Prime Factors, Divisors and Factorials Practice Questions for CAT: 69+ Solved Questions with Step-by-Step Solutions

Solve 69+ Prime Factors, Divisors and Factorials practice questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Prime Factors, Divisors and Factorials

Chapter Roadmap

Prime Factors, Divisors and Factorials

One chapter, one engine: break a number into primes, then compare exponents.

🧱

t1. Prime Factorization and Exponent Matching

Weightage: 4 of 8 chapter PYQs. This is the most important topic in the chapter.

Master: writing numbers as

2^{a} 3^{b} 5^{c} \dots

, comparing powers, finding least exponents, and solving equations like

a^{m} b^{n} = N

🔢

t2. Divisor Counting and Factor Properties

Weightage: 3 of 8 chapter PYQs. Important, but depends heavily on t1.

Master: how exponents create factors, special divisors, and factor-counting patterns.

🏗️

t3. Factorial Divisibility

Weightage: 1 of 8 chapter PYQs. Lower frequency, but conceptually powerful.

Master: how many times a prime appears inside

n!

, and how to test whether one factorial expression divides another.

By the end: you will stop seeing numbers as large objects. You will see them as prime-exponent profiles.

Topic Hero: Prime Factorization and Exponent Matching

Topic t1 • CAT Number System

Prime Factorization and Exponent Matching

The hidden language of divisibility, powers, and integer equations.

4 / 8

chapter PYQs are directly from this topic

Core skill

match prime exponents, not big numbers

One-line idea:

If two positive integers are equal, their prime-exponent profiles are equal.

Example:

72 = 2^{3} \cdot 3^{2}

means the profile is

(2 : 3, 3 : 2)

Prime Factors, Divisors and Factorials: Solved Questions with Step-by-Step Explanations (5 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 $3$ .

Question 3 · Quantitative Ability MCQ

Let $n$ be the least positive integer such that $10 8^{2}$ is a factor of $7 2^{n}$ . Let $m$ be the least positive integer such that $7 2^{n}$ is a factor of $10 8^{m}$ . Then $m + n$ is

A.
6
B.
7
C.
8
D.
9

Correct Answer:

Step-by-Step Solution

Key idea: this is a least-exponent mutual-divisibility question. The trigger is "least positive integer" together with one power being a factor of another power.

Step 1: Prime-factorize the bases.

$72 = 2^{3} 3^{2}$

$108 = 2^{2} 3^{3}$

Step 2: Find the least $n$ such that $10 8^{2}$ divides $7 2^{n}$ .

$10 8^{2} = (2^{2} 3^{3})^{2} = 2^{4} 3^{6}$

$7 2^{n} = (2^{3} 3^{2})^{n} = 2^{3 n} 3^{2 n}$

For divisibility, each exponent in $10 8^{2}$ must be at most the corresponding exponent in $7 2^{n}$ :

$3 n \geq 4 \Rightarrow n \geq 2$

$2 n \geq 6 \Rightarrow n \geq 3$

The binding condition is $n \geq 3$ , so the least $n = 3$ .

Step 3: Use this $n$ to find the least $m$ such that $7 2^{n}$ divides $10 8^{m}$ .

With $n = 3$ :

$7 2^{3} = 2^{9} 3^{6}$

Also:

$10 8^{m} = (2^{2} 3^{3})^{m} = 2^{2 m} 3^{3 m}$

For divisibility:

$2 m \geq 9 \Rightarrow m \geq 5$

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

The binding condition is $m \geq 5$ , so the least $m = 5$ .

Step 4: Add.

$m + n = 5 + 3 = 8$

Answer: 8

Question 4 · Quantitative Ability MCQ

Let $N = 2^{5} \cdot 3^{3} \cdot 5^{2}$ . The smallest positive integer $m$ such that $m N$ is a perfect cube and has exactly $160$ positive divisors is

A.
40
B.
80
C.
160
D.
320

Correct Answer:

Step-by-Step Solution

Key idea: this is a cube-completion question with an exact divisor-count constraint. The final exponents must be multiples of $3$ , and their “plus one” factors must multiply to $160$ .

Step 1: Understand the cube condition.

If $m N$ is a perfect cube, every final prime exponent must be a multiple of $3$ .

For $N = 2^{5} 3^{3} 5^{2}$ , the final exponents of $2, 3, 5$ must be at least:

$2 : at least 6,$

$3 : at least 3,$

$5 : at least 3.$

Step 2: Use the divisor-count condition.

If final exponents are $a, b, c$ , then:

$(a + 1) (b + 1) (c + 1) = 160.$

Since $a, b, c$ are multiples of $3$ , each factor $a + 1, b + 1, c + 1$ is congruent to $1 (mod 3)$ .

Also:

$a + 1 \geq 7, b + 1 \geq 4, c + 1 \geq 4.$

Step 3: Factor $160$ into allowed factors.

Divisors of $160$ that are congruent to $1 (mod 3)$ include:

$1, 4, 10, 16, 25, 40, 160.$

We need three factors, with the first at least $7$ and the other two at least $4$ .

The only possible product is:

$160 = 10 \cdot 4 \cdot 4.$

Step 4: Assign the factors.

The factor $10$ must belong to the prime $2$ , because only $2$ has a minimum factor of at least $7$ .

Thus:

$a + 1 = 10 \Rightarrow a = 9,$

$b + 1 = 4 \Rightarrow b = 3,$

$c + 1 = 4 \Rightarrow c = 3.$

Step 5: Find $m$ .

Starting from $N = 2^{5} 3^{3} 5^{2}$ , we need:

$m = 2^{9 - 5} \cdot 3^{3 - 3} \cdot 5^{3 - 2} = 2^{4} \cdot 5 = 80.$

Answer: $80$ .

Question 5 · Quantitative Ability MCQ

The number of ordered pairs $(a, b)$ of natural numbers, both greater than $1$ , such that

$a^{2} b^{3} = 2^{12} 3^{9}$

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

Correct Answer:

Step-by-Step Solution

Key idea: this is an exponent-allocation question. The RHS contains only primes $2$ and $3$ , so $a$ and $b$ can contain only $2$ and $3$ .

Step 1: Write unknown exponent profiles.

Let

$a = 2^{x} 3^{y}, b = 2^{u} 3^{v},$

where $x, y, u, v$ are non-negative integers.

Step 2: Match exponents after using $a^{2} b^{3}$ .

$a^{2} b^{3} = (2^{x} 3^{y})^{2} (2^{u} 3^{v})^{3} = 2^{2 x + 3 u} 3^{2 y + 3 v} .$

Therefore,

$2 x + 3 u = 12, 2 y + 3 v = 9.$

Step 3: Count non-negative solutions for the prime $2$ .

In $2 x + 3 u = 12$ , $3 u$ must be even, so $u$ must be even.

Possible $u$ values are $0, 2, 4$ :

$u = 0 \Rightarrow x = 6, u = 2 \Rightarrow x = 3, u = 4 \Rightarrow x = 0.$

So there are $3$ possibilities.

Step 4: Count non-negative solutions for the prime $3$ .

In $2 y + 3 v = 9$ , $3 v$ must be odd, so $v$ must be odd.

Possible $v$ values are $1, 3$ :

$v = 1 \Rightarrow y = 3, v = 3 \Rightarrow y = 0.$

So there are $2$ possibilities.

Step 5: Combine independently.

Raw exponent-profile combinations: $3 \times 2 = 6$ .

Step 6: Apply $a > 1$ .

The only forbidden case is $a = 1$ , i.e. $x = 0$ and $y = 0$ .

This happens when $u = 4$ and $v = 3$ , exactly one raw combination.

Here $b$ is still greater than $1$ , so no extra exclusion is needed.

Valid ordered pairs: $6 - 1 = 5$ .

Answer: Option A.

Prime Factors, Divisors and Factorials Practice Questions for CAT: 69+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Prime Factors, Divisors and Factorials

Prime Factors, Divisors and Factorials

Topic Hero: Prime Factorization and Exponent Matching

Prime Factorization and Exponent Matching

Prime Factors, Divisors and Factorials: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More practice questions in this unit

A question from this chapter

Login to view the complete practice questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course