JEE Mains · Maths · STD 11 - 6. permutation and combination
Let \(\mathrm{n}\) be a non-negative integer. Then the number of divisors of the form " \(4 \mathrm{n}+1\) " of the number \((10)^{10} \cdot(11)^{11} \cdot(13)^{13}\) is equal to \(....\)
- A \(924\)
- B \(750\)
- C \(125\)
- D \(654\)
Answer & Solution
Correct Answer
(A) \(924\)
Step-by-step Solution
Detailed explanation
\(\mathrm{N}=2^{10} \times 5^{10} \times 11^{11} \times 13^{13}\) Now, power of \(2\) must be zero, power of \(5\) can be anything, power of \(13\) can be anything. But, power of \(11\) should be even. So, required number of divisors is \(1 \times 11 \times 14 \times 6=924\)
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