2. We subtract it to get 2. Now we have 0 which is directly divisible by 2 being multiplied by 2. Then the remaining number is 0 which is further followed by none. That’s why we add another 0 to the quotient. We get 1 as the remainder.
We now use the concept of remainder theorem which gives if q is the dividend, p is the divisor and r is remainder then we will have a value $k\in \mathbb{Z}$ such that $q=pk+r,0\le r < p$.
Therefore, we find the similar form for 201 where $201=2\times 100+1$.
Here r is the remainder with value 1.
So, the correct answer is “Option A”.

Note: We need to remember the condition for $0\le r < p$. The condition will keep the reminder in a range otherwise we would have found an infinite number of combinations for the theorem $q=pk+r$.