The number of positive divisors of 360 which | vedclass

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Question

(A) First,find the prime factorization of $360$: $360 = 2^3 	imes 3^2 	imes 5^1$. For a divisor to be a multiple of $3$,it must contain at least one

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) First,find the prime factorization of $360$: $360 = 2^3 	imes 3^2 	imes 5^1$. For a divisor to be a multiple of $3$,it must contain at least one factor of $3$. Let the divisor be of the form $2^a 	imes 3^b 	imes 5^c$,where $0 \le a \le 3$,$1 \le b \le 2$,and $0 \le c \le 1$. The number of choices for $a$ is $4$ (i.e.,$0, 1, 2, 3$). The number of choices for $b$ is $2$ (i.e.,$1, 2$). The number of choices for $c$ is $2$ (i.e.,$0, 1$). Total number of divisors = $4 	imes 2 	imes 2 = 16$. Thus,the correct option is $A$.

The number of positive divisors of $360$ which are multiples of $3$ is

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