Sum of Factors
The sum of all factors of N = p^a × q^b × r^c is a product, not a sum: σ(N) = [(p^(a+1) − 1)/(p − 1)] × [(q^(b+1) − 1)/(q − 1)] × [(r^(c+1) − 1)/(r − 1)]. Each bracket is just the sum of a geometric series of one prime’s powers (1 + p + p^2 + ... + p^a), and multiplying the brackets distributes into every possible factor exactly once. The cleanest CAT method is to write each bracket as that small GP sum and multiply: for 12 = 2^2 × 3, σ = (1 + 2 + 4)(1 + 3) = 7 × 4 = 28. The same product structure gives the SUM of odd factors (drop the 2-bracket), the sum of even factors (total minus odd), or the sum of reciprocals of factors, which equals σ(N)/N. This formula is also how aliquot-sum and perfect-number questions are framed.
✅ Solved examples
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📝 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 |