Question

The sum of all two digit numbers which when divided by four give unity as the remainder is
(A). 1200
(B). 1210
(C). 1250
(D). None of these

Answer 1

Hint: This is a question regarding the divisibility and the sum of an AP. The divisibility of 4 is that the last two digits of the number are divisible by 4. Also, the formula of sum of an AP with n terms, first term a and difference d is-
${\mathrm S}_{\mathrm n}=\dfrac{\mathrm n}2\left(2\mathrm a+\left(\mathrm n-1\right)\mathrm d\right)\\$ ....(1)

Complete step-by-step answer:

The two digit numbers divisible by 4 are 12, 16, 20, … 100 and so on. But we need only two digit numbers, so the numbers are 12, 16, 20, …. 96.

We need numbers who give remainder as 1(unity) when divided by 4, so we will add one to all these numbers to get the required series.
The new AP is 13, 17, 21, … 97.
Now we have to find the sum of these numbers. For this AP,
a = 13
d = 4
$a_n$ = 97
$a_n$ = a + (n-1)d
Substituting the values of a, d and an-
97 = 13 + (n-1)4
4(n-1) = 84
n-1 = 21
n =22
Now, using the formula for the sum of AP given in equation (1) is-
${\mathrm S}_{\mathrm n}=\dfrac{22}2\left(2\times13+\left(22-1\right)4\right)\\$ =1210

Hence, the correct answer is B. 1210

Note: To find the correct answer, it is important to first identify the series correction and check if it is an AP, GP or any other form of progression. One should have complete knowledge of all kinds of progressions to solve such problems. Also, if it is required to find a series with a particular remainder, just find the divisible numbers and add the remainder to each term.