Determine the number nearest to 110000 which is exactly divisible by 8, 15 and 21.

Hint: We are asked a number nearest to 110000 which is exactly divisible by 8, 15 and 21. So, the asked number must also be divisible by the LCM of the numbers 8, 15 and 21.
Thus, find the LCM of 8, 15 and 21.
Then, divide 110000 by LCM of 8, 15 and 21 and find the number in integer form.
Hence, the required answer can be found by the product of LCM of 8, 15 and 21 by the quotient of dividing 110000 by LCM of 8, 15 and 2.

Complete step-by-step answer:
We are asked a number nearest to 110000 which is exactly divisible by 8, 15 and 21.
So, the asked number must also be divisible by the LCM of the numbers 8, 15 and 21.
Now, $8 = 2 \times 2 \times 2,15 = 3 \times 5,21 = 3 \times 7$ .

Thus, the LCM of the numbers 8, 15 and 21 is given by LCM $ = 2 \times 2 \times 2 \times 3 \times 5 \times 7 = 840$ .
Now, dividing the number 110000 by 840, we get
 $\dfrac{{110000}}{{840}} = 130.95$
Now, rounding off 130.95 to the nearest integer, we get 131. So, $\dfrac{{110000}}{{840}} \approx 131$ .
Thus, multiply the LCM of numbers 8, 15 and 21 by 131 to get the nearest number which is exactly divisible by 8, 15 and 21.
 \[ \Rightarrow 840 \times 131 = 110040\]
Hence, the nearest number to 110000 which is divisible by 8, 15 and 21 is 110040.
Note: Here, if we multiply LCM 840 by 130 i.e. the number that we get by rounding off 130.95 , we get 109200.
Since 110040 is more close to 110000 as compared to 109200 so if we multiply 840 by 130 we will get the wrong answer.