Question 1

Which of the following numbers is divisible by 4?

Accepted Answer

Step 1: A number is divisible by 4 if its last two digits form a number divisible by 4. Step 2: For 2348, last two digits are 48. Step 3: 48 ÷ 4 = 12 (exact). Therefore, 2348 is divisible by 4.

Question 2

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: For divisibility by 8, check if the last three digits form a number divisible by 8. Step 2: For 45672, last three digits are 672. Step 3: 672 ÷ 8 = 84 (exact). Therefore, 45672 is divisible by 8.

Question 3

Which digit should replace * in 5*3 so that the resulting three-digit number is divisible by 3?

Accepted Answer

Step 1: A number is divisible by 3 if the sum of its digits is divisible by 3. Step 2: Sum of digits = 5 + * + 3 = 8 + *. Step 3: For divisibility by 3, (8 + *) must be divisible by 3. Step 4: If * = 1, sum = 9, which is divisible by 3. Therefore, * = 1.

Question 4

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

Accepted Answer

Step 1: A number divisible by both 6 and 8 must be divisible by their LCM. Step 2: LCM(6, 8) = 24. Step 3: Check 144: 144 ÷ 24 = 6 (exact). Therefore, 144 is divisible by both 6 and 8.

Question 5

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

Question 6

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 gcd(8,9) = 1, LCM(8,9) = 8 × 9 = 72. Step 3: Therefore, the number must be divisible by 72. Step 4: Check: 72 is divisible by 6 (72 ÷ 6 = 12), by 12 (72 ÷ 12 = 6), by 18 (72 ÷ 18 = 4), and by 36 (72 ÷ 36 = 2). All options are divisible by 72, but the strongest statement is divisibility by 72 itself. Among the options, divisibility by 36 is guaranteed.

Question 7

A number is divisible by 6. Which of the following statements is always true?

Accepted Answer

Step 1: A number divisible by 6 must be divisible by both 2 and 3 (since 6 = 2 × 3 and gcd(2,3)=1). Step 2: Divisibility by 2 means the number is even. Step 3: Divisibility by 3 means the sum of digits is divisible by 3. Step 4: Therefore, any number divisible by 6 is always even (divisible by 2) and has a digit sum divisible by 3. Step 5: Checking options: (a) divisible by 12 — not always (e.g., 6, 18, 30 are divisible by 6 but not 12). (b) divisible by 2 — always true. (c) divisible by 9 — not always (e.g., 6, 12, 24 are divisible by 6 but not 9). (d) divisible by 18 — not always (e.g., 6, 12, 24 are divisible by 6 but not 18).

Question 8

A number when divided by 15 leaves remainder 7. What remainder will it leave when divided by 5?

Accepted Answer

Step 1: If a number leaves remainder 7 when divided by 15, it can be written as N = 15k + 7 for some integer k. Step 2: We need to find the remainder when N is divided by 5. Step 3: N = 15k + 7 = 5(3k) + 7. Step 4: Now divide 7 by 5: 7 = 5(1) + 2. Step 5: So N = 5(3k) + 5(1) + 2 = 5(3k+1) + 2. Step 6: Therefore, remainder when N is divided by 5 is 2.

Question 9

How many four-digit numbers are divisible by both 7 and 11, but NOT divisible by 13?

Accepted Answer

Step 1: Numbers divisible by both 7 and 11 are divisible by LCM(7,11) = 77. Step 2: Four-digit numbers range from 1000 to 9999. Step 3: Smallest four-digit multiple of 77: ⌈1000/77⌉ × 77 = 14 × 77 = 1078. Step 4: Largest four-digit multiple of 77: ⌊9999/77⌋ × 77 = 129 × 77 = 9933. Step 5: Count of multiples of 77: 129 - 14 + 1 = 116. Step 6: Now find four-digit multiples of LCM(7,11,13) = 1001. Step 7: Smallest: ⌈1000/1001⌉ × 1001 = 1 × 1001 = 1001. Step 8: Largest: ⌊9999/1001⌋ × 1001 = 9 × 1001 = 9009. Step 9: Count of multiples of 1001: 9 - 1 + 1 = 9. Step 10: Answer = 116 - 9 = 107.

Question 10

A number N has the property that when divided by 8 it leaves remainder 3, and when divided by 5 it leaves remainder 2. If N is a three-digit number, how many such numbers exist?

Accepted Answer

Step 1: N ≡ 3 (mod 8) and N ≡ 2 (mod 5). Step 2: From first condition, N = 8k + 3. Step 3: Substitute into second: 8k + 3 ≡ 2 (mod 5), so 8k ≡ -1 ≡ 4 (mod 5). Step 4: Since 8 ≡ 3 (mod 5), we have 3k ≡ 4 (mod 5). Step 5: Multiply by 2 (inverse of 3 mod 5): k ≡ 8 ≡ 3 (mod 5), so k = 5m + 3. Step 6: Therefore N = 8(5m + 3) + 3 = 40m + 27. Step 7: For three-digit numbers: 100 ≤ 40m + 27 ≤ 999. Step 8: 73 ≤ 40m ≤ 972, so 1.825 ≤ m ≤ 24.3. Step 9: m ∈ {2, 3, 4, ..., 24}, giving 23 numbers.

Question 11

A number when divided by 15 leaves remainder 8. What is the remainder when the square of this number is divided by 15?

Accepted Answer

Step 1: Let N ≡ 8 (mod 15). Step 2: We need to find N² (mod 15). Step 3: N² ≡ 8² ≡ 64 (mod 15). Step 4: 64 = 4 × 15 + 4, so 64 ≡ 4 (mod 15). Step 5: Therefore, the remainder when N² is divided by 15 is 4.

Question 12

A certain number leaves remainder 4 when divided by 9, and remainder 6 when divided by 7. What is the remainder when this number is divided by 63?

Accepted Answer

Step 1: Let N ≡ 4 (mod 9) and N ≡ 6 (mod 7). Step 2: From first condition, N = 9k + 4. Step 3: Substitute into second: 9k + 4 ≡ 6 (mod 7), so 9k ≡ 2 (mod 7). Step 4: Since 9 ≡ 2 (mod 7), we have 2k ≡ 2 (mod 7), so k ≡ 1 (mod 7). Step 5: Therefore k = 7j + 1, and N = 9(7j + 1) + 4 = 63j + 13. Step 6: The remainder when N is divided by 63 is 13.

Question 13

A five-digit number is formed by writing the digits 1, 2, 3, 4, 5 in some order. How many such numbers are divisible by 4?

Accepted Answer

Step 1: A number is divisible by 4 if its last two digits form a number divisible by 4. Step 2: We need to check which two-digit numbers formed from {1,2,3,4,5} are divisible by 4. Step 3: Possible two-digit endings: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96. Step 4: From digits {1,2,3,4,5}, valid endings divisible by 4 are: 12, 24, 32, 52. Step 5: For each valid ending, the remaining 3 digits can be arranged in the first 3 positions in 3! = 6 ways. Step 6: Total = 4 × 6 = 24.

IBPS RRB PO Divisibility Rules

Divisibility Rules— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Divisibility Rules — Practice Questions

60-Second Revision — Divisibility Rules