Determine the number nearest to 110000 but greater 100000 which is exactly divisible by each of 8, 15 and 21.
Answers
SOLUTION :
GIVEN : Three numbers 8, 15 and 21.
First find the L.C.M of 8, 15 and 21
The prime factors of 8, 15 and 21 :
8 = 2 x 2 x 2 = 2³
15 = 3¹ x 5¹
21 = 3¹ x 7¹
L.C.M of 8, 15 and 21 = 2² x 3¹ x 5¹ x 7¹
[LCM of two or more numbers = product of the greatest power of each prime factor involved in the numbers, with highest power.]
L.C.M of 8, 15 and 21 = 840
Divide 110000 by 840, the remainder is 800. Subtract the remainder 800 from 110000.
Now, 110000 - 800 = 109200 is divisible by each of 8, 15 and 21.
Hence, 109200 is the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.
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Answer:
Step-by-step explanation:
Given :-
Number are 8, 15 and 21.
To Find :-
Number nearest to 110000 but greater 100000
Solution :-
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
By dividing 110000 by 840, it is not divisible.
so we will take 800 as remainder.
Therefore, 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 - 800 = 109200
Hence, 109200 is exactly divisible by 8, 15 and 21.