determine the number nearest to 100000 but greater than 10000 which is exactly divisible by each of 8,15, and 21​

Answer:

8=2

3

15=3×3

21=3×7

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

2

3

×3

2

×7=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.

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