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

Answer
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Hint: We are given that a number is divisible are, 8, 15, and 21, which is greater than 100000 and nearest to 110000. So, we will find the LCM of 8,15,21 first and by dividing 110000 and subtracting that reminder from 110000 will give us our result.

Complete step by step answer:

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
 \[8{\text{ }} = {\text{ }}2\; \times {\text{ }}2\; \times {\text{ }}2\]
\[15{\text{ }} = {\text{ }}3\; \times {\text{ }}5\]
 \[21{\text{ }} = {\text{ }}3\; \times {\text{ }}7\]
L.C.M. \[ = {\text{ }}2\; \times {\text{ }}2\; \times {\text{ }}2\; \times {\text{ }}3\; \times {\text{ }}5\; \times {\text{ }}7{\text{ }} = {\text{ }}840\]
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 i.e. divisible by 8, 15 and 21
\[ = {\text{ }}110000{\text{ }} - {\text{ }}800{\text{ }} = {\text{ }}109200\]
Hence 109200 is exactly divisible by 8, 15 and 21.
So, we have our answer as, 109200.

Note: We find in this problem the nearest and as well as the largest number of which is less than 110000 and divisible by 8, 15, and 21. The rule of solving this type of problem is to find the LCM of the given number by which it is divisible and then divide the number given in the question with the LCM found.
LCM, i.e. the least common multiple is the smallest positive number that is a multiple of two or more numbers.