Product of Factors
The factors of N pair up neatly: if f divides N, so does N/f, and each such pair multiplies to N. There are d/2 such pairs (where d = d(N) is the number of factors), so the product of ALL factors is N^(d/2). For 36 = 2^2 × 3^2, d = 9 and the product of all factors is 36^(9/2) = 6^9. The half-power is not a problem even when d is odd: an odd d means N is a perfect square, its square root pairs with itself, and N^(d/2) is still a clean integer because √N is an integer. CAT often asks the reverse — "the product of the factors of N is N^3, how many factors does N have?" — which gives d/2 = 3 ⇒ d = 6. A close cousin counts the number of ways to write N as a product of two factors: d/2 if N is not a perfect square, and (d+1)/2 if it is (the square root being self-paired).
✅ 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 |