Question

determine the number that is nearest to 100000 but greater than 100000 which is excatly divisible by 8,15.and 21

Answer 1

Answer :

Here, 8 = 2 × 2 × 2
15 = 3 × 5
21 = 3 × 7

So, the LCM of 8, 15 and 21 is

= 2 × 2 × 2 × 3 × 5 × 7

= 840

Thus, the required number be of the form 840q, where q is a larger Natural number for which, we will ne able to get the required solution.

After putting a few number of q = 1, 2, ..., we get

840 × 119 = 99960, when q = 119
840 × 120 = 100800, when q = 120

Here, 100800 > 100000.

Therefore, the required number is 100800.

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Answer 2

Answer:

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