A classic problem: how many positive divisors does 720 have? A student who has not seen the trick starts listing: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, \ldots — and gets tangled, misses some, and double-counts others. A student who has internalised the trick answers in one line. The trick is to recognise that the question is not really about divisors — it is about choices of exponents.

Once you see that, "number of divisors" problems, "number of common divisors" problems, "smallest number with exactly N divisors" problems, and many Olympiad questions collapse into short exponent-counting arguments.

The core identity

Let n have the prime factorisation

n = p_1^{a_1} \cdot p_2^{a_2} \cdots p_r^{a_r}.

Then the number of positive divisors of n, written \tau(n) or d(n), is

\tau(n) = (a_1 + 1)(a_2 + 1) \cdots (a_r + 1).

Why: a positive divisor of n has the form p_1^{b_1} p_2^{b_2} \cdots p_r^{b_r} where 0 \leq b_i \leq a_i. For each prime p_i you have a_i + 1 choices for b_i (the values 0, 1, 2, \ldots, a_i). The choices are independent, so by the multiplication principle, the total number of divisors is the product \prod (a_i + 1).

That one line of reasoning — "a divisor is a choice of exponent at each prime" — is what reframes the whole question.

The standard example

720 = 2^4 \cdot 3^2 \cdot 5. So

\tau(720) = (4+1)(2+1)(1+1) = 5 \cdot 3 \cdot 2 = 30.

720 has exactly 30 positive divisors. You did not list a single one; you only looked at the exponents.

Counting divisors as picking exponents, one prime at a timeA slider scans four examples. Example 0: n equals 12 = 2 squared times 3, exponent choices 3 times 2 equals 6 divisors. Example 1: n equals 100 = 2 squared times 5 squared, 3 times 3 = 9 divisors. Example 2: n equals 360 = 2 cubed times 3 squared times 5, 4 times 3 times 2 = 24 divisors. Example 3: n equals 720 = 2 to the fourth times 3 squared times 5, 5 times 3 times 2 = 30 divisors.
Four numbers, one formula. At each step the pattern is the same: a divisor is a choice of exponent between $0$ and $a_i$ for each prime $p_i$. Multiply the counts together.

The recognition rule

You see any of these phrasings in a problem:

Your response should be automatic: factor everything into primes and turn the question into an exponent-counting question. That is the entire thinking-process skill. You will almost never need anything else.

Four variants you should recognise

Variant 1: "Divisors greater than k"

How many positive divisors of 120 are greater than 12?

120 = 2^3 \cdot 3 \cdot 5, so \tau(120) = 4 \cdot 2 \cdot 2 = 16. The divisors of 120 come in pairs (d, 120/d) whose product is 120. If d < \sqrt{120} \approx 10.95, then 120/d > 10.95. So the number of divisors > \sqrt{120} equals the number < \sqrt{120}.

Divisors \leq 10: 1, 2, 3, 4, 5, 6, 8, 10 — that's 8. So divisors \geq 12 = 16 - 8 = 8.

The trick of pairing divisors (d, n/d) is a direct consequence of the exponent-picking view: picking (b_1, b_2, b_3) = (\text{choices}) gives d, and picking (a_i - b_i) gives n/d.

Variant 2: "Numbers with exactly N divisors"

Find the smallest positive integer with exactly 12 divisors.

You want \tau(n) = (a_1 + 1)(a_2 + 1) \cdots = 12. Factor 12 as a product of integers \geq 2:

The smallest is 60. Answer: 60. Notice the strategy: to minimise n, assign the largest exponents to the smallest primes.

Variant 3: "Common divisors of a and b"

How many positive common divisors do 288 and 432 share?

The common divisors of a and b are exactly the divisors of \gcd(a, b). So this question becomes: "how many divisors does \gcd(288, 432) have?"

288 = 2^5 \cdot 3^2, 432 = 2^4 \cdot 3^3. So \gcd = 2^4 \cdot 3^2 = 144. Number of divisors: (4+1)(2+1) = 15.

Two exponent-level steps: compute the minimum exponent at each prime (for the GCD), then apply the divisor-count formula. No listing.

Variant 4: "Count perfect-square divisors"

How many positive divisors of 720 are perfect squares?

720 = 2^4 \cdot 3^2 \cdot 5. A divisor 2^{b_1} \cdot 3^{b_2} \cdot 5^{b_3} is a perfect square iff every exponent is even. So b_1 \in \{0, 2, 4\} (3 choices), b_2 \in \{0, 2\} (2 choices), b_3 \in \{0\} (1 choice — 5's exponent in 720 is 1, and the only even value \leq 1 is 0). Total: 3 \cdot 2 \cdot 1 = 6 square divisors.

The constraint "even exponent" cuts each factor of (a_i + 1) down to \lfloor a_i / 2 \rfloor + 1. Once you notice that, the problem is a one-liner.

Why this specific instinct matters

The divisor function \tau(n) is the gateway to an entire area of number theory called multiplicative functions. Once you have \tau, the same exponent-picking argument gives you:

All of these share the same DNA — compute something at each prime, then multiply. If you train yourself to spot the pattern "question is about divisors / coprimality" → "translate to exponent counting," you have unlocked the whole family of problems.

Indian school and Olympiad context

This recognition appears constantly in:

None of these yield to listing divisors. All yield to choosing exponents.

A gotcha to watch for

The formula \tau(n) = \prod (a_i + 1) counts positive divisors. If a problem asks for "divisors including negatives," double it (each positive d pairs with -d). If a problem asks for "proper divisors" (excluding n itself), subtract 1. If "non-trivial" (excluding 1 and n), subtract 2 — except when n = 1, in which case be careful.

These small adjustments happen after the exponent count. The hard part is always the exponent count itself.

One-line takeaway

"How many divisors?" is a trick question. The real question is always "how many ways to choose an exponent at each prime?" Factor n, read off the exponents, add 1 to each, multiply. Any follow-up ("divisors greater than / perfect-square divisors / common divisors") is a variation on the same theme — a constraint on the choices.

Related: Number Theory Basics · Factor Into Primes First · Smallest n Divisible by — Think LCM · Divisibility Tree Explorer · Sieve of Eratosthenes Animation