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

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Hint – Find out the L.C.M of the given numbers, then divide the given greatest number with this L.C.M to reach the answer.

The number which is exactly divisible by 8, 15, and 21 is also divisible by the L.C.M of numbers 8, 15, and 21. So, we have to take the L.C.M of the given numbers.
So, first factorize the given numbers
The factors of 8 are
$8 = 2 \times 2 \times 2$
The factors of 15 are
$15 = 3 \times 5$
The factors of 21 are
\[21 = 3 \times 7\]
So, the L.C.M of 8, 15, and 21
As we see 3 is the common factor of 15 and 21, so it is written only one time in L.C.M
L.C.M$ = 2 \times 2 \times 2 \times 3 \times 5 \times 7 = 840$
Now if we divide 110000 by 840, we will find out that it is not exactly divisible and we get 800 as remainder, and the quotient is 130.
Number $ = {\text{quotient}}\dfrac{{{\text{remainder}}}}{{{\text{divisor}}}}$
$110000 = 130\dfrac{{800}}{{840}}$
So, we have to subtract the remainder from the number 110000.
Thus the number nearest to 110000 but greater than 100000 which is exactly divisible by 840 and also divisible by 8, 15, and 21 be
$ \Rightarrow 110000 - 800 = 109200$
Hence 109200 is exactly divisible by 8, 15, and 21, which is greater than 100000 and nearest to 110000.

Note: In such types of questions always remember the key concept that first find out the L.C.M of given numbers then divide the highest number with the L.C.M and calculate the value of remainder, then subtract the remainder from the greatest number we will get the required answer.
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