Question 1

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

Accepted Answer

Step 1: Find the smallest number ≥ 100 divisible by 6. 100 ÷ 6 = 16.67, so the smallest is 6 × 17 = 102. Step 2: Find the largest number ≤ 200 divisible by 6. 200 ÷ 6 = 33.33, so the largest is 6 × 33 = 198. Step 3: Count multiples of 6 from 102 to 198: These are 6×17, 6×18, ..., 6×33. Step 4: Number of terms = 33 - 17 + 1 = 17. Therefore, there are 17 numbers.

Question 2

A number is divisible by both 4 and 6. Which of the following is definitely true about this number?

Accepted Answer

Step 1: Find LCM of 4 and 6. Prime factorization: 4 = 2², 6 = 2 × 3. LCM = 2² × 3 = 12. Step 2: If a number is divisible by both 4 and 6, it must be divisible by their LCM, which is 12. Step 3: Check: 12 is divisible by 4 (12÷4=3) and 6 (12÷6=2). Therefore, the number must be divisible by 12.

Question 3

A number 7a8b is divisible by both 4 and 9. What are the values of a and b?

Accepted Answer

Step 1: For divisibility by 4, the last two digits (8b) must form a number divisible by 4. Testing: 80 (÷4=20✓), 81, 82, 83, 84 (÷4=21✓), 85, 86, 87, 88 (÷4=22✓), 89. So b ∈ {0, 4, 8}. Step 2: For divisibility by 9, sum of all digits must be divisible by 9. Sum = 7 + a + 8 + b = 15 + a + b. Step 3: For divisibility by 9, (15 + a + b) must be divisible by 9. So (a + b) ≡ 3 (mod 9). Step 4: If b = 0, a = 3 (sum = 18✓). If b = 4, a = 8 (sum = 27✓). If b = 8, a = 4 (sum = 27✓). Step 5: Among options, a = 3, b = 0 is the answer.

Question 4

Which of the following numbers is divisible by 11?

Accepted Answer

Step 1: Apply divisibility rule for 11: The difference between the sum of digits in odd positions and sum of digits in even positions must be divisible by 11. Step 2: Check option B (5742): Odd positions (right to left): 2 + 7 = 9. Even positions: 4 + 5 = 9. Difference: 9 - 9 = 0, which is divisible by 11. Step 3: Verify: 5742 ÷ 11 = 522. Therefore, 5742 is divisible by 11.

Question 5

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 8 and 5 is 40. Step 4: Verify: 40 ÷ 8 = 5 (remainder 0) ✓, 40 ÷ 5 = 8 (remainder 0) ✓.

Question 6

How many three-digit numbers are divisible by both 6 and 8?

Accepted Answer

Step 1: Find LCM(6,8). 6 = 2 × 3, 8 = 2³. LCM = 2³ × 3 = 24. Step 2: A number is divisible by both 6 and 8 if and only if it is divisible by 24. Step 3: Find three-digit multiples of 24. Smallest three-digit number: 100. Largest three-digit number: 999. Step 4: Smallest three-digit multiple of 24: ⌈100/24⌉ × 24 = 5 × 24 = 120. Step 5: Largest three-digit multiple of 24: ⌊999/24⌋ × 24 = 41 × 24 = 984. Step 6: Count multiples: 24k where 5 ≤ k ≤ 41. Number of values = 41 - 5 + 1 = 37.

Question 7

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 8 (72 ÷ 8 = 9), by 9 (72 ÷ 9 = 8), and by 12 (72 ÷ 12 = 6). All options except 5 divide 72. Step 5: A number divisible by 72 is NOT necessarily divisible by 5 (e.g., 72 itself). Therefore, the answer is that it must be divisible by 12.

Question 8

A number is formed by writing digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in some order. Which of the following is always true about this number?

Accepted Answer

Step 1: The number contains each digit 1-9 exactly once. Step 2: Sum of digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45. Step 3: A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Since 45 = 9 × 5, the sum is divisible by 9. Step 4: Therefore, any such number is divisible by 9. Step 5: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. Since 45 is divisible by 3, any such number is divisible by 3. Step 6: Since the number is divisible by 9, it is automatically divisible by 3. The strongest statement is divisibility by 9.

Question 9

A 6-digit number 2A3B4C is divisible by 7, 11, and 13. If A, B, C are single digits, what is the value of A × B × C?

Accepted Answer

Step 1: The number must be divisible by LCM(7, 11, 13). Since 7, 11, and 13 are all prime, LCM = 7 × 11 × 13 = 1001. Step 2: The 6-digit number 2A3B4C can be written as 203040 + 1000A + 100B + 10C. Step 3: For divisibility by 1001, we need 203040 + 1000A + 100B + 10C ≡ 0 (mod 1001). Step 4: Find 203040 mod 1001. 203040 = 1001 × 202 + 838, so 203040 ≡ 838 (mod 1001). Step 5: We need 838 + 1000A + 100B + 10C ≡ 0 (mod 1001). Since 1000 ≡ -1 (mod 1001), we have 838 - A + 100B + 10C ≡ 0 (mod 1001). Step 6: Rearranging: A - 100B - 10C ≡ 838 (mod 1001). Step 7: Since A, B, C are single digits (0-9), the expression A - 100B - 10C ranges from 0 - 900 - 90 = -990 to 9 - 0 - 0 = 9. Step 8: For this to equal 838 (mod 1001), we need A - 100B - 10C = 838 - 1001 = -163. Step 9: So A - 100B - 10C = -163, which means 100B + 10C - A = 163. Step 10: Testing values: If B = 1, then 100 + 10C - A = 163, so 10C - A = 63. If C = 7, then A = 70 - 63 = 7. Check: 100(1) + 10(7) - 7 = 100 + 70 - 7 = 163 ✓. Step 11: The number is 273744. Verify: 273744 ÷ 1001 = 273.47... Let me recalculate. 273744 ÷ 7 = 39106.29... This doesn't work. Let me reconsider. Step 12: Actually, 1001 × 273 = 273273, and 1001 × 274 = 274274. So we need 2A3B4C = 274274. This gives A = 7, B = 2, C = 7. Verify: 274274 ÷ 1001 = 274 ✓. Step 13: A × B × C = 7 × 2 × 7 = 98.

Question 10

A number when divided by 12 leaves remainder 7. When the same number is divided by 18, what is the remainder? (Given: the number is a 4-digit positive integer less than 2000)

Accepted Answer

Step 1: If a number N leaves remainder 7 when divided by 12, then N = 12k + 7 for some integer k. Step 2: We need to find N mod 18. Since N = 12k + 7, we substitute values of k such that N < 2000. Step 3: For k = 1: N = 19 (remainder when divided by 18 = 1). For k = 7: N = 91 (91 ÷ 18 = 5 remainder 1). For k = 13: N = 163 (163 ÷ 18 = 9 remainder 1). For k = 157: N = 1891 (1891 ÷ 18 = 105 remainder 1). Step 4: The pattern shows remainder is always 1 when N = 12k + 7 is divided by 18 (since 12k + 7 ≡ 12k + 7 (mod 18), and 12k mod 18 cycles through 0, 12, 6, 0... while 7 remains constant, giving remainder 1 for all valid k).

Question 11

A number is divisible by both 8 and 9. If the number is also divisible by 5, what is the smallest such 4-digit number?

Accepted Answer

Step 1: Find LCM(8, 9, 5). Since 8 = 2³, 9 = 3², 5 = 5, and they share no common factors, LCM = 8 × 9 × 5 = 360. Step 2: The number must be a multiple of 360. Step 3: Find the smallest 4-digit multiple of 360. The smallest 4-digit number is 1000. Step 4: Divide 1000 by 360: 1000 ÷ 360 = 2.777... Step 5: Round up to the next integer: 3. Step 6: Multiply: 360 × 3 = 1080. Step 7: Verify: 1080 ÷ 8 = 135 ✓, 1080 ÷ 9 = 120 ✓, 1080 ÷ 5 = 216 ✓.

Question 12

A 5-digit number 3A5B2 is divisible by 36. If A and B are single digits, what is the maximum value of A + B?

Accepted Answer

Step 1: A number is divisible by 36 if and only if it is divisible by both 4 and 9 (since 36 = 4 × 9 and gcd(4,9) = 1). Step 2: 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 → 12 ✓, B = 3 → 32 ✓, B = 5 → 52 ✓, B = 7 → 72 ✓, B = 9 → 92 ✓. So B ∈ {1, 3, 5, 7, 9}. Step 3: Divisibility by 9: Sum of digits 3 + A + 5 + B + 2 = 10 + A + B must be divisible by 9. So A + B ≡ 8 (mod 9), meaning A + B ∈ {8, 17}. Step 4: To maximize A + B, we want A + B = 17. Since A and B are single digits (0-9), and B ∈ {1, 3, 5, 7, 9}, if B = 9, then A = 8. Check: 38592. Sum = 3+8+5+9+2 = 27, divisible by 9 ✓. Last two digits: 92 ÷ 4 = 23 ✓. Step 5: Maximum A + B = 8 + 9 = 17.

Question 13

A number N is formed by writing the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in some order. How many such numbers are divisible by 11?

Accepted Answer

Step 1: The number N has 9 digits using each of 1-9 exactly once. Step 2: For divisibility by 11, the alternating sum of digits must be divisible by 11. If positions are numbered 1-9 from left to right, the alternating sum is: d₁ - d₂ + d₃ - d₄ + d₅ - d₆ + d₇ - d₈ + d₉. Step 3: Sum of all digits 1+2+...+9 = 45. Step 4: Let S_odd = sum of digits in odd positions (1,3,5,7,9) and S_even = sum of digits in even positions (2,4,6,8). Then S_odd + S_even = 45 and the alternating sum = S_odd - S_even. Step 5: For divisibility by 11: S_odd - S_even ≡ 0 (mod 11). Step 6: From S_odd + S_even = 45, we get S_odd = (45 + (S_odd - S_even))/2. For S_odd - S_even = 0: S_odd = S_even = 22.5 (impossible, not integer). For S_odd - S_even = 11: S_odd = 28, S_even = 17. For S_odd - S_even = -11: S_odd = 17, S_even = 28. Step 7: We need to count partitions of {1,2,...,9} into two sets of sizes 5 and 4 with sums 28 and 17 (or vice versa). Step 8: Finding subsets of {1,...,9} with sum 28 and size 5: {9,8,7,3,1}, {9,8,6,4,1}, {9,8,6,3,2}, {9,7,6,5,1}, {9,7,6,4,2}, {9,7,5,4,3}, {8,7,6,5,2}, {8,7,6,4,3}. That's 8 subsets. Step 9: For each valid partition, we can arrange the 5 digits in odd positions in 5! ways and 4 digits in even positions in 4! ways. Total = 8 × 5! × 4! = 8 × 120 × 24 = 23040.

Question 14

A number is formed by concatenating the integers 1, 2, 3, ..., 99 to get 123456789101112...9899. What is the remainder when this number is divided by 8?

Accepted Answer

Step 1: For divisibility by 8, only the last 3 digits of a number matter. Step 2: The concatenated number ends with ...9899, so the last 3 digits are 899. Step 3: Find 899 mod 8. Step 4: 899 = 8 × 112 + 3, so 899 mod 8 = 3. Step 5: Therefore, the remainder when the entire number is divided by 8 is 3.

SBI Clerk Divisibility Rules

Divisibility Rules— Rules & Concept

Key Points to Remember

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Divisibility Rules — Practice Questions

60-Second Revision — Divisibility Rules