Question 1

Which of the following numbers is divisible by 8?

Accepted Answer

Step 1: Divisibility rule for 8 — check if the last 3 digits of the number are divisible by 8. Step 2: Check each option: (a) 624: 624 ÷ 8 = 78 ✓. (b) 732: 732 ÷ 8 = 91.5 ✗. (c) 514: 514 ÷ 8 = 64.25 ✗. (d) 922: 922 ÷ 8 = 115.25 ✗. Therefore, 624 is divisible by 8.

Question 2

A number is divisible by both 4 and 9. Which of the following must also divide this number?

Accepted Answer

Step 1: A number divisible by both 4 and 9 must be divisible by their LCM. Step 2: Since 4 = 2² and 9 = 3², and gcd(4,9) = 1, we have LCM(4,9) = 4 × 9 = 36. Step 3: Therefore, any number divisible by both 4 and 9 must be divisible by 36. Step 4: Checking the options: 36 divides the number, so 12 (a factor of 36) also divides it. The answer is 12.

Question 3

A number when divided by 11 leaves a remainder of 7. When the same number is divided by 13, it leaves a remainder of 4. If the number lies between 500 and 700, what is the sum of digits of that number?

Accepted Answer

Step 1: Let the number be N. We have:
N ≡ 7 (mod 11)
N ≡ 4 (mod 13)

Step 2: From the first condition, N = 11k + 7 for some integer k.

Step 3: Substitute into the second condition:
11k + 7 ≡ 4 (mod 13)
11k ≡ -3 ≡ 10 (mod 13)

Step 4: Find the multiplicative inverse of 11 mod 13.
11 × 6 = 66 = 5(13) + 1, so 11 × 6 ≡ 1 (mod 13)
Inverse of 11 is 6.

Step 5: Multiply both sides by 6:
k ≡ 6 × 10 ≡ 60 ≡ 8 (mod 13)
So k = 13m + 8 for some integer m.

Step 6: Substitute back:
N = 11(13m + 8) + 7 = 143m + 88 + 7 = 143m + 95

Step 7: For N to lie between 500 and 700:
500 < 143m + 95 < 700
405 < 143m < 605
2.83... < m < 4.23...
So m = 3 or m = 4

Step 8: When m = 3: N = 143(3) + 95 = 429 + 95 = 524
When m = 4: N = 143(4) + 95 = 572 + 95 = 667

Step 9: Verify both:
524 ÷ 11 = 47 remainder 7 ✓
524 ÷ 13 = 40 remainder 4 ✓
667 ÷ 11 = 60 remainder 7 ✓
667 ÷ 13 = 51 remainder 4 ✓

Step 10: Both 524 and 667 satisfy the conditions. Sum of digits:
524: 5 + 2 + 4 = 11
667: 6 + 6 + 7 = 19

Since the question asks for 'that number' (singular) between 500 and 700, the first valid number in this range is 524. Sum of digits = 11.

Question 4

Which of the following numbers is NOT divisible by 8?

Accepted Answer

Step 1: Apply divisibility rule for 8: a number is divisible by 8 if its last three digits form a number divisible by 8. Step 2: Check 4328: last three digits are 328. 328 ÷ 8 = 41, so 4328 is divisible by 8. Step 3: Check 5432: last three digits are 432. 432 ÷ 8 = 54, so 5432 is divisible by 8. Step 4: Check 6536: last three digits are 536. 536 ÷ 8 = 67, so 6536 is divisible by 8. Step 5: Check 7644: last three digits are 644. 644 ÷ 8 = 80.5, so 7644 is NOT divisible by 8. Therefore, the answer is 7644.

Question 5

A number leaves remainder 0 when divided by 9. Which statement is always true about this number?

Accepted Answer

Step 1: A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Step 2: If a number is divisible by 9, it is also divisible by 3 (since 9 = 3²). Step 3: Therefore, the number must also be divisible by 3. The answer is: The number is divisible by 3.

Question 6

A number is divisible by both 4 and 9. Which of the following must also divide this number?

Accepted Answer

Step 1: A number divisible by both 4 and 9 must be divisible by their LCM. Step 2: Since 4 = 2² and 9 = 3², and gcd(4,9) = 1, we have LCM(4,9) = 4 × 9 = 36. Step 3: Therefore, any number divisible by both 4 and 9 must be divisible by 36. Step 4: Among the options, 36 is the only number that must divide such a number.

Question 7

A number is divisible by both 4 and 6. Which of the following must also divide this number?

Accepted Answer

Step 1: A number divisible by both 4 and 6 must be divisible by their LCM. Step 2: LCM(4, 6) = 12 (since 4 = 2², 6 = 2 × 3, so LCM = 2² × 3 = 12). Step 3: Therefore, any number divisible by both 4 and 6 must be divisible by 12.

Question 8

Which of the following numbers is divisible by 11?

Accepted Answer

Step 1: Apply divisibility rule for 11: alternating sum of digits (from right to left) must be divisible by 11. Step 2: For 5742: (2 - 4 + 7 - 5) = 0, which is divisible by 11. Step 3: Verify: 5742 ÷ 11 = 522. Therefore, 5742 is divisible by 11.

Question 9

How many numbers between 100 and 200 are divisible by 7?

Accepted Answer

Step 1: Find the smallest number ≥ 100 divisible by 7: 100 ÷ 7 = 14.28..., so first number is 7 × 15 = 105. Step 2: Find the largest number ≤ 200 divisible by 7: 200 ÷ 7 = 28.57..., so last number is 7 × 28 = 196. Step 3: Count: numbers are 7 × 15, 7 × 16, ..., 7 × 28. Step 4: Count = 28 - 15 + 1 = 14.

Question 10

A number is divisible by both 4 and 6. Which of the following must also divide this number?

Accepted Answer

Step 1: A number divisible by both 4 and 6 must be divisible by their LCM. Step 2: Find LCM(4, 6): 4 = 2², 6 = 2 × 3, so LCM = 2² × 3 = 12. Step 3: Any number divisible by 12 is also divisible by 12 itself, and by all its factors (1, 2, 3, 4, 6, 12). Therefore, the answer is 12.

Question 11

A number is divisible by 9. Which statement must be true about this number?

Accepted Answer

Step 1: Divisibility rule for 9: a number is divisible by 9 if and only if the sum of its digits is divisible by 9. Step 2: If a number is divisible by 9, then the sum of its digits must be divisible by 9. Step 3: Since 9 = 3 × 3, any number divisible by 9 is also divisible by 3. Therefore, the sum of digits must be divisible by 3 as well.

Question 12

Which of the following numbers is divisible by 11?

Accepted Answer

Step 1: Apply divisibility rule for 11: alternating sum of digits (from right to left) must be divisible by 11. Step 2: For 5742: (2 - 4 + 7 - 5) = 0, which is divisible by 11. Step 3: Verify: 5742 ÷ 11 = 522. Therefore, 5742 is divisible by 11.

Question 13

A number leaves remainder 0 when divided by 9. What can be said about the sum of its digits?

Accepted Answer

Step 1: A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Step 2: If a number leaves remainder 0 when divided by 9, it is divisible by 9. Step 3: Therefore, the sum of its digits must be divisible by 9.

Question 14

Which of the following numbers is NOT divisible by 3?

Accepted Answer

Step 1: Apply divisibility rule for 3: sum of all digits must be divisible by 3. Step 2: Check 4827: 4 + 8 + 2 + 7 = 21, divisible by 3. Step 3: Check 5634: 5 + 6 + 3 + 4 = 18, divisible by 3. Step 4: Check 7249: 7 + 2 + 4 + 9 = 22, NOT divisible by 3. Step 5: Check 6381: 6 + 3 + 8 + 1 = 18, divisible by 3. Therefore, 7249 is NOT divisible by 3.

Question 15

A number is divisible by 8 if its last three digits form a number divisible by 8. Which of the following is divisible by 8?

Accepted Answer

Step 1: Apply divisibility rule for 8: check if the last three digits form a number divisible by 8. Step 2: For 45,624: last three digits are 624. Step 3: 624 ÷ 8 = 78, so 624 is divisible by 8. Step 4: Therefore, 45,624 is divisible by 8.

Question 16

Which of the following numbers is NOT divisible by 6?

Accepted Answer

Step 1: A number is divisible by 6 if it is divisible by both 2 and 3. Step 2: For 4,374: it is even (divisible by 2). Step 3: Check divisibility by 3: sum of digits = 4 + 3 + 7 + 4 = 18, which is divisible by 3. Step 4: So 4,374 is divisible by 6. Step 5: For 4,375: it is odd (not divisible by 2). Step 6: Therefore, 4,375 is NOT divisible by 6.

Question 17

Which of the following numbers is NOT divisible by 8?

Accepted Answer

Step 1: Apply divisibility rule for 8: a number is divisible by 8 if its last three digits form a number divisible by 8. Step 2: Check 3216: last three digits are 216. 216 ÷ 8 = 27 ✓ (divisible). Step 3: Check 4128: last three digits are 128. 128 ÷ 8 = 16 ✓ (divisible). Step 4: Check 5240: last three digits are 240. 240 ÷ 8 = 30 ✓ (divisible). Step 5: Check 6314: last three digits are 314. 314 ÷ 8 = 39.25 ✗ (NOT divisible). Therefore, the answer is 6314.

Question 18

How many numbers between 100 and 200 are divisible by both 6 and 8?

Accepted Answer

Step 1: Find LCM(6, 8): 6 = 2 × 3, 8 = 2³, so LCM = 2³ × 3 = 24. Step 2: Numbers divisible by both 6 and 8 are multiples of 24. Step 3: Find multiples of 24 between 100 and 200: 120, 144, 168, 192. Step 4: Count them: 4 numbers.

Question 19

A number when divided by 8 leaves remainder 0, and when divided by 5 leaves remainder 0. What is the smallest such positive number?

Accepted Answer

Step 1: The number must be divisible by both 8 and 5. Step 2: Find LCM(8, 5): Since 8 = 2³ and 5 = 5 (coprime), LCM = 8 × 5 = 40. Step 3: The smallest positive number divisible by both is 40.

Question 20

A number is divisible by both 8 and 9. If the number lies between 500 and 700, how many such numbers exist?

Accepted Answer

Step 1: A number divisible by both 8 and 9 must be divisible by LCM(8, 9). Since 8 = 2³ and 9 = 3², LCM(8, 9) = 72. Step 2: Find multiples of 72 between 500 and 700. The smallest multiple ≥ 500 is: 500 ÷ 72 = 6.94..., so 7 × 72 = 504. The largest multiple ≤ 700 is: 700 ÷ 72 = 9.72..., so 9 × 72 = 648. Step 3: Multiples are 7 × 72 = 504, 8 × 72 = 576, 9 × 72 = 648. Therefore, there are 3 such numbers.

Question 21

A five-digit number 4A6B2 is divisible by 4 and 11. What is the value of A + B?

Accepted Answer

Step 1: For divisibility by 4, the last two digits (B2) must form a number divisible by 4. So 10B + 2 ≡ 0 (mod 4). This means 2B + 2 ≡ 0 (mod 4), so 2B ≡ 2 (mod 4), giving B ∈ {1, 5, 9}. Step 2: For divisibility by 11, the alternating sum of digits must be divisible by 11: (4 + 6 + 2) - (A + B) = 12 - A - B ≡ 0 (mod 11). So A + B ≡ 1 (mod 11). Step 3: Since A and B are single digits, A + B ≤ 18, so A + B ∈ {1, 12}. Step 4: If B = 1, then A ∈ {0, 11} → A = 0 gives A + B = 1. If B = 5, then A ∈ {7} gives A + B = 12. If B = 9, then A ∈ {3} gives A + B = 12. Step 5: Testing B = 1, A = 0: 40612. Check: last two digits 12 ÷ 4 = 3 ✓. Alternating sum: 4 - 0 + 6 - 1 + 2 = 11 ✓. Therefore A + B = 1.

Question 22

A number is formed by writing integers from 1 to 40 consecutively: 123456789101112...3839 40. What is the remainder when this number is divided by 8?

Accepted Answer

Step 1: For divisibility by 8, we only need to check the last three digits of the number. Step 2: The number ends with ...3839 40. The last three digits are 940. Step 3: Divide 940 by 8: 940 = 8 × 117 + 4. Step 4: Therefore, the remainder is 4.

Question 23

How many three-digit numbers are divisible by 6 but NOT divisible by 5?

Accepted Answer

Step 1: Three-digit numbers range from 100 to 999. Step 2: Count multiples of 6 in this range: ⌊999/6⌋ - ⌊99/6⌋ = 166 - 16 = 150. Step 3: Count multiples of both 6 and 5 (i.e., multiples of LCM(6,5) = 30): ⌊999/30⌋ - ⌊99/30⌋ = 33 - 3 = 30. Step 4: Numbers divisible by 6 but NOT by 5 = 150 - 30 = 120.

UGC NET Divisibility Rules

UGC NET Divisibility Rules — Past Exam Questions

Divisibility Rules— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Divisibility Rules — Practice Questions

60-Second Revision — Divisibility Rules