Question 1

Which number is divisible by 11?

Accepted Answer

Step 1: A number is divisible by 11 if the difference between the sum of digits in odd positions and the sum of digits in even positions is either 0 or divisible by 11. Step 2: For 5742: Odd positions (right to left): 2 + 7 = 9. Even positions: 4 + 5 = 9. Difference = 9 - 9 = 0. Step 3: Since difference is 0, 5742 is divisible by 11. Step 4: Verify: 5742 ÷ 11 = 522. ✓

Question 2

A number is divisible by 4 if its last two digits form a number divisible by 4. Which of the following is NOT divisible by 4?

Accepted Answer

Step 1: For divisibility by 4, check only the last two digits. Step 2: For 3216: last two digits are 16; 16 ÷ 4 = 4. ✓ Divisible by 4. Step 3: For 5428: last two digits are 28; 28 ÷ 4 = 7. ✓ Divisible by 4. Step 4: For 7634: last two digits are 34; 34 ÷ 4 = 8.5. ✗ NOT divisible by 4. Step 5: For 8920: last two digits are 20; 20 ÷ 4 = 5. ✓ Divisible by 4. Therefore, 7634 is NOT divisible by 4.

Question 3

Which of the following numbers 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: For 4527: digit sum = 4 + 5 + 2 + 7 = 18. Step 3: 18 ÷ 3 = 6. ✓ Therefore, 4527 is divisible by 3. Step 4: Verify: 4527 ÷ 3 = 1509. ✓

Question 4

A number is divisible by 5 if its last digit is 0 or 5. How many two-digit numbers are divisible by 5?

Accepted Answer

Step 1: Two-digit numbers range from 10 to 99. Step 2: Numbers divisible by 5 must end in 0 or 5. Step 3: Numbers ending in 0: 10, 20, 30, 40, 50, 60, 70, 80, 90 = 9 numbers. Step 4: Numbers ending in 5: 15, 25, 35, 45, 55, 65, 75, 85, 95 = 9 numbers. Step 5: Total = 9 + 9 = 18 two-digit numbers divisible by 5.

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 264: 264 ÷ 24 = 11. Therefore, 264 is divisible by both 6 and 8.

Question 6

A number N is divisible by 12. When N is divided by 18, the remainder is 6. What is the remainder when N is divided by 36?

Accepted Answer

Step 1: N is divisible by 12, so N = 12k for some integer k. Step 2: N ≡ 6 (mod 18), so 12k ≡ 6 (mod 18). Step 3: Simplify: 12k ≡ 6 (mod 18) → 6(2k) ≡ 6 (mod 18) → 2k ≡ 1 (mod 3). Step 4: Since 2 × 2 = 4 ≡ 1 (mod 3), we have k ≡ 2 (mod 3), so k = 3m + 2. Step 5: N = 12(3m + 2) = 36m + 24. Step 6: Therefore, N ≡ 24 (mod 36), so the remainder is 24.

Question 7

A number N is such that when divided by 6, 8, and 9, it leaves remainders 5, 7, and 8 respectively. What is the smallest positive value of N?

Accepted Answer

Step 1: Observe the pattern: N ≡ 5 (mod 6), N ≡ 7 (mod 8), N ≡ 8 (mod 9). Notice that in each case, remainder = divisor - 1. Step 2: This means N + 1 ≡ 0 (mod 6), N + 1 ≡ 0 (mod 8), N + 1 ≡ 0 (mod 9). Step 3: So N + 1 is divisible by 6, 8, and 9. Therefore, N + 1 must be divisible by LCM(6, 8, 9). Step 4: Find LCM(6, 8, 9): 6 = 2 × 3, 8 = 2³, 9 = 3². LCM = 2³ × 3² = 8 × 9 = 72. Step 5: N + 1 = 72k for positive integer k. Smallest positive N occurs when k = 1, so N + 1 = 72, thus N = 71.

Question 8

How many 4-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: Find 4-digit multiples of 77: smallest is ⌈1000/77⌉ × 77 = 14 × 77 = 1078; largest is ⌊9999/77⌋ × 77 = 129 × 77 = 9933. Step 3: Count of 4-digit multiples of 77 = 129 - 14 + 1 = 116. Step 4: Now find 4-digit multiples of LCM(7,11,13) = 1001. Smallest is ⌈1000/1001⌉ × 1001 = 1 × 1001 = 1001; largest is ⌊9999/1001⌋ × 1001 = 9 × 1001 = 9009. Step 5: Count of 4-digit multiples of 1001 = 9 - 1 + 1 = 9. Step 6: Numbers divisible by 77 but NOT by 13 = 116 - 9 = 107.

Question 9

A 5-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: List all 2-digit numbers formed from {1,2,3,4,5} that are divisible by 4: 12 (12÷4=3✓), 24 (24÷4=6✓), 32 (32÷4=8✓), 52 (52÷4=13✓). Step 3: Check others: 13, 14, 15, 21, 23, 25, 31, 34, 35, 41, 42, 43, 45, 51, 53, 54 are not divisible by 4. Step 4: Valid last two digits: {12, 24, 32, 52}. Step 5: For each valid ending, the first three positions use the remaining 3 digits in any order. Number of arrangements = 3! = 6. Step 6: Total = 4 × 6 = 24.

SSC CPO Divisibility Rules

SSC CPO Divisibility Rules — Past Exam Questions

Divisibility Rules— Rules & Concept

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60-Second Revision — Divisibility Rules