Question

Hii everyone ❤
Plz help me in my homework
I am not getting how solve the following que
* determine the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8,15,21

Answer 1

The number which is divisible by 8, 15, and 21 is also divisible by the L.C.M of the number 8, 15 and 21. The L. C. M of 8, 15, and 21 is 840

110000 can be divided by 840

Thus, the number less than 110000 and nearest to 110000 that is divisible by 840 is 110000 – 800 = 109200

Now, 109200 is also greater than 100000.

Thus, the number nearest to 110000 and greater than 100000 which is exactly divisible by 8, 15, and 21 is 109200.

Answer 2

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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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