CUET UG Applied Mathematics Booster Test 2 - Number Theory and Modular Fundamentals -

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CUET UG Applied Mathematics Booster Test 2 - Number Theory and Modular Fundamentals

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Number Theory and Modular Fundamentals

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QUESTION 1 OF 20

The classification of integers into equivalence classes modulo m creates exactly how many distinct non-overlapping sets?

Answer Review

1 The classification of integers into equivalence classes modulo m creates exactly how many distinct non-overlapping sets?

A m - 1

B m

C m + 1

D Infinite

💡 Short Explanation

�� Modulo m creates remainder classes. �� Possible remainders are 0 to m − 1. �� Hence exactly m distinct classes exist.

✅ Correct Answer Explanation

In modulo arithmetic, integers are grouped according to their remainders when divided by m. The possible remainders are: Thus, there are exactly m equivalence classes. → Therefore, Option B is correct. → Option A is incorrect because one valid remainder class is omitted. → Option C incorrectly adds one extra class. → Option D is incorrect because the number of classes is finite for fixed m.

❓ Why Other Options Are Incorrect

�� Option A → Ignores one remainder class.
�� Option C → There cannot be more than m remainder classes modulo m.
�� Option D → Equivalence classes modulo m are finite in number.

🎯 Strategy

Used: Option Grouping

Application: Identify all possible remainders under modulo m.

Final Logic: Number of remainder classes equals number of possible remainders.

🧠 Memory Trick

"Modulo m → m remainder groups."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Equivalence Classes Modulo m

2 Match List 1 with List 2.

List 1	List 2
1. 1300 hours	a. 7
2. 1800 hours	b. 1
3. 0000 hours	c. 6
4. 0700 hours	d. 0

A 1-b, 2-c, 3-d, 4-a

B 1-a, 2-b, 3-c, 4-d

C 1-c, 2-d, 3-a, 4-b

D 1-d, 2-a, 3-b, 4-c

💡 Short Explanation

Convert each military time into its hour value. Divide the hour by 12 and take the remainder. Match each remainder with List II.

✅ Correct Answer Explanation

Convert each hour using modulo 12. 1300 hours → 13 mod 12 = 1 → 1 → b 1800 hours → 18 mod 12 = 6 → 2 → c 0000 hours → 0 mod 12 = 0 → 3 → d 0700 hours → 7 mod 12 = 7 → 4 → a Therefore, 1 → b 2 → c 3 → d 4 → a Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because 1300 hours is 1, not 7.
Option C → Incorrect because the modulo 12 values are incorrectly matched.
Option D → Incorrect because multiple military times are paired with wrong clock values.

🎯 Strategy

Used: Substitution

Application: Calculate each hour modulo 12 and directly substitute the result into List II.

Final Logic: Only Option A contains all four correct modulo 12 matches.

🧠 Memory Trick

Clock repeats every 12 hours.

📚 Class XI – Mathematics – Chapter: Number Systems – Topic: Modular Arithmetic

3 Which of the following conditions prove that the relation Rm, defined by
a ≡ b (mod m),
is an equivalence relation?
(1) Rm is irreflexive for all a ∈ Z.
(2) Rm is symmetric: if m divides (a - b), then m divides (b - a).
(3) Rm is transitive: if a ≡ b (mod m) and b ≡ c (mod m), then a ≡ c (mod m).
(4) Rm is reflexive: m divides (a - a) = 0.

A 1, 2

B 1, 2, 3

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

Every equivalence relation has three properties. These are reflexive, symmetric, and transitive. Irreflexive is not a property of equivalence relations.

✅ Correct Answer Explanation

An equivalence relation must satisfy all three conditions: Reflexive a - a = 0 Since every integer divides 0, m divides (a - a). Therefore, the relation is reflexive. Symmetric If m divides (a - b), then b - a = -(a - b). A divisor of (a - b) also divides its negative. Therefore, the relation is symmetric. Transitive If a ≡ b (mod m) and b ≡ c (mod m), then m divides (a - b) and m divides (b - c). Adding the two expressions gives m divides [(a - b) + (b - c)] = a - c. Hence, a ≡ c (mod m). Therefore, Statements 2, 3 and 4 are correct. Statement 1 is incorrect because an equivalence relation is reflexive, not irreflexive.

❓ Why Other Options Are Incorrect

Option A → Missing the transitive and reflexive properties.
Option B → Incorrect because it includes the false irreflexive condition.
Option D → Incorrect because Statement 1 is false.

🎯 Strategy

Used: Option Grouping

Application: Recall the three defining properties of every equivalence relation.

Final Logic: Reflexive + Symmetric + Transitive = Equivalence Relation.

🧠 Memory Trick

RST = Reflexive, Symmetric, Transitive.

📚 Class XII – Mathematics – Chapter: Relations and Functions – Topic: Equivalence Relations

4 Consider the equivalence class of 5 for modulo 7, denoted as. Which of the following positive integers does NOT belong to this class?

A 5

B 12

C 19

D 25

💡 Short Explanation

�� Numbers in the same class leave equal remainders modulo 7. �� Check each number modulo 7. �� Different remainder means different class.

✅ Correct Answer Explanation

Numbers equivalent to 5 modulo 7 satisfy: Check options: → Therefore, 25 does not belong to the equivalence class of 5 modulo 7. → Hence, Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Leaves remainder 5 modulo 7.
�� Option B → Congruent to 5 modulo 7.
�� Option C → Also belongs to the same equivalence class.

🎯 Strategy

Used: Elimination

Application: Compute modulo 7 for each number and remove matching remainders.

Final Logic: Only 25 leaves remainder 4 instead of 5.

🧠 Memory Trick

"Same remainder → same class."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Equivalence Classes

5 Evaluate the expression (16 × 29) mod 7 using the multiplicative property of modulo.

A 2

B 4

C 1

D 6

💡 Short Explanation

�� Reduce numbers modulo 7 first. �� Multiply reduced remainders. �� Take modulo again.

✅ Correct Answer Explanation

Compute: Then: → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect multiplication remainder.
�� Option C → Product modulo 7 is not 1.
�� Option D → Does not satisfy modulo multiplication.

🎯 Strategy

Used: Substitution

Application: Replace each factor with its modulo equivalent before multiplication.

Final Logic: Reduced product gives remainder 2 modulo 7.

🧠 Memory Trick

"Reduce first, multiply later."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Multiplication Property of Modulo Arithmetic

6 If a = 41, b = 21, and m = 5, what are the values of a mod m and b mod m, and does a ≡ b (mod m)?

A 1 and 2, No

B 1 and 1, Yes

C 2 and 2, Yes

D 0 and 1, No

💡 Short Explanation

�� Congruent numbers leave equal remainders. �� Divide both numbers by 5. �� Compare remainders obtained.

✅ Correct Answer Explanation

Compute: Since both leave the same remainder: → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → 21 mod 5 is not 2.
�� Option C → 41 and 21 do not leave remainder 2.
�� Option D → Neither remainder equals 0.

🎯 Strategy

Used: Substitution

Application: Directly calculate modulo values for both integers.

Final Logic: Equal remainders imply congruence modulo 5.

🧠 Memory Trick

"Same remainder ⇒ congruent."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Congruence Relation

7 Find the sum of 132 and 121 modulo 23.

A 0

B 1

C 2

D 23

💡 Short Explanation

�� Add the integers first. �� Apply modulo 23. �� Exact divisibility gives remainder 0.

✅ Correct Answer Explanation

Compute: Now divide by 23: because: → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect remainder calculation.
�� Option C → Product is exactly divisible by 23.
�� Option D → Modulo result must be less than 23.

🎯 Strategy

Used: Substitution

Application: Add numbers and directly evaluate modulo 23.

Final Logic: 253 is perfectly divisible by 23.

🧠 Memory Trick

"Exact division → remainder zero."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Addition Property of Modular Arithmetic

8 Find the remainder when the product (226 × 369 × 122 × 461 × 1025) is divided by 8.

A 2

B 4

C 6

D 0

💡 Short Explanation

�� Reduce each factor modulo 8. �� Presence of a multiple of 8 makes entire product divisible by 8. �� Final remainder becomes 0.

✅ Correct Answer Explanation

Compute modulo 8 values: Thus: and: → Therefore, the correct answer is:

❓ Why Other Options Are Incorrect

�� Option A → Product remainder is not 2.
�� Option C → Incorrect modulo computation.
�� Option D → Product is not exactly divisible by 8.

🎯 Strategy

Used: Substitution

Application: Reduce each factor modulo 8 before multiplying.

Final Logic: Simplified modular product leaves remainder 4.

🧠 Memory Trick

"Reduce factors before multiplying."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Multiplication Modulo Property

9 Find the positive integers less than 20 forming the equivalence class 4 for modulo 6.

A 4, 10, 16

B 4, 8, 12, 16

C 6, 12, 18

D 4, 10, 18

💡 Short Explanation

�� Numbers in the same equivalence class leave equal remainders. �� Check modulo 6 values. �� Keep numbers giving remainder 4.

✅ Correct Answer Explanation

Numbers in equivalence class 4 modulo 6 satisfy: Checking options: → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → 8 and 12 are not congruent to 4 modulo 6.
�� Option C → All numbers are multiples of 6, remainder 0.
�� Option D → 18 leaves remainder 0 modulo 6.

🎯 Strategy

Used: Elimination

Application: Compute modulo 6 values and remove numbers with incorrect remainders.

Final Logic: Only 4, 10, and 16 leave remainder 4 modulo 6.

🧠 Memory Trick

"Same remainder → same class."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Equivalence Classes Modulo m

10 If x (mod 9) = 2, calculate the sum of all possible values of x where 0 < x < 30.

A 62

B 60

C 32

D 33

💡 Short Explanation

�� Values satisfy x ≡ 2 (mod 9). �� List all numbers below 30. �� Add the obtained terms.

✅ Correct Answer Explanation

Since: possible values are: Their sum is: → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Does not equal the total obtained from valid values.
�� Option C → Incomplete summation.
�� Option D → Only partial values considered.

🎯 Strategy

Used: Substitution

Application: Generate all integers satisfying the congruence relation under the given limit.

Final Logic: Adding all valid numbers gives 62.

🧠 Memory Trick

"Add 9 repeatedly."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Congruence Classes

11 Verify that a ≡ b (mod m) if a = 31, b = 21, m = 5. What is the fundamental property establishing this congruence?

A a - b = 10, which is perfectly divisible by 5

B a + b = 52, which is not divisible by 5

C ab = 651, perfectly divisible by 5

D a / b leaves a remainder of 5

💡 Short Explanation

�� Congruence depends on divisibility of the difference. �� Compute a − b. �� Check divisibility by modulus 5.

✅ Correct Answer Explanation

Two integers are congruent modulo m if: Here: Since: 5 divides 10 exactly. Hence: → Therefore, Option A is correct. → Option B is unrelated because congruence does not depend on the sum. → Option C is irrelevant since product divisibility is not the defining condition. → Option D incorrectly interprets congruence through division remainder of a/b.

❓ Why Other Options Are Incorrect

�� Option B → Sum divisibility is not the criterion for congruence.
�� Option C → Product divisibility does not establish congruence modulo m.
�� Option D → Congruence is based on difference, not quotient remainder.

🎯 Strategy

Used: Substitution

Application: Substitute the values into the congruence definition and test divisibility.

Final Logic: Since 5 divides 10, the numbers are congruent modulo 5.

🧠 Memory Trick

"Same remainder ⇒ difference divisible."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Congruence Modulo m

12 If a = ms + r and b = mt + r, establishing that they leave the same remainder when divided by m, what can we deduce about the difference a - b?

A a - b = m(s + t)

B a - b = m(s - t)

C a + b = m(s - t)

D a - b = r(s - t)

💡 Short Explanation

�� Subtract the two equations. �� Common remainder cancels out. �� Remaining expression becomes a multiple of m.

✅ Correct Answer Explanation

Given: Subtracting: → Therefore, Option B is correct. → This proves that a − b is divisible by m, establishing congruence modulo m.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect sign appears after subtraction.
�� Option C → Uses addition instead of subtraction.
�� Option D → Remainder r is cancelled during subtraction.

🎯 Strategy

Used: Substitution

Application: Substitute the given algebraic forms and simplify systematically.

Final Logic: Cancellation of equal remainders produces m(s − t).

🧠 Memory Trick

"Same remainder → difference multiple of m."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Congruence Relation Proof

13 Find the addition modulo 25 if a and b are 15 and 6 respectively.

A 21

B 4

C 9

D 25

💡 Short Explanation

�� Add the integers first. �� Apply modulo 25 afterward. �� Number less than modulus remains unchanged.

✅ Correct Answer Explanation

Compute: Now: since 21 is already less than 25. → Therefore, Option A is correct. → Option B and Option C are incorrect modulo values. → Option D is the modulus itself, not the remainder.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect remainder after modulo operation.
�� Option C → Addition modulo 25 does not produce 9.
�� Option D → Modulo result must be less than 25.

🎯 Strategy

Used: Substitution

Application: Directly compute the sum and compare with modulus 25.

Final Logic: Since 21 < 25, modulo leaves the number unchanged.

🧠 Memory Trick

"Smaller than modulus stays same."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Addition Modulo m

14 Let a = 17, b = 13 and m = 30. Calculate a +_m b.

A 30

B 0

C 17

D 13

💡 Short Explanation

�� Add the integers. �� Apply modulo 30. �� Exact divisibility gives remainder zero.

✅ Correct Answer Explanation

Compute: Now apply modulo 30: → Therefore: → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → 30 is the dividend, not the modulo result.
�� Option C → Only one operand is shown.
�� Option D → Second operand alone cannot be the answer.

🎯 Strategy

Used: Substitution

Application: Perform addition and evaluate modulo directly.

Final Logic: Exact divisibility by 30 gives remainder 0.

🧠 Memory Trick

"Exact division ⇒ modulo zero."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Addition Under Modulo

15 Apply subtraction modulo to positive integers 11 and 3 for modulo 8.

A 0

B 8

C 14

D 3

💡 Short Explanation

�� Subtract the integers first. �� Apply modulo 8. �� Exact divisibility gives remainder 0.

✅ Correct Answer Explanation

Compute: Now apply modulo 8: → Therefore, Option A is correct. → Option B is the modulus, not the modulo result. → Option C is unrelated to subtraction modulo. → Option D is an incorrect remainder.

❓ Why Other Options Are Incorrect

�� Option B → Modulo answers are always less than modulus.
�� Option C → Incorrect arithmetic value.
�� Option D → 8 divided by 8 leaves remainder 0, not 3.

🎯 Strategy

Used: Substitution

Application: Directly subtract and reduce modulo 8.

Final Logic: Exact divisibility produces remainder zero.

🧠 Memory Trick

"Multiple of modulus ⇒ zero remainder."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Subtraction Modulo m

16 If the present time is 9:00 pm, what time will it be after exactly 1250 hours based on modulo 24 arithmetic?

A 11:00 pm

B 10:00 pm

C 9:00 am

D 11:00 am

💡 Short Explanation

�� Time cycles every 24 hours. �� Reduce 1250 modulo 24. �� Add remaining hours to 9:00 pm.

✅ Correct Answer Explanation

Compute: Since: we get: Adding 2 hours to 9:00 pm gives: → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Only 1 hour added instead of 2.
�� Option C → Incorrect AM/PM conversion.
�� Option D → 14-hour shift interpretation is incorrect.

🎯 Strategy

Used: Substitution

Application: Reduce large time duration modulo 24 before adding to current time.

Final Logic: Remaining shift is only 2 hours, leading to 11:00 pm.

🧠 Memory Trick

"Clock repeats every 24 hours."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Clock Arithmetic

17 Evaluate 7^6 (mod 3).

A 0

B 1

C 2

D 3

💡 Short Explanation

�� Reduce base modulo 3 first. �� Apply exponent afterward. �� Final remainder gives the answer.

✅ Correct Answer Explanation

Since: Therefore: → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → 7 is not divisible by 3.
�� Option C → Incorrect power remainder.
�� Option D → Modulo result must be less than 3.

🎯 Strategy

Used: Substitution

Application: Simplify the base using modulo before exponentiation.

Final Logic: Base becomes 1 modulo 3, so every power remains 1.

🧠 Memory Trick

"1 to any power stays 1."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Powers in Modular Arithmetic

18 What is the value of 5^3 mod 3, given 5 ≡ 2 (mod 3)?

A 1

B 2

C 0

D 4

💡 Short Explanation

�� Replace 5 with congruent remainder 2. �� Raise to power 3. �� Reduce modulo 3.

✅ Correct Answer Explanation

Given: Then: Now: → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect modulo reduction of 8.
�� Option C → 8 is not divisible by 3.
�� Option D → Modulo result must be less than 3.

🎯 Strategy

Used: Substitution

Application: Replace the base with its congruent equivalent modulo 3.

Final Logic: 8 leaves remainder 2 upon division by 3.

🧠 Memory Trick

"Replace first, power later."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Power Property of Congruence

📄 Passage

Modular arithmetic, also known as Clock Arithmetic, acts as a powerful tool as it works on the principle of cyclic repetitions of phenomena related to numbers. On a wrist watch, when reckoning hours, we count up to 12 and start again from one. This periodicity forms equivalence classes. If a and m are positive integers, the equivalence class of X w.r.t (mod m) is the set of all integers congruent to X modulo m, defined as [X] = {a ∈ Z | a ≡ X (mod m)}.

Question:
Which of the following numbers belongs to the equivalence class for modulo 3?

A 6

B 8

C 7

D 9

💡 Short Explanation

�� Numbers in the same equivalence class leave identical remainders. �� Check remainders modulo 3. �� Identify the matching value.

✅ Correct Answer Explanation

The equivalence class represented here corresponds to numbers congruent to 2 modulo 3. Check options: → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Leaves remainder 0 modulo 3.
�� Option C → Leaves remainder 1 modulo 3.
�� Option D → Also leaves remainder 0 modulo 3.

🎯 Strategy

Used: Elimination

Application: Compute modulo 3 remainder for each option.

Final Logic: Only 8 belongs to the required equivalence class.

🧠 Memory Trick

"Same remainder → same class."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Equivalence Classes

📄 Passage

Question:
Based on the equivalence class definition, if X = 0 and m = 3, which set correctly represents ?

A {..., -5, -2, 1, 4, 7, ...}

B {..., -4, -1, 2, 5, 8, ...}

C {0, ±3, ±6, ...}

D {0, 1, 2, 3, ...}

💡 Short Explanation

�� Equivalence class [0] contains multiples of 3. �� All numbers are congruent to 0 modulo 3. �� Positive and negative multiples are included.

✅ Correct Answer Explanation

The equivalence class: contains all integers divisible by 3: → Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Numbers leave remainder 1 modulo 3.
�� Option B → Numbers leave remainder 2 modulo 3.
�� Option D → Contains numbers from different equivalence classes.

🎯 Strategy

Used: Elimination

Application: Identify numbers divisible by 3 and remove sets with other remainders.

Final Logic: Equivalence class [0] consists only of multiples of 3.

🧠 Memory Trick

"[0] means all multiples of modulus."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Equivalence Classes Modulo m

CUET UG Applied Mathematics Booster Test 2 - Number Theory and Modular Fundamentals

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