Question

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

Answer 1

Answer:

109200

Step-by-step explanation:

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 = 2 × 2 × 2

15 = 3 × 5

 21 = 3 × 7

L.C.M. = 2 × 2 × 2 × 3 × 5 × 7 = 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 and also divisible by 8, 15 and 21 is 

= 110000 - 840 = 109200

Hence 109200 is exactly divisible by 8, 15 and 21.

 

Answer 2

LCM of 8 = 2 * 2 * 2

LCM of 15 = 3 * 5

LCM of 21 = 3 * 7

LCM of 8,15,21 = 2^3 * 3 * 5 * 7

                         = 840.

Division Method:

840)110000(130

        840

        ----------

          2600

          2520

        -----------

             800

        ------------

Therefore the number nearest to 110000 but greater than 100000 that is divisible by 840 is 110000 - 800 = 109200.

The required number is 109200.

Hope this helps!