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

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(C) The prime factorization of $72$ is $72 = 2^3 	imes 3^2$.Since $n$ is a factor of $72$,$n$ must be of the form $2^a 	imes 3^b$,where $0 \le a \le

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The prime factorization of $72$ is $72 = 2^3 	imes 3^2$.Since $n$ is a factor of $72$,$n$ must be of the form $2^a 	imes 3^b$,where $0 \le a \le 3$ and $0 \le b \le 2$.We are given $xy = n$,where $x, y \in N$. 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 each factor $n$,the number of pairs $(x, y)$ such that $xy=n$ is the number of ways to distribute the prime factors of $n$ into $x$ and $y$.For a factor $n = 2^a 3^b$,the number of pairs $(x, y)$ is $(a+1)(b+1)$.We need to sum this over all factors $n$ of $72$.The sum is $\sum_{a=0}^3 \sum_{b=0}^2 (a+1)(b+1) = (\sum_{a=0}^3 (a+1)) 	imes (\sum_{b=0}^2 (b+1))$.$= (1+2+3+4) 	imes (1+2+3) = 10 	imes 6 = 60$.

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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