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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 $\mathrm{n}_1+\mathrm{n}_2=$
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Verified Answer
The correct answer is:
$75$
$12600=2^3 \times 3^2 \times 5^2 \times 7$
Number of divisors which are multiple of 3
$$
\begin{aligned}
& =(3+1) \cdot 2 \cdot(2+1)(1+1)=4 \times 2 \times 3 \times 2 \\
& \therefore \quad n_1=48
\end{aligned}
$$
Number of divisors which are multiple of 14
$$
\begin{aligned}
& =3 \times(2+1) \times(2+1) \times 1=3 \times 3 \times 3 \\
& \therefore \quad n_2=27 \\
& \text { Total }=n_1+n_2=48+27=75 .
\end{aligned}
$$
Number of divisors which are multiple of 3
$$
\begin{aligned}
& =(3+1) \cdot 2 \cdot(2+1)(1+1)=4 \times 2 \times 3 \times 2 \\
& \therefore \quad n_1=48
\end{aligned}
$$
Number of divisors which are multiple of 14
$$
\begin{aligned}
& =3 \times(2+1) \times(2+1) \times 1=3 \times 3 \times 3 \\
& \therefore \quad n_2=27 \\
& \text { Total }=n_1+n_2=48+27=75 .
\end{aligned}
$$
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