Solve this :

Q. Let the product of all the divisors of 1440 be P. If P is divisible by 24^x, then the maximum value of x is :
(a) 28                                              (b) 30
(c) 32                                              (d) 36

If N is the given number, then, it can be written as:
N = (p1)^(n1) * (p2)^(n2) *...
where, p1, p2 etc are primes.
Now, if k is a divisor of N, so is N/k and their product is k*(N/k) = N.
Thus each such pair of distinct divisors will multiply to give N.
Hence, product of all divisors of N will be:
P = N^(total number of divisors/2)
Also, total number of divisors for N = (p1)^(n1) * (p2)^(n2) *...
is = (n1 + 1)(n2 + 1)...
Hence,
P = N^[(n1 + 1)(n2 + 1).../2].

Here,
N = 1440 = 2^5 * 3^2 *5
Hence,
P = N*[(5+1)(2+1)(1+1)/2]
= [2^5 * 3^2 *5]^[18]
= 2^90 * 3^36 * 5^18
Also, 24 = 2^3 * 3
Hence, we see that 24^(30) is a factor of P, where 30 is maximum possible power of 24.