Determine the number nearest to 110000 but greater 100000 which is exa

1.	Determine the number nearest to 110000 but greater 100000 which is exactly divisible by each of 8, 15 and 21.
Answer» First, let’s find the L.C.M of 8, 15 and 21. By prime factorisation, we have 8 = 2 × 2 × 2 15 = 3 × 5 21 = 3 × 7 ⇒ L.C.M (8, 15 and 21) = 2³ × 3 × 5 × 7 = 840 When 110000 is divided by 840, the remainder that is obtained is 800. So, 110000 – 800 = 109200 should be divisible by each of 8, 15 and 21. Also, we have 110000 + 40 = 110040 is also divisible by each of 8, 15 and 21. ⇒ 109200 and 110040 both are greater than 100000 but 110040 is greater than 110000. Hence, 109200 is the number nearest to 110000 and greater than 100000 which is exactly divisible by each of 8, 15 and 21.

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

Answer»

First, let’s find the L.C.M of 8, 15 and 21.

By prime factorisation, we have

8 = 2 × 2 × 2

15 = 3 × 5

21 = 3 × 7

⇒ L.C.M (8, 15 and 21) = 2³ × 3 × 5 × 7 = 840

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

So, 110000 – 800 = 109200 should be divisible by each of 8, 15 and 21.

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

⇒ 109200 and 110040 both are greater than 100000 but 110040 is greater than 110000.

Hence, 109200 is the number nearest to 110000 and greater than 100000 which is exactly divisible by each of 8, 15 and 21.