Question 1

Which number among the following is NOT divisible by 9?

Accepted Answer

Step 1: A number is divisible by 9 if the sum of its digits is divisible by 9. Step 2: For 3456: 3 + 4 + 5 + 6 = 18, divisible by 9. Step 3: For 2781: 2 + 7 + 8 + 1 = 18, divisible by 9. Step 4: For 1234: 1 + 2 + 3 + 4 = 10, NOT divisible by 9. Step 5: For 5544: 5 + 5 + 4 + 4 = 18, divisible by 9. Therefore, 1234 is NOT divisible by 9.

Question 2

A number is divisible by 11 if the alternating sum of its digits is divisible by 11. Which of the following numbers is divisible by 11?

Accepted Answer

Step 1: Apply divisibility rule for 11: alternating sum of digits (starting from right) 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 (exact). Therefore, 5742 is divisible by 11.

Question 3

Which of the following numbers is divisible by both 6 and 8?

Accepted Answer

Step 1: A number is divisible by 6 if it is divisible by both 2 and 3. Step 2: A number is divisible by 8 if its last three digits form a number divisible by 8. Step 3: For divisibility by both 6 and 8, the number must be divisible by LCM(6,8) = 24. Step 4: Check 264: 264 ÷ 24 = 11 (exact). Therefore, 264 is divisible by both 6 and 8.

Question 4

A number is divisible by 12. Which of the following must also be true?

Accepted Answer

Step 1: 12 = 3 × 4, where 3 and 4 are coprime. Step 2: If a number is divisible by 12, it must be divisible by both 3 and 4. Step 3: A number divisible by 4 is always divisible by 2 (since 2 is a factor of 4). Step 4: Therefore, any number divisible by 12 must be divisible by 2, 3, and 4. Step 5: Among the options, divisibility by 3 is guaranteed. The answer is that it is divisible by 3.

Question 5

Which of the following numbers is divisible by both 6 and 8?

Accepted Answer

Step 1: A number is divisible by 6 if it is divisible by both 2 and 3. Step 2: A number is divisible by 8 if its last three digits form a number divisible by 8. Step 3: For divisibility by both 6 and 8, the number must be divisible by LCM(6,8) = 24. Step 4: Check 288: 288 ÷ 24 = 12. Therefore, 288 is divisible by both 6 and 8.

Question 6

A number when divided by 9 leaves remainder 0. Which of the following is definitely true about this number?

Accepted Answer

Step 1: If a number is divisible by 9 (remainder 0), then by the divisibility rule for 9, the sum of its digits must be divisible by 9. Step 2: If the sum of digits is divisible by 9, then the sum is also divisible by 3 (since 9 = 3 × 3). Step 3: Therefore, the number must also be divisible by 3. Step 4: The number is divisible by 3 and 9, but not necessarily by 6 (which requires divisibility by both 2 and 3).

Question 7

Which of the following numbers is divisible by 3 but NOT divisible by 9?

Accepted Answer

Step 1: A number is divisible by 3 if the sum of its digits is divisible by 3. Step 2: A number is divisible by 9 if the sum of its digits is divisible by 9. Step 3: For 4521: 4 + 5 + 2 + 1 = 12, divisible by 3 but not by 9. Step 4: For 3456: 3 + 4 + 5 + 6 = 18, divisible by 9 (and hence by 3). Step 5: For 2781: 2 + 7 + 8 + 1 = 18, divisible by 9. Step 6: For 5544: 5 + 5 + 4 + 4 = 18, divisible by 9. Therefore, 4521 is divisible by 3 but not by 9.

Question 8

A number is divisible by 6 if it is divisible by both 2 and 3. Which of the following numbers is divisible by 6?

Accepted Answer

Step 1: For divisibility by 6, the number must be even (divisible by 2) and have a digit sum divisible by 3. Step 2: For 3456: It is even (ends in 6) and digit sum = 3 + 4 + 5 + 6 = 18, divisible by 3. Step 3: Therefore, 3456 is divisible by 6. Step 4: Verify: 3456 ÷ 6 = 576 (exact).

Question 9

A number when divided by 9 leaves remainder 0. Which of the following statements is always true about this number?

Accepted Answer

Step 1: If a number is divisible by 9 (remainder 0), then by the divisibility rule for 9, the sum of its digits must be divisible by 9. Step 2: A number divisible by 9 is also divisible by 3 (since 3 is a factor of 9). Step 3: The number must be a multiple of 9. Therefore, the number is divisible by 3.

Question 10

How many numbers between 100 and 300 are divisible by 7 but not by 3?

Accepted Answer

Step 1: Find numbers divisible by 7 between 100 and 300. Smallest: 105 = 7×15, Largest: 294 = 7×42. Count = 42 - 15 + 1 = 28. Step 2: Find numbers divisible by both 7 and 3 (i.e., by 21) between 100 and 300. Smallest: 105 = 21×5, Largest: 294 = 21×14. Count = 14 - 5 + 1 = 10. Step 3: Numbers divisible by 7 but not by 3 = 28 - 10 = 18.

Question 11

How many three-digit numbers are divisible by both 6 and 8 but NOT divisible by 12?

Accepted Answer

Step 1: For a number to be divisible by both 6 and 8, it must be divisible by LCM(6,8). Step 2: Find LCM: 6=2×3, 8=2³, so LCM(6,8)=24. Step 3: Any multiple of 24 is also a multiple of 12 (since 24=2³×3 and 12=2²×3, and 24÷12=2). Step 4: Therefore, every number divisible by both 6 and 8 is divisible by 12. Step 5: The set of numbers divisible by both 6 and 8 but not by 12 is empty. Answer: 0.

Question 12

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

Accepted Answer

Step 1: If a number is divisible by both 4 and 9, we need to find their LCM. Step 2: Since 4 = 2² and 9 = 3², and gcd(4,9) = 1, then LCM(4,9) = 4 × 9 = 36. Step 3: Any number divisible by both 4 and 9 must be divisible by 36. Therefore, the answer is 36.

Question 13

Which of the following numbers is divisible by 6 but not by 12?

Accepted Answer

Step 1: A number is divisible by 6 if it is divisible by both 2 and 3. Step 2: A number is divisible by 12 if it is divisible by both 4 and 3 (or equivalently, by 3 and 4). Step 3: We need a number divisible by 6 but not by 12, meaning divisible by 2 and 3, but not by 4. Step 4: Testing 54: 54 ÷ 6 = 9 ✓, 54 ÷ 12 = 4.5 ✗. Testing 48: 48 ÷ 6 = 8 ✓, 48 ÷ 12 = 4 ✓ (excluded). Testing 66: 66 ÷ 6 = 11 ✓, 66 ÷ 12 = 5.5 ✗. Testing 72: 72 ÷ 6 = 12 ✓, 72 ÷ 12 = 6 ✓ (excluded). Answer is 54 or 66.

Question 14

A six-digit number 5A4B7C is divisible by 9 and 4. If A = 2, what is the value of B + C?

Accepted Answer

Step 1: For divisibility by 9, the sum of all digits must be divisible by 9. Sum = 5 + 2 + 4 + B + 7 + C = 18 + B + C. For divisibility by 9: 18 + B + C ≡ 0 (mod 9), so B + C ≡ 0 (mod 9). Thus B + C = 0, 9, or 18. Step 2: For divisibility by 4, the last two digits (7C) must form a number divisible by 4. Testing: 70(no), 71(no), 72(yes), 73(no), 74(no), 75(no), 76(yes), 77(no), 78(no), 79(no). So C = 2 or C = 6. Step 3: If C = 2, then B + C = 9 gives B = 7. If C = 6, then B + C = 9 gives B = 3. Checking: 72 ÷ 4 = 18 ✓ and 76 ÷ 4 = 19 ✓. Both work. If B + C = 0, then C = 0 and B = 0, but 70 is not divisible by 4. If B + C = 18, then C = 9 and B = 9, but 79 is not divisible by 4. So B + C = 9.

Question 15

Which of the following numbers is divisible by 3 but NOT divisible by 6?

Accepted Answer

Step 1: A number divisible by 6 must be divisible by both 2 and 3. Step 2: A number divisible by 3 but NOT by 6 must be divisible by 3 but NOT by 2 (i.e., it must be odd). Step 3: Check each option: (a) 144 = 2⁴ × 3² (divisible by 6), (b) 135 = 3³ × 5 (odd, divisible by 3, NOT by 2), (c) 126 = 2 × 3² × 7 (divisible by 6), (d) 162 = 2 × 3⁴ (divisible by 6). Step 4: Answer is 135.

Question 16

A five-digit number 3A7B2 is divisible by both 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 must be divisible by 4. Testing: B ∈ {1,3,5,7,9} gives 12, 32, 52, 72, 92. Only 12, 32, 52, 72, 92 are divisible by 4. So B ∈ {1,3,5,7,9}. Step 2: For divisibility by 11, alternating sum (3 - A + 7 - B + 2) must be divisible by 11. So 12 - A - B ≡ 0 (mod 11), meaning A + B ≡ 1 (mod 11). Step 3: Testing B values: If B = 1, then A + 1 ≡ 1 (mod 11), so A = 0. Check: 30712 → (3-0+7-1+2) = 11 ✓. Step 4: A + B = 0 + 1 = 1.

Question 17

A number when divided by 11 leaves remainder 7. What is the remainder when this number is divided by 11 again after adding 44 to it?

Accepted Answer

Step 1: Let the number be N. Given: N ≡ 7 (mod 11). Step 2: After adding 44: N + 44 ≡ 7 + 44 (mod 11). Step 3: 7 + 44 = 51 ≡ 51 - 44 = 7 (mod 11) [since 44 ≡ 0 (mod 11)]. Step 4: Therefore, remainder = 7.

Question 18

How many numbers between 100 and 500 are divisible by 6 but NOT divisible by 9?

Accepted Answer

Step 1: Count numbers divisible by 6 between 100 and 500: ⌊500/6⌋ - ⌊99/6⌋ = 83 - 16 = 67. Step 2: Count numbers divisible by both 6 and 9 (i.e., divisible by LCM(6,9) = 18): ⌊500/18⌋ - ⌊99/18⌋ = 27 - 5 = 22. Step 3: Numbers divisible by 6 but NOT by 9 = 67 - 22 = 45.

Question 19

A number is divisible by both 8 and 9. Which of the following is definitely true about this number?

Accepted Answer

Step 1: A number divisible by both 8 and 9 must be divisible by their LCM. Step 2: Since 8 = 2³ and 9 = 3², and GCD(8,9) = 1, LCM(8,9) = 8 × 9 = 72. Step 3: Therefore, the number must be divisible by 72.

Question 20

A number leaves remainder 5 when divided by 7. What remainder does this number leave when divided by 7 after being multiplied by 3?

Accepted Answer

Step 1: Let N ≡ 5 (mod 7). Step 2: We need to find 3N (mod 7). Step 3: 3N ≡ 3 × 5 ≡ 15 (mod 7). Step 4: 15 = 2 × 7 + 1, so 15 ≡ 1 (mod 7). Step 5: Therefore, remainder = 1.

RRB NTPC Divisibility Rules

Divisibility Rules— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Divisibility Rules — Practice Questions

60-Second Revision — Divisibility Rules