Question

Let the product of all the divisors of 1440 be p.If p is divisible by24^x then the maximum value of x is

Answer 1

The maximum Value of x is

30

• If N is the given number,

then

• N = (p1)^(n1) * (p2)^(n2) *...

where,

• p1, p2 are primes.

Now,

• if k is a divisor of N,
• N/k and their product is k*(N/k) = N.
• each pair of specific divisors will multiply to given N.

Hence,

• product of all divisors of N will be

P = N^(total number of divisors/2)

• total number of divisors for N = (p1)^(n1)
• *(p2)^(n2) *is = (n1 + 1)(n2 + 1)...
• P = N^[(n1 + 1)(n2 + 1).../2].
• N = 1440 = 2^5 * 3^2 *5
• P = N*[(5+1)(2+1)(1+1)/2]
• = [2^5 * 3^2 *5]^[18]
• = 2^90 * 3^36 * 5^18
• 24 = 2^3 * 3
• 24^(30) is a factor of P,
• where 30 is maximum possible power of 24