Q60 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics

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 $\mathrm{n}_1+\mathrm{n}_2=$

  • Correct answer
    A
    $75$
  • B
    $57$
  • C
    $51$
  • D
    $33$

Solution

$12600=2^3 \times 3^2 \times 5^2 \times 7$ 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 $$ \begin{aligned} & =3 \times(2+1) \times(2+1) \times 1=3 \times 3 \times 3 \\ & \therefore \quad n_2=27 \\ & \text { Total }=n_1+n_2=48+27=75 . \end{aligned} $$

Hence, the correct answer is option: 'A'.

Relevant Questions