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

Given: 24, 15 and 36.

To find: Here we have to find the greatest number of 6 digits exactly divisible by 24, 15 and 36.

Solution:

The greatest 6-digit number  $=$  999999

LCM is the least common multiple of any three numbers and to find the greatest 6-digit number we have to check if 999999 is divisible by the LCM of 24, 15 and 36.

Now, calculating LCM of 24, 15 and 36 using prime factorization method:

Writing the numbers as a product of their prime factors:

Prime factorization of 24:

• $2\ \times\ 2\ \times\ 2\ \times\ 3\ =\ 2^3\ \times\ 3^1$

Prime factorization of 15:

• $3\ \times\ 5\ =\ 3^1\ \times\ 5^1$

Prime factorization of 36:

• $2\ \times\ 2\ \times\ 3\ \times\ 3\ =\ 2^2\ \times\ 3^2$

Multiplying highest power of each prime number:

• $2^3\ \times\ 3^2\ \times\ 5^1\ =\ 360$

LCM(24, 15, 36)  $=$  360

Now,

$999999\ =\ 360\ \times\ 2777\ +\ 279$

Here remainder is 279. So,

Required number  $=$  999999  $-$  279  $=$  999720

So, the greatest 6 digit number which is exactly divisible by 24, 15 and 36 is 999720.

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