If N is number of pairs of two divisors of 75600 such that the product of these divisors is 60480, then the value of N is

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answer is 24.

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Detailed Solution

$x = 75600 = 2^{4} 3^{3} 5^{2} 7^{1}$

Let divisor is $2^{a_{1}} {.3}^{b_{1}} {.5}^{c_{1}} {.7}^{d_{1}} and 2^{a_{2}} {.3}^{b_{2}} {.5}^{c_{2}} {.7}^{d_{2}}$

Product of divisors = $2^{6} {.3}^{3} {.5}^{1} {.7}^{1}$

$\Rightarrow a_{1} + a_{2} = 6 \Rightarrow 3 w a y s$

$\Rightarrow b_{1} + b_{2} = 3 \Rightarrow 4 w a y s$

$\Rightarrow c_{1} + c_{2} = 1 \Rightarrow 2 w a y s$

$\Rightarrow d_{1} + d_{2} = 1 \Rightarrow 2 w a y s$

Total way $N = \frac{3 \times 4 \times 2 \times 2}{2} = 24$

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