Number Theory and Pattern Puzzles Practice Questions for XAT: 51+ Solved Questions with Step-by-Step Solutions

Solve 51+ Number Theory and Pattern Puzzles practice questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Number Theory & Pattern Puzzles

Chapter Roadmap

Divisibility and Factors
The DNA of numbers. Prime factorization, counting factors, and sum of factors.

HCF and LCM
The intersection and union of factors. Translating word problems into mathematical logic.

Remainders and Cyclicity
Finding the last digits and remainders of massive powers using cyclic patterns.

Factorials and Trailing Zeros
Legendre's formula and counting the hidden tens in large products.

Pattern Puzzles
Decoding recursive sequences, cyclic operations, and logical constraints.

The Intuition: Integers as Lego Blocks

The Intuition

The Core Philosophy: Every integer greater than 1 is either a prime (a single, unbreakable block) or a composite (a structure built from prime blocks).

Why this matters for exams:

Sharing blocks: This is the Highest Common Factor (HCF).

Combining all blocks: This is the Least Common Multiple (LCM).

Leftover blocks: This is the Remainder.

Number Theory and Pattern Puzzles: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

For which of the following sets of three positive integers ${X, Y, Z}$ will the incorrect formula $X \times Y \times Z = HCF \times LCM$ accidentally yield the correct product?

A.
{2, 3, 4}
B.
{2, 3, 6}
C.
{3, 4, 1}
D.
{2, 4, 8}

Correct Answer:

Step-by-Step Solution

Key idea: This is a Cheat Sheet / Strategy question testing the boundary of the formula $HCF \times LCM = A \times B$ .

Step 1: Recall the rule.

The formula $HCF \times LCM = Product$ is ONLY valid for TWO numbers. For three numbers, it is generally false.

Step 2: Test the options to find the exception.

We need a set where $X \cdot Y \cdot Z = HCF (X, Y, Z) \times LCM (X, Y, Z)$ .

Option A: {2, 3, 4}

HCF = 1, LCM = 12. Product = 24. Formula = 12. (False)

Option B: {2, 3, 6}

HCF = 1, LCM = 6. Product = 36. Formula = 6. (False)

Option C: {3, 4, 1}

HCF = 1, LCM = 12. Product = 12. Formula = 1 * 12 = 12. (True!)

Option D: {2, 4, 8}

HCF = 2, LCM = 8. Product = 64. Formula = 16. (False)

Step 3: Understand why C works.

When one of the numbers is 1, the HCF of the set becomes 1.

The LCM of the set becomes the LCM of the other two numbers.

So the formula gives $1 \times LCM (X, Y)$ .

The actual product is $X \cdot Y \cdot 1 = X \cdot Y$ .

These are equal if and only if $X \cdot Y = LCM (X, Y)$ , which happens when $X$ and $Y$ are coprime.

In {3, 4, 1}, 3 and 4 are coprime.

Answer: {3, 4, 1}.

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Let $Z (N)$ be the number of trailing zeros of $N!$ when expressed in base 6. How many positive integers $N \leq 100$ satisfy the condition that $Z (N)$ is a perfect square?

A.
12
B.
14
C.
16
D.
18

Correct Answer:

Step-by-Step Solution

Key idea: This is a synthesis of Legendre's formula for trailing zeros in a non-decimal base and perfect square constraints. We must first determine the formula for $Z (N)$ in base 6, then count how many $N$ yield a perfect square result.

Step 1: Determine $Z (N)$ for base 6.

Base $6 = 2 \times 3$ . The number of trailing zeros is determined by the limiting prime factor in the prime factorization of $N!$ .

The exponent of a prime $p$ in $N!$ is $E_{p} (N!) = \sum_{k = 1}^{\infty} ⌊ \frac{N}{p ^{k}} ⌋$ .

For base 6, we need pairs of $(2, 3)$ . Since 3 is larger than 2, $E_{3} (N!) < E_{2} (N!)$ for all $N \geq 1$ . Thus, the number of trailing zeros is determined by the exponent of 3.

$Z (N) = E_{3} (N!) = ⌊ \frac{N}{3} ⌋ + ⌊ \frac{N}{9} ⌋ + ⌊ \frac{N}{27} ⌋ + \dots$

Step 2: Analyze the range $N \leq 100$ .

We need to find $N$ such that $Z (N) = k^{2}$ for some integer $k$ .

Let's estimate the maximum value of $Z (100)$ .

$Z (100) = ⌊ \frac{100}{3} ⌋ + ⌊ \frac{100}{9} ⌋ + ⌊ \frac{100}{27} ⌋ + ⌊ \frac{100}{81} ⌋ = 33 + 11 + 3 + 1 = 48$ .

So possible perfect squares for $Z (N)$ are $0, 1, 4, 9, 16, 25, 36, 49$ (but max is 48, so up to 36).

Step 3: Count $N$ for each perfect square value $S \in {0, 1, 4, 9, 16, 25, 36}$ .

Note that $Z (N)$ is a non-decreasing step function. It stays constant for intervals of 3, except at multiples of 9, 27, etc., where it jumps more.

$Z (N) = 0$ : $⌊ N /3 ⌋ = 0 ⟹ N \in {1, 2}$ . (2 values)
$Z (N) = 1$ : $⌊ N /3 ⌋ + \dots = 1$ .

$N = 3 ⟹ 1 + 0 = 1$ .

$N = 4 ⟹ 1 + 0 = 1$ .

$N = 5 ⟹ 1 + 0 = 1$ .

$N = 6 ⟹ 2 + 0 = 2$ .

So $N \in {3, 4, 5}$ . (3 values)

$Z (N) = 4$ :

Check $N$ around $3 \times 4 = 12$ .

$N = 9 ⟹ 3 + 1 = 4$ .

$N = 10 ⟹ 3 + 1 = 4$ .

$N = 11 ⟹ 3 + 1 = 4$ .

$N = 12 ⟹ 4 + 1 = 5$ .

So $N \in {9, 10, 11}$ . (3 values)

$Z (N) = 9$ :

Check $N$ around $3 \times 9 = 27$ ? No, approx $N \approx 3 \times 9 = 27$ is too high because of higher powers.

Let's test values.

$N = 12 \to 5$ .

$N = 15 \to 5 + 1 = 6$ .

$N = 18 \to 6 + 2 = 8$ .

$N = 19 \to 6 + 2 = 8$ .

$N = 20 \to 6 + 2 = 8$ .

$N = 21 \to 7 + 2 = 9$ .

$N = 22 \to 7 + 2 = 9$ .

$N = 23 \to 7 + 2 = 9$ .

$N = 24 \to 8 + 2 = 10$ .

So $N \in {21, 22, 23}$ . (3 values)

$Z (N) = 16$ :

$N = 30 \to 10 + 3 = 13$ .

$N = 33 \to 11 + 3 = 14$ .

$N = 36 \to 12 + 4 = 16$ .

$N = 37 \to 12 + 4 = 16$ .

$N = 38 \to 12 + 4 = 16$ .

$N = 39 \to 13 + 4 = 17$ .

So $N \in {36, 37, 38}$ . (3 values)

$Z (N) = 25$ :

$N = 45 \to 15 + 5 = 20$ .

$N = 48 \to 16 + 5 = 21$ .

$N = 51 \to 17 + 5 = 22$ .

$N = 54 \to 18 + 6 = 24$ .

$N = 55 \to 18 + 6 = 24$ .

$N = 56 \to 18 + 6 = 24$ .

$N = 57 \to 19 + 6 = 25$ .

$N = 58 \to 19 + 6 = 25$ .

$N = 59 \to 19 + 6 = 25$ .

$N = 60 \to 20 + 6 = 26$ .

So $N \in {57, 58, 59}$ . (3 values)

$Z (N) = 36$ :

$N = 72 \to 24 + 8 = 32$ .

$N = 75 \to 25 + 8 = 33$ .

$N = 78 \to 26 + 8 = 34$ .

$N = 81 \to 27 + 9 + 3 = 39$ (Jump!).

Wait, let's check near 81 carefully.

$N = 80 \to 26 + 8 + 2 = 36$ .

$N = 79 \to 26 + 8 + 2 = 36$ .

$N = 78 \to 26 + 8 + 2 = 36$ ? No. $⌊ 78/27 ⌋ = 2$ . $⌊ 78/9 ⌋ = 8$ . $⌊ 78/3 ⌋ = 26$ . Sum = 36.

$N = 77 \to 25 + 8 + 2 = 35$ .

So $N \in {78, 79, 80}$ . (3 values)

Total count = $2 + 3 + 3 + 3 + 3 + 3 + 3 = 20$ ?

Let me re-evaluate $Z (N) = 0$ . $N = 1, 2$ . Correct.

Let me re-evaluate ranges.

Usually, for large $N$ , $Z (N)$ increases by 1 every 3 numbers, but jumps by extra amounts at multiples of 9, 27, etc.

The number of solutions for $Z (N) = k$ is typically 3, unless $k$ is hit exactly at a jump point where it might skip or have fewer/more.

Actually, $Z (N)$ takes every integer value?

$Z (1) = 0, Z (2) = 0$ . (2 vals)

$Z (3) = 1, Z (4) = 1, Z (5) = 1$ . (3 vals)

$Z (6) = 2, Z (7) = 2, Z (8) = 2$ . (3 vals)

$Z (9) = 4$ . Skips 3.

So $Z (N) = 3$ has 0 solutions.

$Z (10) = 4, Z (11) = 4$ . (3 vals for 4? No, $Z (9) = 4$ too. So 9,10,11. 3 vals).

$Z (12) = 5, 13 = 5, 14 = 5$ . (3 vals).

$Z (15) = 6, 16 = 6, 17 = 6$ . (3 vals).

$Z (18) = 8$ . Skips 7.

$Z (19) = 8, 20 = 8$ . ( $Z (18) = 8$ too. So 18,19,20. 3 vals).

$Z (21) = 9, 22 = 9, 23 = 9$ . (3 vals).

Squares: 0, 1, 4, 9, 16, 25, 36.

Values skipped: 3, 7, ...

Are any squares skipped?

Sequence of Z(N): 0,0, 1,1,1, 2,2,2, 4,4,4, 5,5,5, 6,6,6, 8,8,8, 9,9,9...

Squares present:

0: Yes (2 values)

1: Yes (3 values)

4: Yes (3 values)

9: Yes (3 values)

16: Yes (3 values)

25: Yes (3 values)

36: Yes (3 values)

Next square 49. $Z (100) = 48$ . $Z (101) = 48 + 0 = 48$ ? No.

$Z (100) = 48$ .

$Z (101) = 48$ .

$Z (102) = 49$ ? $⌊ 102/3 ⌋ = 34, ⌊ 102/9 ⌋ = 11, ⌊ 102/27 ⌋ = 3, ⌊ 102/81 ⌋ = 1$ . Sum=49.

But $N \leq 100$ . So 49 is not reached.

Total = $2 + 3 + 3 + 3 + 3 + 3 + 3 = 20$ .

Wait, did I miss any?

Options are 12, 14, 16, 18. My count 20 is not an option.

Let me re-read carefully. "Positive integers $N \leq 100$ ".

Did I calculate $Z (N)$ correctly?

Base 6. Limiting factor 3.

$Z (N) = E_{3} (N!)$ .

Let's re-verify the "skipped" values.

Jumps occur at multiples of 9 (jump +1 extra), 27 (jump +1 extra), 81 (jump +1 extra).

Normal increase is +1 every 3 N.

At $N = 9$ , $Z (8) = 2, Z (9) = 4$ . Jumped 3.

At $N = 18$ , $Z (17) = 6, Z (18) = 8$ . Jumped 7.

At $N = 27$ , $Z (26) = ?$ .

$Z (26) = 8 + 2 + 0 = 10$ ? No.

$Z (26) = ⌊ 26/3 ⌋ + ⌊ 26/9 ⌋ = 8 + 2 = 10$ .

$Z (27) = 9 + 3 + 1 = 13$ . Jumped 11, 12.

Squares:

0: $N = 1, 2$ . (2)

1: $N = 3, 4, 5$ . (3)

4: $N = 9, 10, 11$ . (3)

9: $N = 21, 22, 23$ . (3)

16: $N = 36, 37, 38$ . (3)

25: $N = 57, 58, 59$ . (3)

36: $N = 78, 79, 80$ . (3)

Is it possible that some squares are skipped?

Let's check 16 again.

$Z (35) = 11 + 3 + 1 = 15$ .

$Z (36) = 12 + 4 + 1 = 17$ ?

$⌊ 36/3 ⌋ = 12$ . $⌊ 36/9 ⌋ = 4$ . $⌊ 36/27 ⌋ = 1$ . Sum=17.

Ah! $Z (36) = 17$ .

$Z (35) = 15$ .

So 16 is SKIPPED.

Let's check 25 again.

$Z (56) = 18 + 6 + 2 = 26$ ?

$⌊ 56/3 ⌋ = 18$ . $⌊ 56/9 ⌋ = 6$ . $⌊ 56/27 ⌋ = 2$ . Sum=26.

$Z (55) = 18 + 6 + 2 = 26$ ?

$⌊ 55/3 ⌋ = 18$ . $⌊ 55/9 ⌋ = 6$ . $⌊ 55/27 ⌋ = 2$ . Sum=26.

$Z (54) = 18 + 6 + 2 = 26$ ?

$⌊ 54/3 ⌋ = 18$ . $⌊ 54/9 ⌋ = 6$ . $⌊ 54/27 ⌋ = 2$ . Sum=26.

$Z (53) = 17 + 5 + 1 = 23$ .

So 24, 25 are SKIPPED.

Let's check 36 again.

$Z (77) = 25 + 8 + 2 = 35$ .

$Z (78) = 26 + 8 + 2 = 36$ .

$Z (79) = 26 + 8 + 2 = 36$ .

$Z (80) = 26 + 8 + 2 = 36$ .

$Z (81) = 27 + 9 + 3 + 1 = 40$ .

So 36 is HIT. (3 values: 78, 79, 80).

Let's check 4, 9, 1, 0 again.

0: Hit.

1: Hit.

4: Hit.

9: Hit.

16: Skipped.

25: Skipped.

36: Hit.

Total = $2 (f or 0) + 3 (f or 1) + 3 (f or 4) + 3 (f or 9) + 3 (f or 36) = 14$ .

Answer: 14

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

The sum of 4 distinct positive integers is 104. What is the maximum possible value of their Highest Common Factor (HCF)?

A.
10
B.
8
C.
13
D.
26

Correct Answer:

Step-by-Step Solution

Key idea: This is a Maximizing GCD with Distinctness question. We need to find the largest HCF that satisfies the sum constraint while keeping the integers distinct.

Step 1: Express the integers in terms of their HCF.

Let the HCF be $h$ . The 4 integers can be written as $h \cdot x_{1}, h \cdot x_{2}, h \cdot x_{3}, h \cdot x_{4}$ , where $x_{i}$ are distinct positive integers.

Step 2: Set up the sum equation.

The sum is $h (x_{1} + x_{2} + x_{3} + x_{4}) = 104$ .

This means $h$ must be a divisor of 104, and $h = 104/ S$ , where $S = x_{1} + x_{2} + x_{3} + x_{4}$ .

Step 3: Minimize the sum of the multipliers $S$ .

To maximize $h$ , we must minimize $S$ . Since the $x_{i}$ must be distinct positive integers, the smallest possible values are 1, 2, 3, and 4.

Minimum $S = 1 + 2 + 3 + 4 = 10$ .

Step 4: Find the maximum valid $h$ .

Since $S \geq 10$ , we have $h \leq 104/10 = 10.4$ .

We need the largest divisor of 104 that is less than or equal to 10.4.

The divisors of 104 are: 1, 2, 4, 8, 13, 26, 52, 104.

The largest divisor $\leq 10.4$ is 8.

Step 5: Verify if $h = 8$ is possible.

If $h = 8$ , then $S = 104/8 = 13$ .

We need 4 distinct positive integers that sum to 13. For example, 1, 2, 3, and 7 sum to 13.

Thus, the maximum HCF is 8.

Answer: 8

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

If the Highest Common Factor (HCF) of two distinct positive integers is 12, which of the following pairs CANNOT be the two integers?

A.
(12, 36)
B.
(24, 36)
C.
(12, 48)
D.
(24, 48)

Correct Answer:

Step-by-Step Solution

Key idea: This is an HCF Constraint question. We need to identify the pair whose HCF does not match the given value.

Step 1: Understand the constraint.

If the HCF of two numbers is 12, both numbers must be multiples of 12. Furthermore, when divided by 12, the resulting quotients must be coprime (their HCF must be 1).

Step 2: Evaluate each option by finding the HCF.

(12, 36): $12 = 12 \times 1$ , $36 = 12 \times 3$ . Quotients 1 and 3 are coprime. HCF is 12.
(24, 36): $24 = 12 \times 2$ , $36 = 12 \times 3$ . Quotients 2 and 3 are coprime. HCF is 12.
(12, 48): $12 = 12 \times 1$ , $48 = 12 \times 4$ . Quotients 1 and 4 are coprime. HCF is 12.
(24, 48): $24 = 12 \times 2$ , $48 = 12 \times 4$ . Quotients 2 and 4 share a factor of 2. The actual HCF is $12 \times 2 = 24$ .

Step 3: Conclude.

The pair (24, 48) has an HCF of 24, not 12. Therefore, it cannot be the pair.

Answer: (24, 48)

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Let $S$ be the set of all integers from 1 to 100. How many elements $x \in S$ satisfy the condition that $g cd (x, 100) = g cd (x + 1, 100)$ ?

A.
0
B.
10
C.
20
D.
40

Correct Answer:

Step-by-Step Solution

Key idea: This is a Coprime Constraints question that looks like a tedious counting problem but collapses via a simple logical contradiction.

Step 1: Analyze the GCD condition.

Let $d = g cd (x, 100)$ . We are given that $g cd (x + 1, 100) = d$ as well.

By definition of GCD, $d$ must divide both $x$ and $x + 1$ .

Step 2: Apply the properties of divisibility.

If $d$ divides $x$ and $d$ divides $x + 1$ , then $d$ must divide their difference:

$d ∣ ((x + 1) - x) ⟹ d ∣ 1$ .

The only positive integer that divides 1 is 1 itself.

Therefore, we must have $d = 1$ .

Step 3: Interpret the result.

The condition $g cd (x, 100) = g cd (x + 1, 100)$ can ONLY be satisfied if both GCDs are exactly 1.

This means we need $g cd (x, 100) = 1$ AND $g cd (x + 1, 100) = 1$ .

In other words, both $x$ and $x + 1$ must be coprime to 100.

Step 4: Check for parity constraints.

$100 = 2^{2} \times 5^{2}$ . For a number to be coprime to 100, it must not be divisible by 2 (i.e., it must be odd).

However, $x$ and $x + 1$ are consecutive integers. One of them MUST be even.

The even number will be divisible by 2, and since 2 divides 100, its GCD with 100 will be at least 2.

Therefore, it is impossible for BOTH $x$ and $x + 1$ to be coprime to 100.

Step 5: Conclusion.

There are no such integers $x$ . The count is 0.

Answer: 0.

Number Theory and Pattern Puzzles Practice Questions for XAT: 51+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Number Theory & Pattern Puzzles

The Intuition: Integers as Lego Blocks

Number Theory and Pattern Puzzles: 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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