Total number of divisors of 480 that are of t | vedclass

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(A) First,find the prime factorization of $480$: $480 = 48 	imes 10 = (16 	imes 3) 	imes (2 	imes 5) = 2^4 	imes 3^1 	imes 2^1 	imes 5^1 = 2^5

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) First,find the prime factorization of $480$: $480 = 48 	imes 10 = (16 	imes 3) 	imes (2 	imes 5) = 2^4 	imes 3^1 	imes 2^1 	imes 5^1 = 2^5 	imes 3^1 	imes 5^1$. Any divisor of $480$ is of the form $d = 2^a 	imes 3^b 	imes 5^c$,where $0 \leq a \leq 5$,$0 \leq b \leq 1$,and $0 \leq c \leq 1$. We want the divisor to be of the form $4n + 2$,which means $d$ must be divisible by $2$ but not by $4$. This implies $a$ must be exactly $1$ (so that $d = 2^1 	imes 3^b 	imes 5^c = 2(3^b 	imes 5^c)$). For $b$,we have $2$ choices ($0$ or $1$). For $c$,we have $2$ choices ($0$ or $1$). Total number of such divisors = $1 	imes 2 	imes 2 = 4$. The divisors are: $2^1 	imes 3^0 	imes 5^0 = 2$ $2^1 	imes 3^1 	imes 5^0 = 6$ $2^1 	imes 3^0 	imes 5^1 = 10$ $2^1 	imes 3^1 	imes 5^1 = 30$ All these are of the form $4n + 2$ $(2=4(0)+2, 6=4(1)+2, 10=4(2)+2, 30=4(7)+2)$.

Total number of divisors of $480$ that are of the form $4n + 2$,where $n \geq 0$,is equal to:

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