Math, asked by AninditaMohanty5876, 10 months ago

Find the number nearest to 110000 but greater than 10000 which is exactly divisible by each of 815 and 21

Answers

Answered by bunnybublu777
0

Answer:

Solution:-

The number which is divisible by 8, 15 and 21 is also divisible by the L.C.M. of the number.

The L.C.M. of 8, 15 and 21 is =

 8 = 2 × 2 × 2

15 = 3 × 5

 21 = 3 × 7

L.C.M. = 2 × 2 × 2 × 3 × 5 × 7 = 840

If we divide 110000 by 840, we will find out that it is not exactly divisible and we get 800 as remainder.

Thus the number nearest to 110000 but greater than 100000 which is exactly divisible by 840 and also divisible by 8, 15 and 21 is 

= 110000 - 840 = 109200

Hence 109200 is exactly divisible by 8, 15 and 21.

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