Among the positive divisors of the number 126 | vedclass

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Question

(A) The prime factorization of $12600$ is $12600 = 2^3 	imes 3^2 	imes 5^2 	imes 7^1$. For $n_1$ (divisors that are multiples of $3$): $A$ divisor

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) The prime factorization of $12600$ is $12600 = 2^3 	imes 3^2 	imes 5^2 	imes 7^1$. For $n_1$ (divisors that are multiples of $3$): $A$ divisor is a multiple of $3$ if it contains at least one factor of $3$. The number of choices for the exponent of $2$ is $(3+1) = 4$. The number of choices for the exponent of $3$ is $2$ (must be $1$ or $2$). The number of choices for the exponent of $5$ is $(2+1) = 3$. The number of choices for the exponent of $7$ is $(1+1) = 2$. Thus,$n_1 = 4 	imes 2 	imes 3 	imes 2 = 48$. For $n_2$ (divisors that are multiples of $14$): $A$ divisor is a multiple of $14 = 2^1 	imes 7^1$ if it contains at least one factor of $2$ and at least one factor of $7$. The number of choices for the exponent of $2$ is $3$ (must be $1, 2,$ or $3$). The number of choices for the exponent of $3$ is $(2+1) = 3$. The number of choices for the exponent of $5$ is $(2+1) = 3$. The number of choices for the exponent of $7$ is $1$ (must be $1$). Thus,$n_2 = 3 	imes 3 	imes 3 	imes 1 = 27$. Therefore,$n_1 + n_2 = 48 + 27 = 75$.

Among the positive divisors of the number $12600$,if $n_1$ is the number of divisors which are multiples of $3$ and $n_2$ is the number of divisors which are multiples of $14$,then $n_1 + n_2 =$

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