Math, asked by pakhma9lakzsonika, 1 year ago

Determine the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8,15and 21.

Answers

Answered by Golda
1282
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.
 
 
Answered by ishitha0000001
183

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