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

TO FIND: The number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

L.C.M Of 8, 15 and 21. 

$8=2^{3}$

$15=3 \times 5$

$21=3 \times 7$

L.C.M of 8,15 and $21=2^{3} \times 3 \times 5 \times 7=840$

When 110000 is divided by 840, the remainder is obtained as 800.

Now, 110000 − 800 = 109200 is divisible by each of 8, 15 and 21.

Also, 110000 + 40 = 110040 is divisible by each of 8, 15 and 21.

109200 and 110040 are greater than 100000.

Hence, 110040 is the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.