Number Systems, Digit Problems and Divisibility Practice Questions for GATE DA

Solve 103+ Number Systems, Digit Problems and Divisibility practice questions for GATE DA with answers and detailed solutions. Free sample questions below.

Fundamental Rules of Number Formation

When forming an $n$ -digit number, the most significant digit (the leading digit) can never be $0$ .

Basic Counting Rules

1. Leading Zero Constraint: For an

n

-digit number, the first position has restricted choices. If forming a number from a set containing

0

, the first position cannot be

0

2. Repetition Allowed: If digits can be repeated, each subsequent position has the full set of available choices.

3. Repetition Not Allowed: If digits cannot be repeated, each subsequent position has one fewer choice than the previous.

Example

Form a 3-digit number from the set ${0, 1, 2, 3, 4}$ without repetition.

Hundreds place: $4$ choices (cannot be $0$ ).
Tens place: $4$ choices (includes $0$ , but one non-zero digit is used).
Units place: $3$ choices.

Total numbers = $4 \times 4 \times 3 = 48$ .

Applying Divisibility Rules to Formed Numbers

Applying Divisibility Rules

Fix the digits that satisfy the divisibility rule first, then fill the remaining positions.

Div by 2
Unit:

0, 2, 4, 6, 8

Div by 4
Last 2 digits

Div by 5
Unit:

0

5

Worked Example

How many 4-digit numbers from ${1, 2, 3, 4, 5}$ without repetition are divisible by $5$ ?

Step 1 Fix unit digit:

5

(

1

choice).

Step 2 Thousands place:

4

choices.

Step 3 Hundreds place:

3

choices.

Step 4 Tens place:

2

choices.

Total = $1 \times 4 \times 3 \times 2 = 24$

Number Systems, Digit Problems and Divisibility: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Quantitative Aptitude MCQ

Consider the formation of 4-digit numbers using the digits ${0, 1, 2, 3, 4, 5}$ without repetition. Which of the following statements is TRUE?

A.
The thousands place has 6 possible choices.
B.
The total number of such 4-digit numbers is 360.
C.
If the number must be divisible by 5, the units digit must be 0 or 5.
D.
The total number of such 4-digit numbers with repetition allowed is 625.

Correct Answer:

Step-by-Step Solution

Key idea: Evaluate each statement by applying the fundamental rules of number formation, especially the leading zero constraint.

Step 1: Statement A: The thousands place cannot be 0, so it has only 5 choices (1, 2, 3, 4, 5), not 6. False.

Step 2: Statement B: Total numbers = $5 \times 5 \times 4 \times 3 = 300$ , not 360. (360 would be $6 \times 5 \times 4 \times 3$ , ignoring the leading zero constraint). False.

Step 3: Statement C: A number is divisible by 5 if its units digit is 0 or 5. This is a standard divisibility rule. True.

Step 4: Statement D: With repetition allowed, total numbers = $5 \times 6 \times 6 \times 6 = 1080$ (thousands has 5 choices, others have 6). 625 is $5^{4}$ , which is incorrect. False.

Answer: C

Question 2 · Quantitative Aptitude MCQ

Consider the formation of 3-digit numbers using the digits ${0, 1, 2, 3, 4}$ without repetition. Which of the following statements is TRUE?

A.
The total number of such 3-digit numbers is 60.
B.
The number of such 3-digit numbers divisible by 5 is 12.
C.
The number of such 3-digit numbers divisible by 2 is 24.
D.
The number of such 3-digit numbers with an even digit sum is 24.

Correct Answer:

Step-by-Step Solution

Key idea: Evaluate each statement by applying the fundamental rules of number formation, checking the leading zero constraint and specific divisibility/parity rules.

Step 1: Evaluate Statement A (Total numbers).

Hundreds: 4 choices (1, 2, 3, 4).
Tens: 4 choices.
Units: 3 choices.
Total = $4 \times 4 \times 3 = 48$ . Statement A (60) is FALSE.

Step 2: Evaluate Statement B (Divisible by 5).

Units must be 0 (since 5 is not in the set).
Units: 1 choice (0).
Hundreds: 4 choices.
Tens: 3 choices.
Total = $1 \times 4 \times 3 = 12$ . Statement B is TRUE.

Step 3: Evaluate Statement C (Divisible by 2).

Units must be 0, 2, or 4.
Units=0: $4 \times 3 = 12$ .
Units=2: Hundreds 3, Tens 3 $\to 9$ .
Units=4: Hundreds 3, Tens 3 $\to 9$ .
Total = $12 + 9 + 9 = 30$ . Statement C (24) is FALSE.

Step 4: Evaluate Statement D (Even digit sum).

3 evens: {0,2,4} $\to 2 \times 2 \times 1 = 4$ .
1 even, 2 odds: Evens {0,2,4}, Odds {1,3}.
Even=0: $2 \times 2 \times 1 = 4$ .
Even=2 or 4: $2 \times (3 \times 2) = 12$ .
Total = $4 + 4 + 12 = 20$ . Statement D (24) is FALSE.

Answer: B

Question 3 · Quantitative Aptitude MCQ

Consider the formation of 5-digit numbers using the digits ${0, 1, 2, 3, 4, 5}$ without repetition. Which of the following statements is TRUE?

A.
The total number of such 5-digit numbers is 720.
B.
The number of such 5-digit numbers divisible by 10 is 120.
C.
The number of such 5-digit numbers divisible by 5 is 240.
D.
The number of such 5-digit numbers with an even digit sum is 240.

Correct Answer:

Step-by-Step Solution

Key idea: Evaluate each statement by applying the fundamental rules of number formation, checking the leading zero constraint and specific divisibility/parity rules.

Step 1: Evaluate Statement A (Total numbers).

Ten-thousands: 5 choices (1, 2, 3, 4, 5).
Remaining 4 positions: $5 \times 4 \times 3 \times 2 = 120$ .
Total = $5 \times 120 = 600$ . Statement A (720) is FALSE.

Step 2: Evaluate Statement B (Divisible by 10).

Units must be 0.
Units: 1 choice (0).
Ten-thousands: 5 choices.
Remaining 3 positions: $4 \times 3 \times 2 = 24$ .
Total = $1 \times 5 \times 24 = 120$ . Statement B is TRUE.

Step 3: Evaluate Statement C (Divisible by 5).

Units must be 0 or 5.
Units=0: $5 \times 4 \times 3 \times 2 = 120$ .
Units=5: Ten-thousands 4 choices, remaining $4 \times 3 \times 2 = 24 \to 96$ .
Total = $120 + 96 = 216$ . Statement C (240) is FALSE.

Step 4: Evaluate Statement D (Even digit sum).

Evens: ${0, 2, 4}$ (3). Odds: ${1, 3, 5}$ (3).
Even sum requires 3 evens + 2 odds.
Sets = $(3 3) \times (2 3) = 3$ .
Arrangements = $5! = 120$ .
Total = $3 \times 120 = 360$ . Statement D (240) is FALSE.

Answer: B

Question 4 · Quantitative Aptitude MCQ

What is the smallest 3-digit number that can be formed using distinct digits from ${1, 2, 3, 4, 5, 6}$ such that the sum of its digits is odd?

A.
123
B.
125
C.
124
D.
126

Correct Answer:

Step-by-Step Solution

Key idea: To minimize the number, choose the smallest available digits from left to right, while ensuring the digit sum is odd.

Step 1: Hundreds digit = 1 (smallest available)

Step 2: Tens digit = 2 (next smallest)

Step 3: Units digit must make the sum odd:

Current sum: $1 + 2 = 3$ (odd)
To keep the total sum odd, units digit must be even (odd + even = odd)
Smallest even digit from remaining ${3, 4, 5, 6}$ is 4

Step 4: Number = 124, sum = $1 + 2 + 4 = 7$ (odd) ✓

Verify other options:

123: sum = 6 (even) ✗
125: sum = 8 (even) ✗
126: sum = 9 (odd) ✓, but 126 > 124

Answer: C (124)

Question 5 · Quantitative Aptitude MCQ

How many 3-digit numbers can be formed using the digits ${1, 2, 3, 4, 5}$ without repetition such that the number is divisible by $5$ ?

A.
6
B.
12
C.
24
D.
60

Correct Answer:

Step-by-Step Solution

Key idea: Fix the most restricted position first (units digit for divisibility by 5), then fill the remaining positions.

Step 1: Divisibility by 5 requires the units digit to be 5 (since 0 is not in the set). This fixes 1 choice for the units place.

Step 2: The hundreds place can be filled by any of the remaining 4 digits.

Step 3: The tens place can be filled by any of the remaining 3 digits.

Step 4: Multiply the choices: $1 \times 4 \times 3 = 12$ .

Answer: 12

Question 6 · Quantitative Aptitude MCQ

Directions for Assertion-Reason questions:

(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.

(D) Assertion is false but Reason is true.

Assertion (A): The number of 4-digit numbers formed using ${0, 1, 2, 3, 4, 5}$ without repetition that are divisible by $5$ is $108$ .

Reason (R): A number is divisible by $5$ if and only if its units digit is $0$ or $5$ .

A.
A
B.
B
C.
C
D.
D

Correct Answer:

Step-by-Step Solution

Key idea: Verify the assertion by calculation, verify the reason by definition, then check if the reason fully explains the assertion.

Step 1: Verify Reason (R): A number is divisible by 5 iff its units digit is 0 or 5. This is a standard, true divisibility rule. R is True.

Step 2: Verify Assertion (A): Calculate the count.

Case 1: Units = 0. Thousands: 5 choices (1-5). Hundreds: 4. Tens: 3. Total = $5 \times 4 \times 3 = 60$ .
Case 2: Units = 5. Thousands: 4 choices (1-4, cannot be 0). Hundreds: 4. Tens: 3. Total = $4 \times 4 \times 3 = 48$ .
Total = $60 + 48 = 108$ . A is True.

Step 3: Does R explain A? R gives the divisibility condition, which is the starting point. However, to get 108, you must also handle the leading zero constraint (thousands place cannot be 0). R alone does not account for this. Thus, R is not the complete explanation.

Answer: B

Question 7 · Quantitative Aptitude MCQ

How many 3-digit numbers can be formed using the digits ${0, 1, 2, 3, 4, 5}$ without repetition such that the number is divisible by $5$ ?

A.
40
B.
30
C.
36
D.
20

Correct Answer:

Step-by-Step Solution

Key idea: This requires casework based on the divisibility rule for 5, combined with the leading zero constraint.

Step 1: A number is divisible by 5 if its units digit is 0 or 5. We must split this into two cases.

Step 2: Case 1: Units digit is 0.

Units: 1 choice (0).
Hundreds: 5 choices (1, 2, 3, 4, 5).
Tens: 4 choices.
Total for Case 1 = $1 \times 5 \times 4 = 20$ .

Step 3: Case 2: Units digit is 5.

Units: 1 choice (5).
Hundreds: 4 choices (cannot be 0, and 5 is used).
Tens: 4 choices (includes 0, but two non-zero digits are used).
Total for Case 2 = $1 \times 4 \times 4 = 16$ .

Step 4: Add the cases. Total numbers = $20 + 16 = 36$ .

Answer: 36

Question 8 · Quantitative Aptitude MCQ

Directions for Assertion-Reason questions:

(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.

(D) Assertion is false but Reason is true.

Assertion (A): The number of 4-digit numbers formed using ${0, 1, 2, 3, 4, 5, 6}$ without repetition that are divisible by $25$ is $220$ .

Reason (R): A number is divisible by $25$ if its last two digits are $25$ or $50$ .

A.
A
B.
B
C.
C
D.
D

Correct Answer:

Step-by-Step Solution

Key idea: Verify the assertion by calculation, verify the reason by definition, then check if the reason fully explains the assertion.

Step 1: Verify Reason (R): A number is divisible by 25 if its last two digits are 00, 25, 50, or 75. Since repetition is not allowed and 7 is not in the set, the only valid endings are 25 and 50. R is TRUE.

Step 2: Verify Assertion (A): Calculate the count for divisibility by 25.

Case 1: Ends in 25. First two digits from {0, 1, 3, 4, 6}. Thousands: 4 choices (not 0). Hundreds: 4 choices. Total = $4 \times 4 = 16$ .
Case 2: Ends in 50. First two digits from {1, 2, 3, 4, 6}. Thousands: 5 choices. Hundreds: 4 choices. Total = $5 \times 4 = 20$ .
Total = $16 + 20 = 36$ .

Step 3: Compare with A. A claims the answer is 220. Since $36 \neq = 220$ , A is FALSE. (Note: 220 is the count for divisibility by 5, not 25).

Answer: D

Question 9 · Quantitative Aptitude MCQ

How many 3-digit numbers can be formed using the digits ${0, 1, 2, 3, 4}$ without repetition such that the number is divisible by $4$ ?

A.
15
B.
12
C.
18
D.
24

Correct Answer:

Step-by-Step Solution

Key idea: This requires casework based on the divisibility rule for 4, combined with the leading zero constraint.

Step 1: A number is divisible by 4 if its last two digits form a number divisible by 4. The valid pairs from ${0, 1, 2, 3, 4}$ without repetition are: 04, 12, 20, 24, 32, 40.

Step 2: Calculate the number of valid hundreds digits for each pair, remembering the hundreds digit cannot be 0.

Ends in 04: Hundreds can be 1, 2, 3 (3 choices).
Ends in 12: Hundreds can be 3, 4 (2 choices, 0 is invalid).
Ends in 20: Hundreds can be 1, 3, 4 (3 choices).
Ends in 24: Hundreds can be 1, 3 (2 choices, 0 is invalid).
Ends in 32: Hundreds can be 1, 4 (2 choices, 0 is invalid).
Ends in 40: Hundreds can be 1, 2, 3 (3 choices).

Step 3: Sum the valid combinations. Total = $3 + 2 + 3 + 2 + 2 + 3 = 15$ .

Answer: 15

Question 10 · Quantitative Aptitude MCQ

Directions for Assertion-Reason questions:

(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.

(D) Assertion is false but Reason is true.

Assertion (A): The number of 3-digit numbers formed using ${0, 1, 2, 3, 4}$ without repetition that are divisible by $4$ is $15$ .

Reason (R): A number is divisible by $4$ if its units digit is divisible by $4$ .

A.
A
B.
B
C.
C
D.
D

Correct Answer:

Step-by-Step Solution

Key idea: Verify the assertion by calculation, verify the reason by definition, then check if the reason fully explains the assertion.

Step 1: Verify Reason (R): A number is divisible by 4 if the number formed by its last two digits is divisible by 4. The reason claims it depends only on the units digit, which is the rule for divisibility by 2, not 4. R is FALSE.

Step 2: Verify Assertion (A): Calculate the count for divisibility by 4.

Valid last two digits: 04, 12, 20, 24, 32, 40.
04: 3 choices for hundreds.
12: 2 choices (no 0).
20: 3 choices.
24: 2 choices (no 0).
32: 2 choices (no 0).
40: 3 choices.
Total = $3 + 2 + 3 + 2 + 2 + 3 = 15$ . A is TRUE.

Step 3: Since A is true and R is false, the correct option is C.

Answer: C

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Number Systems, Digit Problems and Divisibility Practice Questions for GATE DA

Fundamental Rules of Number Formation

Fundamental Rules of Number Formation

Basic Counting Rules

Example

Applying Divisibility Rules to Formed Numbers

Applying Divisibility Rules

Worked Example

Number Systems, Digit Problems and Divisibility: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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