GATE CS 2015 Set 2 Q.19 (CS Q.9): Combinatorics, Counting, Recurrence Relations and Generating Functions

The number of divisors of 2100 is _____.

Your answer

Official answer and solution

Solution

Find the prime factorization of 2100 and use the divisor formula $d(n) = \prod (a_i + 1)$ .

Prime factorize 2100:

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

Any positive divisor $d$ of 2100 is of the form $2^a \cdot 3^b \cdot 5^c \cdot 7^d$ , where:

a \in \{0, 1, 2\},\quad b \in \{0, 1\},\quad c \in \{0, 1, 2\},\quad d \in \{0, 1\}.

The total number of divisors is the product of the possibilities for each exponent:

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

Final answer

Common trap

Omitting 1 or 2100 from the count, or forgetting that the exponent 0 corresponds to a valid power.

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