Find the sum of all two digit numbers greater than 50 which when divided by 7 leave remainder of 4 - Maths -

Question

Find the sum of all two digit numbers greater than 50 which
when divided by 7 leave remainder of 4 ?

Ankush Jain · Accepted Answer

The first and last numbers between 50 and 99 which on divisibility by 7 leaves the remainder 4 are 53 and 95 respectively.

Since, all the numbers between 50 and 99 which are divisible by 7 and leaves the remainder 4 are in AP, where, a = 53, l = 95 and d = 7.

Now, l = a + (n – 1)d

$\Rightarrow 53 + (n - 1) 7 = 95 \Rightarrow (n - 1) 7 = 95 - 53 = 42 \Rightarrow n - 1 = 6 ∴ n = 7$

Hence, there are 7 numbers between 53 and 95 which are exactly divisible by 7 and leaves the remainder 4 in each case.

Now, sum of all 7 numbers is find as:

$S_{7} = \frac{7}{2} [2 \times 53 + (7 - 1) 7] = \frac{7}{2} [106 + 42] = \frac{7}{2} [148] = 7 \times 74 = 518$

Find the sum of all two digit numbers greater than 50 which when divided by 7 leave remainder of 4.?