Number of Factors and Their Sum

Once you have a number's prime form, you can instantly count its factors, sum them, and count how many ways it splits into two factors — all popular placement questions.

Formulas (for N = p^a × q^b × r^c …)

Quantity	Formula
Number of factors	(a+1)(b+1)(c+1)…
Sum of factors	Π (p^(a+1) − 1)/(p − 1)
Product of factors	N^(number of factors / 2)
Ways as product of two	(number of factors)/2, rounded up if a perfect square
Number of odd factors	ignore the power of 2, multiply the rest

Example: 72 = 2³ × 3². Factors = 4 × 3 = 12. Sum = (2⁴−1)/(2−1) × (3³−1)/(3−1) = 15 × 13 = 195. ✓ (1+2+3+4+6+8+9+12+18+24+36+72 = 195.)
Ways as product of two = 12/2 = 6. Odd factors = factors of 3² = 3.

Exam Tricks & Tips

• 🎯 Add 1 to each exponent and multiply — that is the total factor count.
• 🎯 Odd factors: drop the 2-power entirely, then (b+1)(c+1)… for the odd primes.
• 🎯 Even factors = total factors − odd factors.
• 🎯 Ways to write N as a product of two factors = half the factor count (round up for perfect squares, which have an odd factor count).
• 🎯 A number is a perfect square exactly when its factor count is odd.
• ❌ Don't forget to add 1 to each exponent — using a·b·c instead of (a+1)(b+1)(c+1) is the classic error.

Expected exam pattern: number of factors, sum of factors, number of ways to express as a product of two, or count of even/odd factors. 1 to 2 questions, ~1 minute.

Quick recap: Prime-factorise; factor count = product of (exponent+1); sum uses the geometric-series formula per prime; ways-as-product = half the factor count.

Q1. Number of factors of N = p^a × q^b?
A1. (a+1)(b+1).

Q2. How many factors does 72 have?
A2. 72 = 2³×3², so 4×3 = 12.

Q3. Sum of factors of 72?
A3. 15 × 13 = 195.

Q4. Product of all factors of N?
A4. N^(number of factors / 2).

Q5. Ways to write 72 as a product of two factors?
A5. 12/2 = 6.

Q6. Number of odd factors of 72?
A6. Ignore 2³; factors of 3² = 3.

Q7. How can you tell a number is a perfect square from its factors?
A7. Its number of factors is odd.

Q8. Even factors of 72?
A8. Total 12 − odd 3 = 9.

Q9. Sum-of-factors formula per prime?
A9. (p^(a+1) − 1)/(p − 1).

Q10. Number of factors of 360 = 2³×3²×5?
A10. 4×3×2 = 24.

Q11. Common mistake in factor counting?
A11. Multiplying the exponents instead of (exponent + 1) terms.

Q12. Ways to write a perfect square N as a product of two factors?
A12. (factor count + 1)/2, since the count is odd (the square root pairs with itself).

Once a number is in prime-factorised form, its entire factor structure — how many factors, their sum, their product, how many are even/odd — falls out of small formulas. This is one of the highest-yield 20 minutes you can spend for placements.

What this topic tests: computing the count, sum and product of divisors from a prime factorisation, plus common variants (even factors, perfect squares, pairs).

The method

Beginner — the count of factors

Write N = p^a · q^b · r^c …. Any factor chooses each prime's power from 0 to its max, so:
Number of factors = (a+1)(b+1)(c+1)…
360 = 2³·3²·5¹ → (3+1)(2+1)(1+1) = 4·3·2 = 24 factors.

Intermediate — sum of factors

Sum of factors = [(p^(a+1) − 1)/(p − 1)] × [(q^(b+1) − 1)/(q − 1)] × …, one geometric-series bracket per prime.
For 360: 2 → (2⁴−1)/1 = 15; 3 → (3³−1)/2 = 13; 5 → (5²−1)/4 = 6. Sum = 15·13·6 = 1170.

Advanced — the useful variants

• Product of all factors = N^(d/2), where d is the number of factors (pair each factor f with N/f).
• Even factors: count factors that include at least one 2. For 360, even factors = total − odd factors. Odd factors ignore the 2s: (2+1)(1+1) from 3²·5 = 6 → even = 24 − 6 = 18.
• Perfect-square factors: each prime's power must be even, so choose from {0,2} for 2³ (2 options), {0,2} for 3² (2), {0} for 5¹ (1) → 4 perfect-square factors.
• Number of ways to write N as a product of two factors: d/2 (if N is not a perfect square); if it is, (d+1)/2.

Worked example

How many factors of 720 are perfect squares? 720 = 2⁴·3²·5. Even exponents available: for 2⁴ → {0,2,4} (3 choices), 3² → {0,2} (2), 5¹ → {0} (1). Total = 3·2·1 = 6.

Where it appears

TCS NQT advanced quant, Amazon, Microsoft, Goldman Sachs, and inside HCF/LCM and probability questions ("pick a factor at random").

Speed tricks and shortcuts

• Factor count is the workhorse — memorise (a+1)(b+1)… cold.
• Odd factors = drop all the 2s, then apply the count formula.
• A number has an odd number of factors iff it is a perfect square (the middle factor pairs with itself).
• Mnemonic: "Add one to each power and multiply."

Key formulas

• If N = p^a · q^b · r^c, number of factors = (a+1)(b+1)(c+1).
• Sum of factors = (p^{a+1}−1)/(p−1) · (q^{b+1}−1)/(q−1) · …
• Product of all factors = N^{(number of factors)/2}.
• Number of ways to write N as a product of two factors = (number of factors)/2 (or (d+1)/2 if perfect square).
• Number of even/odd factors: separate the power of 2.
• Perfect square ⇒ odd number of factors.

Usage: prime-factorise N first, then apply the (a+1)(b+1)… count.

Worked Example

Problem: Find (a) the number of factors of 360 and (b) the sum of all its factors.

Solution: First prime-factorise: 360 = 2³ × 3² × 5¹.

(a) Number of factors = product of (each exponent + 1):
(3 + 1)(2 + 1)(1 + 1) = 4 × 3 × 2 = 24 factors.

(b) Sum of factors = product of each prime's geometric series (p⁰ + p¹ + … + p^a):

• For 2³: 1 + 2 + 4 + 8 = 15
• For 3²: 1 + 3 + 9 = 13
• For 5¹: 1 + 5 = 6

Sum = 15 × 13 × 6 = 1170.

Using the closed form (p^(a+1) − 1)/(p − 1): (2⁴−1)/1 × (3³−1)/2 × (5²−1)/4 = 15 × 13 × 6 = 1170, matching.

Answer: 24 factors; their sum is 1170.