Latest IBM Aptitude Question SOLUTION: The sum of the first 100 numbers, 1 to 100 is divisible by

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(#M40005533) IBM Numerical Ability question problems on numbers Keep an EYE

The sum of the first 100 numbers, 1 to 100 is divisible by

Asked In IBM (12 years ago)

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pratik kumar  (12 years ago)

Sum of first 100 number is = 100/2(100+1) = 5050.

Ans. is B. 2.

Note:
1. Why not divisible by 4. The last two digit (here, 50) is not divisible by 4
2. Why not divisible by 8. The last three digits should be 0's or divisble by 8. Doesnt work here.
3. Why not by 6. Divisibility by 6 means, it should be divided by 2 and 3. Yes, we knwo, itsi dviisble by 2, what about 3. For it to be divisible by e, the sum of the numbers should be divisible by 3. Here it doesnt work. So, 6 is off.

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stuti  (10 years ago)

y not by 5...usasay bhi toh divisible hai na..!!

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jyoti agarwal  (7 years ago)

n(n+1)/2=100(100+1)/2
=5050
hence 5050 is divisible by 2

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