Number System — RRB NTPC Study Notes

Overview

The Number System is the foundation of mathematics and a high-weightage topic in RRB NTPC. Expect 3–5 direct questions from this area, plus many indirect applications in Time-Speed-Distance, Ratio-Proportion, and Data Interpretation. Mastery here builds confidence for the entire Math section.

This topic tests your understanding of different number categories (natural, whole, integers, rationals), your speed with divisibility checks, and your ability to find factors, multiples, LCM and HCF efficiently. Questions range from direct definitions to multi-step problems requiring divisibility rules and prime factorization. The key is to know the properties cold and practice mental arithmetic so you can solve under time pressure.

Strong performance in Number System gives you a solid platform: many algebra and arithmetic problems ultimately depend on number properties. Treat this as your warm-up and scoring zone.

Key Concepts

• **Natural Numbers (N)**: Counting numbers starting from 1: {1, 2, 3, 4, ...}. Used when counting discrete objects. Zero is not a natural number.

• **Whole Numbers (W)**: Natural numbers plus zero: {0, 1, 2, 3, ...}. Represents quantities that can be zero (e.g., items in an empty basket).

• **Integers (Z)**: All positive and negative whole numbers including zero: {..., -3, -2, -1, 0, 1, 2, 3, ...}. No fractions or decimals.

• **Rational Numbers (Q)**: Any number expressible as p/q where p and q are integers and q ≠ 0. Includes terminating decimals (0.5 = 1/2) and repeating decimals (0.333... = 1/3). Every integer is rational (5 = 5/1).

• **Divisibility**: A number a is divisible by b if a ÷ b leaves no remainder. Divisibility tests help you check this mentally without performing full division.

• **Factors**: Numbers that divide a given number exactly. For 12: factors are 1, 2, 3, 4, 6, 12. Every number has at least two factors: 1 and itself.

• **Multiples**: Results of multiplying a number by integers. Multiples of 5: 5, 10, 15, 20, ... A number has infinitely many multiples.

• **Prime vs Composite**: A prime has exactly two factors (1 and itself); examples: 2, 3, 5, 7, 11. A composite has more than two factors; examples: 4, 6, 8, 9. Note: 1 is neither prime nor composite; 2 is the only even prime.

Formulas / Key Facts

**Divisibility Rules (must memorize):**

• **By 2**: Last digit is 0, 2, 4, 6, or 8.
• **By 3**: Sum of all digits is divisible by 3. (Example: 123 → 1+2+3=6, divisible by 3.)
• **By 4**: Last two digits form a number divisible by 4. (Example: 316 → 16 is divisible by 4.)
• **By 5**: Last digit is 0 or 5.
• **By 6**: Divisible by both 2 and 3 simultaneously.
• **By 8**: Last three digits form a number divisible by 8.
• **By 9**: Sum of all digits is divisible by 9.
• **By 10**: Last digit is 0.
• **By 11**: Difference between sum of digits at odd places and sum at even places is 0 or divisible by 11. (Example: 1331 → (1+3)-(3+1)=0, divisible by 11.)

**Number of Factors Formula**: If N = p₁^a × p₂^b × p₃^c (prime factorization), then number of factors = (a+1)(b+1)(c+1).

**Sum of Factors Formula**: For N = p^a, sum = (p^(a+1) - 1)/(p - 1). For composite factorizations, multiply the sum formulas of each prime power.

**Co-prime Numbers**: Two numbers whose HCF is 1 (no common factor except 1). Example: 8 and 15 are co-prime.

**Perfect Number**: A number equal to the sum of its proper divisors (excluding itself). Example: 6 = 1+2+3.

**Number of trailing zeros in N!**: Count how many times 5 appears in prime factorization of N! = floor(N/5) + floor(N/25) + floor(N/125) + ...

Worked Examples

**Example 1: Check divisibility of 4872 by 8**

Step 1: Extract last three digits: 872. Step 2: Check 872 ÷ 8 = 109 exactly. **Answer**: Yes, 4872 is divisible by 8.

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**Example 2: Find the number of factors of 360**

Step 1: Prime factorization: 360 = 2³ × 3² × 5¹. Step 2: Apply formula: (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24. **Answer**: 360 has 24 factors.

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**Example 3: What is the smallest number that must be added to 1056 to make it divisible by 23?**

Step 1: Divide 1056 by 23: 1056 ÷ 23 = 45 remainder 21. Step 2: To make it divisible, we need to reach the next multiple of 23. Step 3: Required addition = 23 - 21 = 2. **Answer**: Add 2 (1056 + 2 = 1058, which is 46 × 23).

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**Example 4: How many numbers between 1 and 100 are divisible by both 3 and 5?**

Step 1: Numbers divisible by both 3 and 5 are divisible by LCM(3,5) = 15. Step 2: Multiples of 15 between 1 and 100: 15, 30, 45, 60, 75, 90. Step 3: Count = floor(100/15) = 6. **Answer**: 6 numbers.

Common Mistakes

**Mistake 1**: Confusing "divisible by 6" with "divisible by 2 or 3" → **Fix**: Divisibility by 6 requires BOTH conditions simultaneously (even number AND digit-sum divisible by 3).

**Mistake 2**: Thinking 1 is prime → **Fix**: By definition, primes have exactly two distinct factors. 1 has only one factor (itself), so it is neither prime nor composite.

**Mistake 3**: Applying the divisibility rule for 4 to the last digit only → **Fix**: Check the last TWO digits. Similarly, for 8 check last THREE digits.

**Mistake 4**: Miscalculating the sum of digits for divisibility by 9 or 3 → **Fix**: Write down the sum step-by-step. For 4857: 4+8+5+7=24; 24÷3=8 (divisible by 3), but 24÷9 is not whole (not divisible by 9).

**Mistake 5**: Forgetting that 2 is the only even prime → **Fix**: Students often exclude 2 from prime lists or call it composite. Remember: 2 is prime, and it's the smallest prime.

Quick Reference

• Natural: {1,2,3,...}; Whole: {0,1,2,...}; Integers: {...,-1,0,1,...}; Rationals: p/q form.
• Divisibility by 3 → sum of digits divisible by 3; by 9 → sum divisible by 9.
• Divisibility by 4 → last 2 digits; by 8 → last 3 digits.
• Number of factors of p₁^a × p₂^b = (a+1)(b+1).
• HCF = product of lowest powers of common primes; LCM = product of highest powers.
• 1 is neither prime nor composite; 2 is the only even prime.