Question 1

A number is divisible by both 4 and 9. Which of the following is definitely true about this number?

Accepted Answer

Step 1: A number divisible by 4 must have its last two digits divisible by 4 (divisibility rule for 4). Step 2: A number divisible by 9 must have the sum of its digits divisible by 9 (divisibility rule for 9). Step 3: If a number is divisible by both 4 and 9, and since 4 and 9 are coprime (HCF = 1), the number must be divisible by their product: 4 × 9 = 36. Step 4: Therefore, any number divisible by both 4 and 9 is divisible by 36. Therefore, answer is B.

Question 2

A number is divisible by both 4 and 9. Which of the following statements about this number must be true?

I. The number is divisible by 36.
II. The number is divisible by 6.
III. The sum of its digits is divisible by 9.

Accepted Answer

Step 1: A number divisible by both 4 and 9 must be divisible by their LCM. Since 4 = 2² and 9 = 3², and gcd(4,9) = 1, LCM(4,9) = 4 × 9 = 36. Therefore, the number is divisible by 36. Statement I is TRUE.

Step 2: If a number is divisible by 36, it is divisible by all factors of 36. Since 6 is a factor of 36 (36 = 6 × 6), the number must be divisible by 6. Statement II is TRUE.

Step 3: A number is divisible by 9 if and only if the sum of its digits is divisible by 9 (divisibility rule for 9). Since our number is divisible by 9, the sum of its digits must be divisible by 9. Statement III is TRUE.

Step 4: All three statements are true. Therefore, answer is D.

Question 3

A number when divided by 12 leaves a remainder of 5. When the same number is divided by 18, it leaves a remainder of 11. Find the smallest positive integer that satisfies both conditions.

Accepted Answer

Step 1: Let the number be N. From the first condition, N ≡ 5 (mod 12), so N = 12k + 5 for some integer k ≥ 0. Step 2: Substitute into the second condition: 12k + 5 ≡ 11 (mod 18), which gives 12k ≡ 6 (mod 18). Step 3: Divide by GCD(12, 18) = 6: 2k ≡ 1 (mod 3). Step 4: Solve 2k ≡ 1 (mod 3). Testing k = 0, 1, 2: k = 2 works since 2(2) = 4 ≡ 1 (mod 3). Step 5: So k = 2 + 3m for integer m ≥ 0. The smallest is k = 2. Step 6: N = 12(2) + 5 = 29. Verify: 29 ÷ 12 = 2 remainder 5 ✓; 29 ÷ 18 = 1 remainder 11 ✓. Therefore, answer is B.

RRB Group D Divisibility Rules

Divisibility Rules— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Divisibility Rules — Practice Questions

60-Second Revision — Divisibility Rules