Find the greatest number of 6 digits exactly divisible by 24, 15 and 3

1.	Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.
Answer» We know that, the greatest 6 digit number is 999999. Let’s assume that 999999 is divisible by 24, 15 and 36 exactly. Then, the LCM (24, 15 and 36) should also divide 999999 exactly. Finding the prime factors of 24, 15, and 36, we get 24 = 2 × 2 × 2 × 3 15 = 3 × 5 36 = 2 × 2 × 3 × 3 ⇒ L.C.M of 24, 15 and 36 = 360 Since, \(\frac{(999999)}{360}\) = 2777 × 360 + 279 Here, the remainder is 279. So, the greatest number which is divisible by all three should be = 999999 – 279 = 9997201 ∴ 9997201 is the greatest 6 digit number which is exactly divisible by 24, 15 and 36.

Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.

Answer»

We know that, the greatest 6 digit number is 999999.

Let’s assume that 999999 is divisible by 24, 15 and 36 exactly.

Then, the LCM (24, 15 and 36) should also divide 999999 exactly.

Finding the prime factors of 24, 15, and 36, we get

24 = 2 × 2 × 2 × 3

15 = 3 × 5

36 = 2 × 2 × 3 × 3

⇒ L.C.M of 24, 15 and 36 = 360

Since, \(\frac{(999999)}{360}\) = 2777 × 360 + 279

Here, the remainder is 279.

So, the greatest number which is divisible by all three should be = 999999 – 279 = 9997201

∴ 9997201 is the greatest 6 digit number which is exactly divisible by 24, 15 and 36.