Question 1

Which of the following numbers is NOT 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: Check 3216: last two digits = 16; 16 ÷ 4 = 4 (divisible). Step 3: Check 4528: last two digits = 28; 28 ÷ 4 = 7 (divisible). Step 4: Check 5634: last two digits = 34; 34 ÷ 4 = 8.5 (NOT divisible). Step 5: Check 7840: last two digits = 40; 40 ÷ 4 = 10 (divisible). Therefore, 5634 is NOT divisible by 4.

Question 2

A number is divisible by 3 if the sum of its digits is divisible by 3. Which of the following numbers is divisible by 3?

Accepted Answer

Step 1: Apply the divisibility rule for 3: sum of digits must be divisible by 3. Step 2: For 4527: 4 + 5 + 2 + 7 = 18, and 18 ÷ 3 = 6 (divisible by 3). Step 3: Verify: 4527 ÷ 3 = 1509 (exact). Therefore, 4527 is divisible by 3.

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 11 if the difference between the sum of digits in odd positions and the sum of digits in even positions is divisible by 11. Which of the following is divisible by 11?

Accepted Answer

Step 1: Apply the divisibility rule for 11. Step 2: For 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 (exact). Therefore, 5742 is divisible by 11.

Question 5

How many three-digit numbers are divisible by 6 but not by 9?

Accepted Answer

Step 1: Three-digit numbers range from 100 to 999. Step 2: Count multiples of 6: ⌊999/6⌋ - ⌊99/6⌋ = 166 - 16 = 150. Step 3: A number divisible by both 6 and 9 must be divisible by LCM(6,9) = 18. Step 4: Count multiples of 18: ⌊999/18⌋ - ⌊99/18⌋ = 55 - 5 = 50. Step 5: Numbers divisible by 6 but not by 9 = 150 - 50 = 100.

Question 6

A five-digit number 4A6B2 is divisible by 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→12✓, B=3→32✓, B=5→52✓, B=7→72✓, B=9→92✓. Step 2: For divisibility by 11, alternating sum (4-A+6-B+2) = (12-A-B) must be divisible by 11. So 12-A-B ≡ 0 (mod 11), meaning A+B = 1 or A+B = 12. Step 3: Since A and B are single digits and B ∈ {1,3,5,7,9}, if A+B=1, then A=0 and B=1. If A+B=12, then A+B=12 with B odd. Testing B=1: A=11 (invalid). Testing B=3: A=9 ✓. Step 4: Verify 49632: divisible by 4 (32÷4=8✓) and by 11 (4-9+6-3+2=0✓). Therefore A+B = 12.

Question 7

A number is divisible by 15. Which of the following statements must be true?

Accepted Answer

Step 1: A number divisible by 15 must be divisible by all factors of 15. Step 2: 15 = 3 × 5, so factors are 1, 3, 5, 15. Step 3: Therefore, any number divisible by 15 is divisible by both 3 and 5. Step 4: Divisibility by 3 requires digit sum divisible by 3. Divisibility by 5 requires last digit to be 0 or 5. Step 5: The only statement that must be true is that the number is divisible by both 3 and 5 (equivalently, by 15). Among the options, the correct one identifies a necessary property.

Question 8

A number when divided by 12 leaves remainder 7. What remainder will it leave when divided by 4?

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: Rewrite: N = 12k + 7 = 4(3k) + 7 = 4(3k + 1) + 3. Step 3: Therefore, N leaves remainder 3 when divided by 4.

Question 9

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 gcd(8,9) = 1, LCM(8,9) = 8 × 9 = 72. Step 3: Any number divisible by 72 is divisible by all factors of 72, including 36. Therefore, the answer is 36.

Question 10

A six-digit number 5a7b8c is divisible by 4, 5, and 11 simultaneously. If a, b, c are single digits, what is the maximum value of a + b + c?

Accepted Answer

Step 1: Divisibility by 5 requires c ∈ {0, 5}. Step 2: Divisibility by 4 requires the last two digits (8c) to form a number divisible by 4. If c=0: 80÷4=20 ✓. If c=5: 85÷4=21.25 ✗. So c=0. Step 3: Divisibility by 11 requires alternating sum: (5+7+8) - (a+b+0) ≡ 0 (mod 11) → 20 - a - b ≡ 0 (mod 11) → a + b ≡ 20 ≡ 9 (mod 11). Step 4: Since a, b are single digits (0-9), we have 0 ≤ a+b ≤ 18. Step 5: Values of a+b satisfying a+b ≡ 9 (mod 11) in range [0,18]: a+b ∈ {9}. (Note: 20 is outside range, and -2 is negative.) Step 6: To maximize a+b+c = a+b+0, we need a+b=9. Step 7: Maximum a+b+c = 9+0 = 9.

Question 11

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

Accepted Answer

Step 1: A number divisible by both 6 and 8 must be divisible by LCM(6,8). Step 2: 6 = 2×3, 8 = 2³, so LCM(6,8) = 2³×3 = 24. Step 3: Four-digit multiples of 24: smallest is 1008 (1000÷24≈41.67, so 42×24=1008), largest is 9984 (9999÷24≈416.6, so 416×24=9984). Step 4: Count of multiples of 24: (416-42+1) = 375. Step 5: Now, which of these are divisible by 12? A number divisible by 24 is automatically divisible by 12 (since 24=12×2). Step 6: Therefore, ALL four-digit multiples of 24 are divisible by 12. Step 7: The answer is 0, because there are NO four-digit numbers divisible by both 6 and 8 but NOT divisible by 12.

Question 12

A five-digit number 3a4b2 is divisible by both 8 and 11. What is the value of a + b?

Accepted Answer

Step 1: For divisibility by 8, the last three digits (b2) must form a number divisible by 8. Since the last three digits are 4b2, we need 4b2 ≡ 0 (mod 8). Step 2: Testing: 402÷8=50.25 (no), 412÷8=51.5 (no), 422÷8=52.75 (no), 432÷8=54 (yes), 442÷8=55.25 (no), 452÷8=56.5 (no), 462÷8=57.75 (no), 472÷8=59 (yes), 482÷8=60.25 (no), 492÷8=61.5 (no). So b ∈ {3, 7}. Step 3: For divisibility by 11, alternating sum of digits must be divisible by 11: (3 + 4 + 2) - (a + b) = 9 - a - b ≡ 0 (mod 11). Step 4: So a + b ≡ 9 (mod 11). Step 5: If b = 3: a + 3 ≡ 9 (mod 11), so a = 6. If b = 7: a + 7 ≡ 9 (mod 11), so a = 2. Step 6: Check both: 36432 (a=6, b=3): 36432÷11=3312 ✓, 32472 (a=2, b=7): 32472÷11=2952 ✓. Step 7: Both work, but the question asks for a + b. For a=6, b=3: a+b=9. For a=2, b=7: a+b=9. Therefore, a + b = 9.

SSC MTS Divisibility Rules

SSC MTS Divisibility Rules — Past Exam Questions

Divisibility Rules— Rules & Concept

Key Points to Remember

Exam-Specific Tips

60-Second Revision — Divisibility Rules