Find the maximum value of n such that 50! is perfectly divisible by 25...
2520=7 X 32 X 23 X 5.
The value of n would be given by the value of the number of 7s in 50! This Value Is Equal To[50/7] + [50/49] = 7 + 1 = 8 Option (b) is correct.
Find the maximum value of n such that 50! is perfectly divisible by 25...
We know that 2520 = 2^3 x 3^2 x 5 x 7, and we need to find the maximum value of n such that 50! is divisible by 2520^n.
To do this, we can count the number of factors of 2, 3, 5, and 7 in the prime factorization of 50!. We can then divide each count by the corresponding exponent in the prime factorization of 2520, and take the minimum of these quotients.
For example, the number of factors of 2 in 50! is:
50/2 + 50/4 + 50/8 + 50/16 + 50/32 = 25 + 12 + 6 + 3 + 1 = 47
Dividing by the exponent of 2 in 2520 (which is 3), we get:
47/3 = 15 with a remainder of 2
Similarly, we can find the quotients for the other prime factors:
- The number of factors of 3: 22
- The number of factors of 5: 12
- The number of factors of 7: 8
Dividing by the exponents in 2520, we get:
- 22/2 = 11
- 12/1 = 12
- 8/1 = 8
Therefore, the maximum value of n is the minimum of these quotients, which is 8. So 50! is perfectly divisible by 2520^8.