Question

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

Answer 1

50/7 = 7 1/7. 100/7 = 14 2/7

53/7

= 7 4/7. 95/7 = 13 4/7

Sum = 53 + 60 + 67 + 74 + 81 + 88 + 95

Sum = 3× 148 + 74 = 518

hope this will help

Answer 2

First of all two digit multiple if 7, which is greater than 50 are 56, 63, 70, 77, 84, 91, 98.

So, the two digit no. greater than 50 leaving remainder 4 when divided by 7 are 53, 60, 67, 74, 81, 88, 95

So. the sum is (53+60+67+74+81+88+95)
=518..