Factors vs Multiples
A Factor divides a number $N$ completely without remainder ($F \mid N$), while a Multiple is any product $k \cdot N$ where $k \in \mathbb{Z}^+$.
Prime Building Block Model
Every integer $N > 1$ is uniquely built from prime numbers raised to positive integer powers:
- • $360 = 2^3 \times 3^2 \times 5^1$
- • Power choices for 2: $\{2^0, 2^1, 2^2, 2^3\}$ (4 options)
- • Power choices for 3: $\{3^0, 3^1, 3^2\}$ (3 options), 5: $\{5^0, 5^1\}$ (2 options)
- • Total factors = $4 \times 3 \times 2 = \mathbf{24}$!
Factor properties appear extensively in CAT algebra/number theory questions, integer solution equations ($1/x + 1/y = 1/N$), and MBA CET speed drills:
Total Factors $T(N)$
For $N = p_1^a \cdot p_2^b \cdot p_3^c$. Example: $72 = 2^3 \cdot 3^2 \implies (4)(3) = 12$.
Sum of Factors $S(N)$
Sum of all divisors of $N$. Example: $12 = 2^2 \cdot 3 \implies \frac{7}{1} \cdot \frac{8}{2} = 28$.
Product of Factors $P(N)$
Product of all divisors of $N$. Example: $12 \implies 12^{6/2} = 12^3 = 1728$.
Even vs Odd Factors Breakdown ($N = 2^a \cdot p_2^b \cdot p_3^c$)
Odd Factors
$\text{Odd} = (b+1)(c+1)$Ignore $2^a$ completely ($2^0$ only)
Even Factors
$\text{Even} = a \cdot (b+1)(c+1)$Use exponent $a$ (not $a+1$)
Co-Prime Factor Pairs
$\text{Pairs} = 2^{k-1}$Where $k$ is number of distinct prime factors
SHORTCUT RULES & SPECIAL FACTOR THEOREMS
Counting Perfect Square Divisors
Only even powers of primes ($p^0, p^2, p^4, \dots$) form perfect square factors!
Ways to Express $N = x \times y$
Perfect squares have an odd total number of factors, creating a self-pair $(\sqrt{N} \times \sqrt{N})$.
SOLVED EXAMPLES (LEVEL 0 TO HARD)
Q: Find the total number of factors, even factors, and odd factors of 360.
Step 1: Prime factorization of $360 = 2^3 \times 3^2 \times 5^1$.
Step 2: Total factors $T(360) = (3+1)(2+1)(1+1) = 4 \times 3 \times 2 = 24$.
Step 3: Odd factors = $(2+1)(1+1) = 3 \times 2 = 6$.
Step 4: Even factors = $3 \times (2+1)(1+1) = 3 \times 3 \times 2 = 18$ (Check: $18 + 6 = 24$).
Q: Find the sum of all factors of 240.
Step 1: Prime factorization of $240 = 2^4 \times 3^1 \times 5^1$.
Step 2: $S(240) = \left(\frac{2^5 - 1}{2 - 1}\right) \times \left(\frac{3^2 - 1}{3 - 1}\right) \times \left(\frac{5^2 - 1}{5 - 1}\right)$.
Step 3: $S(240) = (31) \times (8 / 2) \times (24 / 4) = 31 \times 4 \times 6 = 31 \times 24 = 744$.
Q: How many positive integer pairs $(x, y)$ satisfy the equation $\frac{1}{x} + \frac{1}{y} = \frac{1}{12}$?
Step 1: Rearrange equation: $xy - 12x - 12y = 0 \implies (x - 12)(y - 12) = 12^2 = 144$.
Step 2: Let $X = x - 12, Y = y - 12$. Since $x, y > 0$, positive solution pairs $(X, Y)$ equal the number of factors of $144$.
Step 3: Prime factorize $144 = 2^4 \times 3^2 \implies T(144) = (4+1)(2+1) = 5 \times 3 = 15$.
Mistake 1
Incomplete Prime Factorization
Using composite bases like $4^2 \times 3^2$ instead of prime bases $2^4 \times 3^2$, leading to wrong factor counts.
Mistake 2
Wrong Exponent for Even Factors
Adding $+1$ to exponent of 2 when calculating even factors. Use $a \cdot (b+1)(c+1)$, not $(a+1)$.
Mistake 3
Forgetting Negative Factor Pairs
When asked for total integral solutions (not just positive), remember to double the factor count for negative pairs!
1. How many factors does 180 have?
Answer: $180 = 2^2 \times 3^2 \times 5^1 \implies T(180) = (3)(3)(2) = 18$ factors.
2. In how many ways can 100 be expressed as a product of two factors?
Answer: $100 = 2^2 \times 5^2 \implies T(100) = 9$. Since 100 is a square, ways = $(9+1)/2 = 5$.
3. Find the number of co-prime factor pairs of 2310.
Answer: $2310 = 2 \times 3 \times 5 \times 7 \times 11 \implies k = 5$ distinct primes. Pairs = $2^{5-1} = 2^4 = 16$.
FREQUENTLY ASKED QUESTIONS
❓ How do you calculate the total number of factors of a composite number N?
First write N in prime factorized form: N = p1^a * p2^b * p3^c. The total number of factors T(N) = (a + 1)(b + 1)(c + 1).
❓ What is the formula for the sum of all factors of N?
For N = p1^a * p2^b * p3^c, Sum of factors S(N) = [(p1^(a+1) - 1)/(p1 - 1)] * [(p2^(b+1) - 1)/(p2 - 1)] * [(p3^(c+1) - 1)/(p3 - 1)].
❓ How do you find the number of even factors and odd factors?
For N = 2^a * p2^b * p3^c: Odd factors = (b+1)(c+1) [ignoring the power of 2]. Even factors = a * (b+1)(c+1) [taking a instead of a+1].