Question

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

Answer 1

The correct option is B 30

$1440=2^5{.3}^2{.5}^1$
No. of divisors = (5+1).(2+1).(1+1) = 36
Product of divisors = 1.2.3. 480.720.1440. Here all the 36 divisors are written in the increasing order. They can be clubbed into 18 pairs, as shown below.
(1.1440). (2.720).(3.480) -----etc.
$\therefore$ Product of divisors =$(1440{)}^{18}=2^{90}{.3}^{36}{.5}^{18}=(2^3.3{)}^{30}{.3}^6{.5}^{18}={24}^{30}{.3}^6{.5}^{18}$ which is divisible by ${24}^x$
$\therefore$ Maximum value of x =30