TS EAMCET · Maths · Permutation Combination
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 \(\mathrm{n}_1+\mathrm{n}_2=\)
- A \(75\)
- B \(57\)
- C \(51\)
- D \(33\)
Answer & Solution
Correct Answer
(A) \(75\)
Step-by-step Solution
Detailed explanation
\(12600=2^3 \times 3^2 \times 5^2 \times 7\) 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…
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