The number of divisors of 2100 is ______.

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Official answer: 36

Combinatorics

To find the number of positive divisors of $2100$ , find its prime factorization:

2100 = 21 \times 100 = (3 \times 7) \times (2^2 \times 5^2) = 2^2 \times 3^1 \times 5^2 \times 7^1

For a number $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ , the number of positive divisors is:

d(n) = (a_1 + 1)(a_2 + 1)(a_3 + 1)(a_4 + 1)

Substituting the exponents:

d(2100) = (2 + 1)(1 + 1)(2 + 1)(1 + 1) = 3 \times 2 \times 3 \times 2 = 36

Therefore, the number of divisors of $2100$ is 36.

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This Combinatorics previous year question is Q9 of the GATE CSE 2015 Set 2 paper. Find more GATE CSE PYQs on Combinatorics in Engineering Mathematics, or browse all GATE CSE PYQs.

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Combinatorics has 19 GATE CSE PYQs across 15 papers from 2012 to 2025. 5 of them are 1-mark NAT questions like this one, and it is the only one in GATE CSE 2015 Set 2.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.