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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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