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How many divisors does 360 have? Listing them takes time and it is easy to miss one. A prime factorisation answers it in one line, and the same idea counts even divisors, square divisors and factor pairs, and adds them up. This post explains why the method works, works through the question types JEE likes, and gives a routine to make it automatic.
Where does this come from in the syllabus?
The method rests on the Fundamental Theorem of Arithmetic: every composite number can be written as a product of primes in a way that is unique apart from order. It is Theorem 1.1 in NCERT's Class 10 chapter on real numbers (2026-27 reprint). Counting divisors then uses the multiplication principle, which the JEE (Main) 2026 syllabus lists as "the fundamental principle of counting" in its permutations and combinations unit. Check your exam's current syllabus.
Why does (a + 1)(b + 1)(c + 1) count the divisors?
Take 360 = 2³ × 3² × 5. Every divisor has the form 2ⁱ × 3ʲ × 5ᵏ, where i is 0, 1, 2 or 3 (4 choices), j is 0, 1 or 2 (3 choices), and k is 0 or 1 (2 choices). Because the factorisation is unique, different choices give different divisors. So 360 has 4 × 3 × 2 = 24 divisors.
The "+1" in each bracket is the choice of exponent 0, which is how a divisor leaves a prime out.
How do you add all the divisors?
Multiply out (1 + 2 + 4 + 8)(1 + 3 + 9)(1 + 5). Each term of the expansion picks one power of each prime, so each term is exactly one divisor, and every divisor appears once. The sum is 15 × 13 × 6 = 1,170.
In general, each bracket is a geometric series: 1 + p + p² + … + pᵃ = (pᵃ⁺¹ − 1)/(p − 1).
Worked examples: divisors of 360 with conditions
Even divisors. The power of 2 must be at least 1: 3 × 3 × 2 = 18.
Odd divisors. The power of 2 must be 0: 1 × 3 × 2 = 6. Their sum is (1 + 3 + 9)(1 + 5) = 13 × 6 = 78. Check by listing: 1 + 3 + 5 + 9 + 15 + 45 = 78.
Sum of even divisors. 1,170 − 78 = 1,092. Directly: (2 + 4 + 8) × 13 × 6 = 14 × 78 = 1,092.
Perfect-square divisors. Every exponent must be even: i is 0 or 2, j is 0 or 2, k is 0. That gives 2 × 2 × 1 = 4: 1, 4, 9 and 36.
Divisors of the form 4m + 2. These have exactly one factor of 2: 1 × 3 × 2 = 6. They are 2, 6, 10, 18, 30 and 90.
Ways to write 360 as a product of two factors. Divisors pair up as d and 360/d, and 360 is not a square, so 24/2 = 12 unordered pairs.
Coprime factor pairs. Each prime power must go wholly to one factor: 2 choices for each of 3 primes gives 8 ordered pairs, or 4 unordered: 1 × 360, 5 × 72, 8 × 45 and 9 × 40.
Worked example: the reverse question
What is the smallest number with exactly 12 divisors?
Write 12 as a product of the (exponent + 1) values: 12, 6 × 2, 4 × 3 or 3 × 2 × 2. The exponent sets are (11), (5, 1), (3, 2) and (2, 1, 1). Give the largest exponents to the smallest primes:
2¹¹ = 2,048
2⁵ × 3 = 96
2³ × 3² = 72
2² × 3 × 5 = 60
The smallest is 60. Check: 60 has 3 × 2 × 2 = 12 divisors.
Which mistakes cost the most marks?
Factorising incompletely. Writing 360 = 36 × 10 and multiplying their divisor counts (9 × 4 = 36) is wrong, because 36 and 10 share the prime 2.
Forgetting the "+1". 2³ allows 4 choices, not 3.
Halving the count for a perfect square. A square such as 36 has an odd number of divisors (9); it has (9 + 1)/2 = 5 unordered factor pairs, including 6 × 6.
Mixing up count and sum formulas under time pressure.
A practice routine
Understanding why the formula works makes it easier to bend it to new conditions. In two experiments with school students, Rittle-Johnson, Siegler and Alibali (2001) found that conceptual and procedural knowledge "develop iteratively": understanding predicted gains in skill, and skill gains predicted better understanding.
Say the exponent rule aloud before each problem: "power of 2 must be at least 1, others free".
Check one answer by listing every week, choosing a number under 100.
Practise the reverse question with 6, 8, 10 and 12 divisors.
Mix with other counting problems, such as those in our post on distributing objects into groups.
If you want a worked solution that shows each choice of exponent, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Try it now: count the divisors of 720, the odd ones among them, and the sum of the odd ones, then check the last by listing.
