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What is the best way to do it?

**15.** The number nearest to 100000 but greater than 100000 which is exactly divisible by 8, 15 and 21 is

(A) 100000 (B) 100200

(C) 100800 (D) none of these

First of all take the LCM of 8, 15 and 21, which will be 840.

Now divide 100000 by 840, you will get remainder 40. So we have to compensate 40.

Now your requirement is the number which is greater than 100000, so you will get it by

100000+840-40 = **100800 **is the rquired number, which is divisible by 8, 15 and 21.

Thus option C is correct.

Regards

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