Number of Factors
Every factor of N = p^a × q^b × r^c is built by choosing a power of each prime independently: 0 to a copies of p, 0 to b copies of q, 0 to c copies of r. That gives (a+1) choices for p, (b+1) for q, (c+1) for r, so the total number of factors is d(N) = (a+1)(b+1)(c+1). The "+1" matters — it accounts for choosing zero copies (which keeps 1 and the prime out). CAT extends this with conditioned counts. For ONLY odd factors, drop the power of 2 entirely and multiply the remaining (exponent+1) terms. For ONLY even factors, every such factor must contain at least one 2, so subtract the odd-factor count from the total. For factors divisible by a given m (where m | N), divide N by m and count the factors of the quotient. The reflex: never list, always factorise then multiply.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Core divisor formulas
| Prime factorisation | N = p^a × q^b × r^c (p, q, r distinct primes) |
|---|---|
| Number of factors | d(N) = (a+1)(b+1)(c+1) |
| Sum of factors | σ(N) = Π (p^(a+1) − 1)/(p − 1) |
| Product of factors | P(N) = N^(d/2), where d = d(N) |
| Number of odd factors | product of (exponent+1) for odd primes only |
Counting power-tools
| Ways = product of two factors | d/2 if N is not a perfect square; (d+1)/2 if it is |
|---|---|
| Number of even factors | d(N) − d(N with all 2s removed) |
| Factors that are perfect squares | Π (⌊exponent/2⌋ + 1) |
| Factors divisible by m | d(N) counted after dividing N by m (if m | N) |
| Number of co-prime pairs / sum of reciprocals | Σ(1/factor) = σ(N)/N |