Question 1

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, as well as all its factors (1, 2, 3, 4, 6, 12). Therefore, the answer is 12.

Question 2

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. Therefore, 5742 is divisible by 11.

Question 3

A number when divided by 8 leaves remainder 0. Which statement is definitely true about this number?

Accepted Answer

Step 1: If a number is divisible by 8 (remainder 0), then it is also divisible by all factors of 8. Step 2: Factors of 8 are: 1, 2, 4, 8. Step 3: Therefore, the number must be divisible by 4. Step 4: Check: any multiple of 8 (e.g., 16, 24, 32) is also a multiple of 4.

Question 4

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

Accepted Answer

Step 1: Find the smallest multiple of 7 ≥ 100: 100 ÷ 7 = 14.28..., so first multiple is 7 × 15 = 105. Step 2: Find the largest multiple of 7 ≤ 200: 200 ÷ 7 = 28.57..., so last multiple is 7 × 28 = 196. Step 3: Count multiples from 7 × 15 to 7 × 28: 28 - 15 + 1 = 14. Therefore, there are 14 numbers.

Question 5

A number is divisible by 9. Which of the following is definitely true?

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 it is also divisible by 3 (since 3 is a factor of 9). Step 3: This is because if the sum of digits is divisible by 9, it is also divisible by 3. Therefore, the number must be divisible by 3.

Question 6

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 4356: sum = 4 + 3 + 5 + 6 = 18, which is divisible by 9, so 4356 is divisible by 9. For 4357: sum = 4 + 3 + 5 + 7 = 19, not divisible by 3. For 4358: sum = 4 + 3 + 5 + 8 = 20, not divisible by 3. For 4359: sum = 4 + 3 + 5 + 9 = 21, divisible by 3 but NOT by 9. Therefore, 4359 is the answer.

Question 7

A number when divided by 11 leaves remainder 7. What will be the remainder when the square of this number is divided by 11?

Accepted Answer

Step 1: Let the number be n. Given: n ≡ 7 (mod 11). Step 2: We need to find n² (mod 11). Step 3: n² ≡ 7² (mod 11) ≡ 49 (mod 11). Step 4: 49 = 11×4 + 5, so 49 ≡ 5 (mod 11). Step 5: Therefore, remainder is 5.

Question 8

A number is divisible by both 8 and 9. Which of the following must also divide 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, any number divisible by both 8 and 9 must be divisible by 72. Step 4: Check options: 72 divides the number, so 36 (which is 72÷2) also divides it, and 24 (which is 72÷3) also divides it. Step 5: However, 48 does not necessarily divide all such numbers (e.g., 72 is not divisible by 48). The answer is 36.

Question 9

How many numbers between 100 and 500 are divisible by 7 but not by 3?

Accepted Answer

Step 1: Find count of numbers divisible by 7 in range [100, 500]. Smallest: 105 = 7×15, Largest: 497 = 7×71. Count = 71 - 15 + 1 = 57. Step 2: Find count of numbers divisible by both 7 and 3 (i.e., by 21). Smallest: 105 = 21×5, Largest: 483 = 21×23. Count = 23 - 5 + 1 = 19. Step 3: Numbers divisible by 7 but not by 3 = 57 - 19 = 38.

Question 10

A number is divisible by 8 and also divisible by 15. Which of the following must be true about this number?

Accepted Answer

Step 1: If a number is divisible by 8, it is divisible by 2³. Step 2: If a number is divisible by 15, it is divisible by 3 × 5. Step 3: Since 8 and 15 are coprime (GCD(8,15) = 1), the number must be divisible by LCM(8,15) = 8 × 15 = 120. Step 4: Therefore, the number must be divisible by 120, which means it is divisible by all factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.

Question 11

A number N is divisible by 7. When N is divided by 14, the remainder is 0. When N is divided by 21, the remainder is 0. What is the smallest positive value of N that satisfies all three conditions?

Accepted Answer

Step 1: N divisible by 7 means N = 7k. Step 2: N divisible by 14 means N = 14m. Step 3: N divisible by 21 means N = 21p. Step 4: The smallest N satisfying all three is LCM(7, 14, 21). Step 5: 7 = 7, 14 = 2 × 7, 21 = 3 × 7. LCM = 2 × 3 × 7 = 42. Step 6: Verify: 42 ÷ 7 = 6 ✓, 42 ÷ 14 = 3 ✓, 42 ÷ 21 = 2 ✓.

Question 12

A number when divided by 11 leaves remainder 7. Another number when divided by 11 leaves remainder 4. If the sum of these two numbers is divided by 11, what will be the remainder?

Accepted Answer

Step 1: Let the first number be 11a + 7 and second number be 11b + 4. Step 2: Sum = (11a + 7) + (11b + 4) = 11(a + b) + 11 = 11(a + b + 1). Step 3: When divided by 11, remainder is 0.

Question 13

A five-digit number 3a7b2 is divisible by both 4 and 9. If a and b are single digits, what is the maximum value of a + b?

Accepted Answer

Step 1: For divisibility by 4, last two digits (b2) must form a number divisible by 4. So b ∈ {1, 3, 5, 7, 9} giving 12, 32, 52, 72, 92 all divisible by 4. Step 2: For divisibility by 9, sum of digits 3 + a + 7 + b + 2 = 12 + a + b must be divisible by 9. So a + b ∈ {6, 15}. Step 3: Maximum a + b = 15 (when 12 + a + b = 27). Step 4: Check: if a + b = 15, then b ≤ 9 and a ≤ 9, so max is a = 9, b = 6 or a = 8, b = 7, etc. With b = 6, last two digits = 62 (not divisible by 4). With b = 7, last two digits = 72 (divisible by 4). So a = 8, b = 7 works. Thus a + b = 15.

Question 14

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

Accepted Answer

Step 1: Numbers divisible by 6 between 100 and 500: First = 102, Last = 498. Count = (498 - 102)/6 + 1 = 396/6 + 1 = 66 + 1 = 67. Step 2: Numbers divisible by both 6 and 9 (i.e., divisible by LCM(6,9) = 18) between 100 and 500: First = 108, Last = 486. Count = (486 - 108)/18 + 1 = 378/18 + 1 = 21 + 1 = 22. Step 3: Numbers divisible by 6 but NOT by 9 = 67 - 22 = 45.

Question 15

A six-digit number 5a4b8c is divisible by 11. The digits a, b, c are such that a + b + c = 15. If the number is also divisible by 3, what is the value of a + b + c?

Accepted Answer

Step 1: For divisibility by 11, alternating sum of digits must be divisible by 11: (5 - a + 4 - b + 8 - c) = (17 - a - b - c) ≡ 0 (mod 11). Step 2: Given a + b + c = 15, so 17 - 15 = 2 ≡ 0 (mod 11)? No, 2 is not divisible by 11. Step 3: Re-examine: alternating sum = 5 - a + 4 - b + 8 - c = 17 - (a+b+c). For divisibility by 11: 17 - (a+b+c) ≡ 0 (mod 11). If a+b+c = 15, then 17 - 15 = 2, which is NOT ≡ 0 (mod 11). Step 4: For the condition to hold, a+b+c must equal 6 or 17. Since a,b,c are single digits (max sum = 27), a+b+c = 6 makes 17 - 6 = 11 ≡ 0 (mod 11). Step 5: Check divisibility by 3: sum of all digits = 5+a+4+b+8+c = 17 + (a+b+c) = 17 + 6 = 23. 23 ÷ 3 leaves remainder 2, so NOT divisible by 3. Step 6: If a+b+c = 17, then 17 - 17 = 0 ≡ 0 (mod 11) ✓. Sum of digits = 17 + 17 = 34. 34 ÷ 3 leaves remainder 1, NOT divisible by 3. Step 7: The question contains a contradiction; however, the intended answer is likely a+b+c = 6 (satisfying divisibility by 11) or the question expects re-interpretation. Given the phrasing 'If the number is also divisible by 3', the answer is a+b+c = 6.

IBPS Clerk Divisibility Rules

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Divisibility Rules — Practice Questions

60-Second Revision — Divisibility Rules