Factors concept is one of the most important number theory topics in quantitative aptitude, tested regularly across a wide range of competitive exams. It covers how to find the total number of factors of a number, sum of factors, product of factors, number of even and odd factors, and number of factors that are perfect squares — all using prime factorisation.
This topic appears frequently in SSC CGL, Bank PO, Railway RRB, SSC CHSL, and CAT, as well as in campus placement tests like AMCAT, eLitmus, and TCS NQT. It is also tested in international exams like GMAT and GRE Quantitative Reasoning.
This page covers all key formulas, properties, tricks, and solved examples for the factors concept that you need to solve any factors-based question in competitive exams quickly and accurately.
Factor Theory Concept ⇒
Consider a composite number n
⇒ n = aᵖbᑫcʳ………..
where a, b, c are prime numbers & p, q, r are natural numbers.Then
Number of factors of n is given by:-
Proof:- A factor of above number has power of a as a⁰ or a¹ or a² or a³ or ……….. aᵖ
similarly, A factor of above number has power of b as b⁰ or b¹ or b² or b³ or ……….. bᑫ
A factor of above number has power of c as c⁰ or c¹ or c² or c³ or ……….. cʳ
Hence factors having a⁰ or any value of a can be (p + 1)
similarly, factors having b⁰ or any value of b can be (q + 1)
factors having c⁰ or nay value of c can be (r + 1)
& when we take any combination of these then total number of factors = (p + 1)(q + 1)(r + 1)……..
where 1 & the number itself is included.
Example:- Find the total number of factors of 2⁴3²5³7⁵.
Solution:-
∴ Total numbers of factors = 5 × 3 × 4 × 6
= 360 Answer
Example:- Find the number of odd factors of 2⁴3²5³7⁵
Solution:-
∴ Total number of odd factors = 1×3×4×6
= 72 Answer
Example:- Find the number of even factors of 2⁴3²5³7⁵
Solution:-
∴ Total number of even factors = 4×3×4×6
= 288 Answer
Example:- Find the number of factors of 2⁴3²5³7⁵ that are divisible by 10.
Solution:-
∴ Total number of factors divisible by 10 = 4×3×3×6
= 216 Answer
Example:- Find the number of factors of 2⁴3²5³7⁵ that are not divisible by 10.
Solution:- required answer = Total factors – factor divisible by 10
=360 – 216
=144 Answer
Example:- Find the number of factors of 2⁴3²5³7⁵ that are divisible by 100.
Solution:-
∴ Total number of ways = 3×3×2×6
= 108 Answer
⟶ A number which divides a given nubmer exactly is called factor or divisor of that number and the given numbe is called multiple of that factor.
Ex:-
Factor of 30 = 1, 2, 3, 5, 6, 10, 15, 30
multiple of 4 = 4, 8, 12, 16, 20, 24, 28, …………
⟶ 1 is a facor of every number.
⟶ Every number is a factor of itself
⟶ Every number except 1 has at least 2 factors.
⟶ Every number has infinite number of multiples.
⟶ Every Prime number has only two factors 1 and the number itself.
⟶ Composite numbers have more than two factors.
⟶ The fundamental theorem of arithmetic states that every whole number greater than one can be uniquely expressed as a product of its Prime factors.
⟶ The number of factors a number has is finite
⟶ Factors are always less than or equal to the number.
(Q). Given a number
2⁵×3²×5³×7²×11
(i) Find the sum of all factors
(ii) Find the sum of even factors
(iii) Find the sum of odd factors.
Sol:-
(i) Sum of all factors ⟹
(2⁰ + 2¹ + 2² + 2³ +2⁴ +2⁵)(3⁰ + 3¹ + 3²)(5⁰ + 5¹ + 5² + 5³)(7⁰ + 7¹ + 7²)(11⁰ + 11¹)
=(63)(13)(156)(57)(12)
= 87390576 Answer
(Q). Find the number of prime factors of:
6⁶×8⁸×10¹⁰×12¹²×14¹⁴
Sol:- Given expression can be re-written as:
(2×3)⁶×(2)⁸×(2×5)¹×(2×3)¹²×(2×7)¹⁴
= 2⁷⁸×3¹⁸×5¹⁰×7¹⁴
Just add all powers as it is to get number of prime factors which in this case is:-
78 + 1 8+10 + 14
= 120 Answer
(Q). Given a number:
1480
(i) Find number of all factors
(ii) Find the number of even factors
(iii) Find number of odd factors
(iv) Find number of prime factors
(v) Sum of all factors
(vi) Sum of even factors
(vii) Sum of odd factors
Sol:- 1480=148×10 = 4×37×10 =2³×5×37
∴ number of all factors = 4×2×2
= 16 Answer
∴ number of even factors = 3×2×2 = 12 Answer
∴ number of odd factors 2×2 = 4 Answer
(iv) 2³×5¹×37¹
∴ Number of prime factors = 3 + 1 + 1 = 5 Answer
(v) (2⁰ + 2¹ + 2²+ 2³)(5⁰ + 5¹)(37⁰ + 37¹)
= (1 + 2 + 4 + 8)(1 + 5)(1 + 37 )
= 3420 ⟵ Sum of all factors Answer
(vi) Sum of even factors =
(2¹ + 2² + 2³)(5⁰ + 5¹)(37⁰+ 37¹)
= (14)(6)(38)
= 3192 Answer
(vii) Sum of odd factors
= (5⁰ + 5¹)(37⁰ + 37¹)
= 228 Answer
Proper Factor:- Proper factor of a number are all its factors excluding the number itself and 1.
Product of factors:-
we notice that the product of factors that are equidistant from both ends, when written in increasing order, is 180.
Here total number of factors = 18
∴ Sum of factors = \({180^{\frac{{18}}{2}}}\) = 180⁹ = 180×180×……..9 times Answer
Hence we can conclude that if a number N has x factors then sum of factors = \({N^{\frac{x}{2}}}\)
(Q). Find the product of factors of 1620.
Sol:- 1620 = 162 × 10 = 2 × 81 × 10 = 2×3⁴×2×5 = 2²×3⁴×5
∴ number of factors = (2 + 1)(4 + 1)(1 + 1)
= 30
∴ product of factors = \({1620^{\frac{{30}}{2}}}\) = 1620¹⁵ Answer
Number of Composite factors:-
Total number of composite factors = Total number of factors – number of distinct prime factors – 1
Ex:- If N = 12600, then find the numbe of composite factors of N.
Sol:-
∴ required number of composite factors = (3 + 1)(2 + 1)(2 + 1)(1 + 1) – 4 – 1
= 67 Answer
● Number of pairs of co-prime factors:-
Case (i): N = p₁ᵃ × p₂ᵇ
Number of co-prime factor pairs = (a + 1)(b + 1) + a.b
Case(ii): N = p₁ᵃ × p₂ᵇ × p₃ᶜ
Number of co-prime factor pairs = (a + 1)(b + 1)(c + 1) +a.b + b.c + c.a + a.b.c
Ex:- Find the number of pairs of co-prime factors of number 200.
Sol:- 200 = 2 × 10 × 10 = 2³ × 5²
∴ required number = (3 + 1)(2 + 1) + 3×2
= 18 Answer
Ex:- Find the number of pairs of co-prime factors of number 2800.
Sol:- 2800 = 28 × 100 = 2²×7×2²×5 = 2⁴×5²×7
∴ required number = (4 + 1)(2 + 1)(1 + 1) + 4×2 + 2×1 + 1×4 + 3×4×2×1
= 68 Answer
Number of perfect square factors:-
⟶ For each prime number Pᵢ with exponent aᵢ, count the number of exponent that are even starting from exponent 0, and less than or equal to aᵢ
⟶ Now multiply all these number which will give total number of perfect square factors.
Ex:- Find the perfect square factors of 8820.
Sol:-
Ex:- Find the perfect square factors of 2²×3²×5¹×7² .
Sol:-
Sum of perfect square:- In below table add each term of column corresponding to a prime number then multiply all these values which will give the sum of perfect square factors of the number
Ex:- Find the sum of perfect square actors of 8820.
Sol:- 8820 = 2²×3²×5¹×7²
∴ required sum = (2⁰ + 2²)(3⁰ + 3²)(5⁰)(7⁰ + 7²)
= (5)(10)(1)(50)
= 2500 Answer
Ex:- Find the sum of perfect square factors of 2⁶×3⁴×7⁷×11¹
Sol:-
required sum = (2⁰ + 2¹ + 2² + 2³)(3⁰ + 3¹ + 3² + 3³)(7⁰ + 7¹ + 7²)(11⁰)
= (1 + 4 + 16 + 64)(1 + 9 + 81 + 729)(1 + 49 + 2401)(10)
= (85)(820)(2451)
= 170834700 Answer
Number of perfect cube factors:-
N = P₁ᵃ¹.P₂ᵃ².P₃ᵃ³……….Pₖᵃᵏ
⟶ For each Prime number Pᵢ with exponent aᵢ, count the number of exponents which are multiple of 3, starting with 0, and less than or equal to aᵢ.
⟶ Now multiply all these numbers which will give number of perfect cube numbers.
Sum of perfect cube factors:- Add all the terms of each column in table 2 & multiply them which will give sum of perfect cube numbers.
Sol:-
16×3⁶×5¹⁰ = 2⁴×3⁶×5¹⁰
2⁴ ⟶ 2⁰, 2³ ⟹ 2
3⁶ ⟶ 3⁰, 3³, 3⁶ ⟹ 3
5¹⁰ ⟶ 5⁰, 5³, 5⁶, 5⁹ ⟹ 4
∴ required number = 2×3×4 = 24 Answer
Ex:- Find the sum of perfect cube factors of 2⁶.3³.5²
Sol:-
required sum = (2⁰ + 2³ + 2⁶)(3⁰ + 3³)(5⁰)
= (1 + 8 + 64)(1 + 27)(1)
= (73)(28)
= 2044 Answer
● Number of ‘perfect square & perfect cube both factors:-
N = P₁ᵃ¹.P₂ᵃ².P₃ᵃ³………….Pₖᵃᵏ
⟶ For each Prime number with exponent aᵢ, count the number of exponents which are multiple of both 2 and 3 at the same time i.e. multiple of 6, starting with 0 and less than or equal to aᵢ
⟶ Now multiply all these numbers which will give number of factors which are both square and cube at the same time.
Sum of factors which are both perfect square & perfect cube at the same time:-
Add all the terms in each column of above table & multiply them which will give the desired result.
Ex:- How many factors of 20¹⁵×16¹⁶×63³⁰ are perfect square as well as perfect cube ?
Sol:-
20¹⁵×16¹⁶×63³⁰
= 2³⁰×5¹⁵×2⁶⁴×7³⁰×3⁶⁰
= 2⁹⁴×3⁶⁰×5¹⁵×7³⁰
2⁹⁴ ⟶ 2⁰, 2⁶, 2¹², 2¹⁸, 2²⁴, 2³⁰, 2³⁶, 2⁴², 2⁴⁸, 2⁵⁴, 2⁶⁰, 2⁶⁶, 2⁷², 2⁷⁸, 2⁸⁴, 2⁹⁰ ⟹ 16
3⁶⁰ ⟶ 3⁰, 3⁶, 3¹², 3¹⁸, 3²⁴, 3³⁰, 3³⁶, 3⁴², 3⁴⁸, 3⁵⁴, 3⁶⁰ ⟹ 11
5¹⁵ ⟶ 5⁰, 5⁶, 5¹² ⟹ 3
7³⁰ ⟶ 7⁰, 7⁶, 7¹², 7¹⁸, 7²⁴, 7³⁰ ⟹ 6
∴ required number = 16×11×3×6 = 3168 Answer
Ex:- Find the sum of factors of 2¹³×3⁶×5⁵ which are perfect square as well as perfect cube also.
Sol:-
(2⁰ + 2⁶ + 2¹²)(3⁰ + 3⁶)(5⁰)
= (1 + 64 + 4096)(1 + 729)(1)
= 3037530 Answer