Prime Factors, Divisors and Factorials Previous Year Questions (PYQs) for CAT: 5+ Solved Questions with Step-by-Step Solutions

Solve 5+ Prime Factors, Divisors and Factorials previous year 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 NAT

The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is

Correct Answer:

Step-by-Step Solution

Key idea: The phrase "exactly two distinct factors other than 1 and itself" means the number has 1, itself, and exactly 2 more divisors — so the total number of divisors is $4$ . This is a divisor-count question in disguise: we must find how many numbers below $50$ satisfy $d (N) = 4$ .

Why this method applies: Any time a question talks about "factors other than 1 and itself", rewrite it as a statement about total divisor count. Here, $2$ other factors $+ 1 + N$ itself $= 4$ divisors total.

Step 1 — Which shapes give exactly 4 divisors?

Using $d (N) = (a + 1) (b + 1) \dots$ , the only ways to write $4$ as a product of factors bigger than $1$ are:

$4 = 4 \times 1 or 4 = 2 \times 2$

This means $N$ is either:

$N = p^{3}$ (one prime, exponent $3$ , since $3 + 1 = 4$ ), or
$N = p \times q$ (two distinct primes, each exponent $1$ , since $2 \times 2 = 4$ )

Step 2 — Count $N = p^{3} < 50$ :

$2^{3} = 8, 3^{3} = 27, 5^{3} = 125 (too big)$

So $2$ numbers: $8, 27$ .

Step 3 — Count $N = pq < 50$ ( $p < q$ , both prime):

With $p = 2$ : $2 \times 3 = 6, 2 \times 5 = 10, 2 \times 7 = 14, 2 \times 11 = 22, 2 \times 13 = 26, 2 \times 17 = 34, 2 \times 19 = 38, 2 \times 23 = 46$ (next, $2 \times 29 = 58$ , too big) → $8$ numbers.

With $p = 3$ : $3 \times 5 = 15, 3 \times 7 = 21, 3 \times 11 = 33, 3 \times 13 = 39$ (next, $3 \times 17 = 51$ , too big) → $4$ numbers.

With $p = 5$ : $5 \times 7 = 35$ (next, $5 \times 11 = 55$ , too big) → $1$ number.

Total for this shape: $8 + 4 + 1 = 13$ .

Step 4 — Add both shapes:

$13 + 2 = 15$

Common trap: Students often only search for $p \times q$ pairs and forget the $p^{3}$ case, which silently loses $2$ valid numbers ( $8$ and $27$ ).

Answer: 15

Question 2 · Quantitative Ability MCQ

Let n be the least positive integer such that 168 is a factor of $113 4^{n}$ . If m is the least positive integer such that $113 4^{n}$ is a factor of $16 8^{m}$ , then m + n equals

A.
9
B.
15
C.
12
D.
24

Correct Answer:

Step-by-Step Solution

This is a prime-factorization exponent-matching question with mutual divisibility — recognizable because it asks for the "least n" and "least m" linking two different numbers through divisibility, not equality.

Step 1: Write both numbers as prime powers.

$1134 = 2 \times 3^{4} \times 7, 168 = 2^{3} \times 3 \times 7$

Step 2: Find least $n$ such that $168 ∣ 113 4^{n}$ .

$113 4^{n} = 2^{n} \times 3^{4 n} \times 7^{n}$

For $168 = 2^{3} \times 3 \times 7$ to divide this, we need every prime's exponent in $168$ to be $\leq$ the corresponding exponent in $113 4^{n}$ :

$n \geq 3 (for 2^{3}), 4 n \geq 1 \Rightarrow n \geq 1, n \geq 1 (for 7)$

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

Step 3: Now compute $113 4^{3} = 2^{3} \times 3^{12} \times 7^{3}$ .

Step 4: Find least $m$ such that $113 4^{3} ∣ 16 8^{m}$ .

$16 8^{m} = 2^{3 m} \times 3^{m} \times 7^{m}$

We need:

$3 m \geq 3 \Rightarrow m \geq 1, m \geq 12 (for 3^{12}), m \geq 3 (for 7^{3})$

The binding condition is $m \geq 12$ , so least $m = 12$ .

Step 5: $m + n = 12 + 3 = 15$ .

Trap to avoid: the binding constraint always comes from the prime with the largest required exponent — here it's the prime 2 for $n$ (from $2^{3}$ in 168) and the prime 3 for $m$ (from $3^{12}$ in $113 4^{3}$ ), not the same prime both times.

Answer: m + n = 15 (Option B).

Question 3 · Quantitative Ability NAT

If $1 2^{12 x} \times 4^{24 x + 12} \times 5^{2 y} = 8 4^{z} \times 2 0^{12 x} \times 24 3^{3 z - 6}$ , where $x, y$ and $z$ are natural numbers, then $x + y + z$ equals

Correct Answer:

112

Step-by-Step Solution

This is a prime-exponent equality question — recognizable because both sides of the equation are products of powers of the same small set of primes (2, 3, 5, 7), with unknowns x, y, z sitting inside the exponents.

Step 1: Break every base into primes.

$12 = 2^{2} \times 3, 4 = 2^{2}, 8 = 2^{3}, 20 = 2^{2} \times 5, 243 = 3^{5}$

Step 2: Rewrite the left-hand side using these primes.

$1 2^{12 x} \times 4^{24 x + 12} \times 5^{2 y} = 2^{24 x} \cdot 3^{12 x} \times 2^{48 x + 24} \times 5^{2 y} = 2^{72 x + 24} \cdot 3^{12 x} \cdot 5^{2 y}$

Step 3: Rewrite the right-hand side using these primes.

$8^{4 z} \times 2 0^{12 x} \times 24 3^{3 x - 6} = 2^{12 z} \times 2^{24 x} \cdot 5^{12 x} \times 3^{15 x - 30} = 2^{24 x + 12 z} \cdot 3^{15 x - 30} \cdot 5^{12 x}$

Step 4: Since both sides represent the same number, and prime factorization is unique, the exponent of each prime must match separately.

For prime 3: $12 x = 15 x - 30 \Rightarrow 3 x = 30 \Rightarrow x = 10$

For prime 2: $72 x + 24 = 24 x + 12 z \Rightarrow 48 x + 24 = 12 z \Rightarrow z = 4 x + 2 = 4 (10) + 2 = 42$

For prime 5: $2 y = 12 x \Rightarrow y = 6 x = 60$

Step 5: Add up.

$x + y + z = 10 + 60 + 42 = 112$

Trap to avoid: solve for x first using whichever prime appears on only one side of the multiplication that doesn't depend on the others (here, prime 3 gives a clean one-variable equation) — trying to solve all three exponent equations simultaneously without this order leads to confusion.

Answer: x + y + z = 112.

Question 4 · Quantitative Ability MCQ

The number of divisors of $(2^{6} \times 3^{5} \times 5^{3} \times 7^{2})$ , which are of the form $(3 r + 1)$ , where r is a non-negative integer, is

A.
36
B.
56
C.
24
D.
42

Correct Answer:

Step-by-Step Solution

This is a divisor-counting-with-a-remainder-condition question — recognizable because it doesn't ask for the total number of divisors, but only those divisors that leave a specific remainder pattern (here, divisors of the form $3 r + 1$ , meaning divisors that are $\equiv 1 (mod 3)$ ).

Step 1: Write the number in prime-power form.

$N = 2^{6} \times 3^{5} \times 5^{3} \times 7^{2}$

Every divisor has the form $d = 2^{a} 3^{b} 5^{c} 7^{e}$ with $0 \leq a \leq 6$ , $0 \leq b \leq 5$ , $0 \leq c \leq 3$ , $0 \leq e \leq 2$ .

Step 2: Notice what happens if $b \geq 1$ . Then $3 ∣ d$ , so $d \equiv 0 (mod 3)$ — never $\equiv 1$ . So for $d$ to be of the form $3 r + 1$ , we must have $b = 0$ (no factor of 3 at all).

Step 3: With $b = 0$ , work out $d mod 3$ for the remaining primes.

$2 \equiv - 1 (mod 3) \Rightarrow 2^{a} \equiv (- 1)^{a}$

$5 \equiv - 1 (mod 3) \Rightarrow 5^{c} \equiv (- 1)^{c}$

$7 \equiv 1 (mod 3) \Rightarrow 7^{e} \equiv 1 always$

So $d \equiv (- 1)^{a + c} (mod 3)$ . We need $d \equiv 1$ , so $(- 1)^{a + c} = 1$ , which means $a + c$ must be even.

Step 4: Count valid $(a, c)$ pairs. $a \in {0, ..., 6}$ : 4 even values (0,2,4,6), 3 odd values (1,3,5). $c \in {0, ..., 3}$ : 2 even values (0,2), 2 odd values (1,3).

$a + c even = (both even) + (both odd) = (4 \times 2) + (3 \times 2) = 8 + 6 = 14$

Step 5: The exponent $e$ (for prime 7) doesn't affect the mod-3 condition at all — it can be any of its 3 values (0, 1, or 2) freely.

$Total valid divisors = 14 \times 3 = 42$

Trap to avoid: it's tempting to think "form $3 r + 1$ " only concerns divisibility by 3 loosely (like odd/even), but it's a strict modular condition — you must check every prime's contribution mod 3, and any factor of 3 itself instantly disqualifies the divisor.

Answer: 42 (Option D).

Question 5 · Quantitative Ability MCQ

For some natural number n, assume that $(15, 000)!$ is divisible by $(n!)!$ . The largest possible value of n is

A.
4
B.
7
C.
6
D.
5

Correct Answer:

Step-by-Step Solution

Key idea: This is a nested factorial divisibility question. The core tool is the fast rule: for positive integers $a, b$ ,

$b! ∣ a! ⟺ b \leq a .$

Why this applies: The question asks when $(n!)!$ divides $(15000)!$ . Here the "smaller factorial input" is $n!$ itself (not $n$ ), and the "larger factorial input" is $15000$ . So by the fast rule, we need:

$n! \leq 15000$

Step 1 — Trap check: Do NOT compare $n$ with $15000$ directly. The comparison must be between $n!$ and $15000$ , because it is $(n!)!$ dividing $(15000)!$ — the object playing the role of " $b$ " is $n!$ , not $n$ .

Step 2 — Find the largest $n$ with $n! \leq 15000$ :

$5! = 120, 6! = 720, 7! = 5040, 8! = 40320$

$7! = 5040 \leq 15000$ , but $8! = 40320 > 15000$ .

Step 3 — Conclusion:

The largest $n$ satisfying $n! \leq 15000$ is $n = 7$ .

Answer: 7

Prime Factors, Divisors and Factorials Previous Year Questions (PYQs) for CAT: 5+ 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

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