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Educational materials for IPMAT and IIMB UG preparation
10ย years, 4ย months ago
10ย years, 4ย months ago
Hi Aarti,
In order to find the solution of this problem consider 4 types of number: 4k, 4k+1, 4k+2, 4k+3
Sm = m(m+1)(2m+1)/6
Now take the first number,
It contains 4, 8, 12, 16, 20...., 100
In this set all the multiples of 8 would satisfy the condition, ie, 8,16,24.....,96. Total of 12 numbers
Now if you insert 4k+1 and 4k+2 seperatly in this formula, the resulting answer won't be divisible by 4.
Now insert 4k+3 in this formula, you will get numbers 7,15,23.... till 100 in this set. Total numbers in this set = 12
Total numbers in both the set = 12+12 = 24
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