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Q.

The sum of integers from 1 to 100 should be divisible by 2 and 3 is

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a

4233 

b

816 

c

3417 

d

1683 

answer is D.

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Detailed Solution

 

The number divisible by 2 or 3 are 6 multiples

Sum 6 multiples from 1 to 100

6+12+18+............+96

=6 1+2+3+....+16

=6× 16 16+1 2

∵∑n= n n+1 2

=816

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