If n is a factor of 72,such that xy = n,then | vedclass

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(C) The number of ordered pairs $(x, y)$ such that $xy = n$ where $n$ is a factor of $72$ is equivalent to finding the number of pairs $(x, y)$ such t

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The number of ordered pairs $(x, y)$ such that $xy = n$ where $n$ is a factor of $72$ is equivalent to finding the number of pairs $(x, y)$ such that $xy$ divides $72$.Let $72 = 2^3 	imes 3^2$.We want to find the number of pairs $(x, y)$ such that $xy = 2^a 3^b$ where $0 \le a \le 3$ and $0 \le b \le 2$.Let $x = 2^{a_1} 3^{b_1}$ and $y = 2^{a_2} 3^{b_2}$.Then $xy = 2^{a_1+a_2} 3^{b_1+b_2} = 2^a 3^b$.For a fixed $n = 2^a 3^b$,the number of solutions to $a_1 + a_2 = a$ is $(a+1)$ and $b_1 + b_2 = b$ is $(b+1)$.Total pairs = $\sum_{a=0}^{3} \sum_{b=0}^{2} (a+1)(b+1) = (1+2+3+4)(1+2+3) = 10 	imes 6 = 60$.

If $n$ is a factor of $72$,such that $xy = n$,then the number of ordered pairs $(x, y)$ is: (where $x, y \in N$)

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