Prime Factors, Divisors and Factorials Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Prime Factors, Divisors and Factorials short notes for CAT: 30 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Why Prime Factorization Works

Numbers have a hidden prime recipe

Prime factorization turns a number into its exact ingredient list.

360

big number

→

2^{3} \cdot 3^{2} \cdot 5^{1}

prime profile

CAT mindset: Big numbers are not scary. Their prime exponents are small and manageable.

N = 2^{a} 3^{b} 5^{c}

, then

a, b, c

tell us exactly how much

2, 3, 5

are available inside

N

The Prime-Exponent Equality Rule

Core Rule

Equal numbers have equal prime exponents

2^{a} 3^{b} 5^{c} = 2^{p} 3^{q} 5^{r},

then

a = p, b = q, c = r .

Match

2

-powers separately.

Match

3

-powers separately.

If a prime appears on one side but not the other, its exponent on the missing side is

0

This is the foundation behind almost every PYQ in this topic.

The 4-Step Exponent Matching Method

Never expand. Match exponents.

Factorize bases: write

12 = 2^{2} \cdot 3

20 = 2^{2} \cdot 5

144 = 2^{4} \cdot 3^{2}

Distribute outer powers:

(2^{2} \cdot 3)^{x} = 2^{2 x} \cdot 3^{x}

Collect each prime: find total exponent of

2

, total exponent of

3

, and so on.

Match or compare: equality gives equations; divisibility gives inequalities.

This method is slower-looking on paper but faster in the exam because it avoids messy arithmetic.

Powers Multiply Exponents

Outer power means multiplier

(p^{a} q^{b})^{k} = p^{ak} q^{bk}

Base

144 = 2^{4} \cdot 3^{2}

Raise to

145

14 4^{145} = 2^{580} \cdot 3^{290}

Exam shortcut: First factorize the base, then multiply exponents. Do not compute the huge power.

Prime Factors, Divisors and Factorials: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Ability MCQ

Let $a, b, m, n$ be natural numbers such that $a > 1$ , $b > 1$ , $m > 1$ , and $n > 1$ . If

$a^{m} b^{n} = 2^{90} 3^{60},$

what is the largest possible value of $n - m$ ?

A.
88
B.
89
C.
90
D.
87

Correct Answer:

Step-by-Step Solution

Key idea: this is an extremal exponent-allocation question, recognisable because unknown bases and exponents multiply to a fixed prime-power product and we must maximise $n - m$ . The extra layer is the condition $m > 1$ .

Step 1: Read the available prime exponents.

The right side is already prime-factorized:

$2^{90} 3^{60} .$

So the largest exponent available is $90$ , and the smaller is $60$ .

Step 2: Maximise $n$ .

Since $b > 1$ , $b$ must contain at least one prime factor. The largest possible $n$ cannot exceed the largest available exponent, $90$ . To reach $n = 90$ , choose

$b = 2.$

Then $b^{n} = 2^{90}$ , using all the available powers of $2$ .

Step 3: Handle the leftover primes.

After taking $b^{n} = 2^{90}$ , the leftover part is

$3^{60} .$

We need

$a^{m} = 3^{60} .$

Because $m > 1$ , $m$ must be a divisor of $60$ greater than $1$ . The smallest such value is

$m = 2.$

Then

$a = 3^{30},$

which is valid because $a > 1$ .

Step 4: Compute $n - m$ .

$n - m = 90 - 2 = 88.$

Answer: 88.

Trap: $89$ comes from using $m = 1$ , but the question explicitly says $m > 1$ .

Question 2 · Quantitative Ability MCQ

Let $A = 2^{4} \cdot 3^{3}$ and $B = 2^{3} \cdot 3^{2}$ . Let $n$ be the least integer greater than $1$ such that $B$ divides $A^{n}$ . Once this $n$ is fixed, let $m$ be the least positive integer such that $A^{n}$ divides $B^{m}$ .

Find $m + n$ .

A.
3
B.
4
C.
5
D.
6

Correct Answer:

Step-by-Step Solution

Key idea: this is a mutual-divisibility least-power question, recognisable because one number is raised to a least power to become divisible by another, then the process is reversed. The added layer is that $n$ must be greater than $1$ .

Step 1: Write $A^{n}$ in prime powers.

$A^{n} = (2^{4} 3^{3})^{n} = 2^{4 n} 3^{3 n} .$

Step 2: Find the least integer $n > 1$ such that $B ∣ A^{n}$ .

Since

$B = 2^{3} 3^{2},$

we need

$4 n \geq 3 and 3 n \geq 2.$

Both are already true for $n = 1$ . But the question asks for the least integer greater than $1$ , so

$n = 2.$

Step 3: Fix $A^{n}$ using $n = 2$ .

$A^{2} = 2^{8} 3^{6} .$

Step 4: Find the least $m$ such that $A^{2} ∣ B^{m}$ .

$B^{m} = (2^{3} 3^{2})^{m} = 2^{3 m} 3^{2 m} .$

We need

$3 m \geq 8 \Rightarrow m \geq 3,$

and

$2 m \geq 6 \Rightarrow m \geq 3.$

Therefore

$m = 3.$

Step 5: Add.

$m + n = 3 + 2 = 5.$

Answer: 5.

Trap: if you ignore the condition $n > 1$ , you take $n = 1$ , get $m = 2$ , and wrongly obtain $3$ .

More short notes in this unit

chapter

Prime Factors, Divisors and Factorials Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Prime Factors, Divisors and Factorials short notes for CAT: 30 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questio

A question from this chapter

Question 1

Let a, b, m, n be natural numbers such that a > 1, b > 1, m > 1, and n > 1 . If a^{m} b^{n} = 2^{90} 3^{60}, what is the largest possible value of n - m ?

Question 2

Let A = 2^{4} \cdot 3^{3} and B = 2^{3} \cdot 3^{2} . Let n be the least integer greater than 1 such that B divides A^{n} . Once this n is fixed, let m be the least positive integer such that A^{n} divides B^{m} . Find m + n .

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