Introduction to Number System and Divisibility Rules for RRB Exams
The Number System is the foundational bedrock of Quantitative Aptitude in Railway Recruitment Board (RRB) examinations, including RRB NTPC, RRB Group D, and RRB Technician Grade I & III. Understanding the classification of numbers, remainder theorems, unit digits, and divisibility rules enables aspirants to solve complex calculations rapidly without lengthy manual multiplication or division.
Mastering divisibility rules and number properties allows you to save critical seconds during computer-based tests (CBT). In this comprehensive guide, we break down every essential formula, property, and divisibility shortcut with clear step-by-step examples and practice questions designed specifically according to recent RRB exam patterns.
Topic Weightage and Importance
In RRB examinations, Mathematics accounts for 30 to 35 questions out of 100 in CBT 1 and CBT 2. Questions directly from Number System and Divisibility Rules usually account for 3 to 5 questions per shift. Furthermore, concepts of divisibility are applied across Simplification, HCF & LCM, and Data Interpretation.
| Exam Name | Expected Number of Questions | Difficulty Level |
|---|---|---|
| RRB NTPC CBT 1 | 3 - 4 Questions | Easy to Moderate |
| RRB NTPC CBT 2 | 4 - 5 Questions | Moderate to High |
| RRB Group D | 3 - 5 Questions | Easy to Moderate |
| RRB Technician (Grade I & III) | 2 - 4 Questions | Moderate |
Key Concepts and Formulas
1. Classification of Numbers
- Natural Numbers ($N$): Counting numbers $1, 2, 3, 4, \text{...}$
- Whole Numbers ($W$): Natural numbers including zero: $0, 1, 2, 3, \text{...}$
- Integers ($Z$): All positive and negative whole numbers: $ \text{...}, -2, -1, 0, 1, 2, \text{...}$
- Prime Numbers: Numbers greater than 1 with exactly two factors: 1 and the number itself (e.g., $2, 3, 5, 7, 11, 13, 17, 19$). Note: $2$ is the only even prime number.
- Composite Numbers: Numbers having more than two factors (e.g., $4, 6, 8, 9, 10$). Note: $1$ is neither prime nor composite.
- Co-prime Numbers: Two numbers $a$ and $b$ are co-prime if their Highest Common Factor $ \text{HCF}(a, b) = 1$.
2. Essential Divisibility Rules
| Divisor | Divisibility Rule |
|---|---|
| 2 | The last digit of the number must be even ($0, 2, 4, 6, 8$). |
| 3 | The sum of all digits must be divisible by $3$. |
| 4 | The number formed by the last two digits must be divisible by $4$. |
| 5 | The last digit must be either $0$ or $5$. |
| 6 | The number must be divisible by both $2$ and $3$. |
| 7 | Subtract twice the last digit from the remaining truncated number; the result must be divisible by $7$. |
| 8 | The number formed by the last three digits must be divisible by $8$. |
| 9 | The sum of all digits must be divisible by $9$. |
| 10 | The last digit must be $0$. |
| 11 | The absolute difference between the sum of digits at odd positions and the sum of digits at even positions must be $0$ or divisible by $11$. |
| 12 | The number must be divisible by both co-prime factors $3$ and $4$. |
| 72 | The number must satisfy divisibility rules of both $8$ and $9$. |
| 88 | The number must satisfy divisibility rules of both $8$ and $11$. |
3. Core Mathematical Formulas
- Division Algorithm: $ \text{Dividend} = ( \text{Divisor} \times \text{Quotient}) + \text{Remainder}$
- Sum of First $n$ Natural Numbers: $S_n = \frac{n(n + 1)}{2}$
- Sum of Squares of First $n$ Natural Numbers: $S_n^2 = \frac{n(n + 1)(2n + 1)}{6}$
- Sum of Cubes of First $n$ Natural Numbers: $S_n^3 = \frac{n^2(n + 1)^2}{4}$
- Number of Factors of a number $N = p^a \times q^b \times r^c$: $ \text{Total Factors} = (a + 1)(b + 1)(c + 1)$
Solved Examples (Step-by-Step)
Example 1: Finding an Unknown Digit for Divisibility by 9
Question: If the 7-digit number $5432x17$ is completely divisible by $9$, find the value of digit $x$.
Solution:
Step 1: Recall the rule for divisibility by $9$. The sum of all digits must be a multiple of $9$.
Step 2: Calculate the sum of digits: $5 + 4 + 3 + 2 + x + 1 + 7 = 22 + x$.
Step 3: Find the smallest single-digit value for $x$ such that $(22 + x)$ is divisible by $9$.
The smallest multiple of $9$ greater than $22$ is $27$.
$$22 + x = 27 ag{1}$$
$$x = 27 - 22 = 5$$
Answer: The value of $x$ is 5.
Example 2: Combined Divisibility by 72
Question: Find the value of $(x + y)$ if the 8-digit number $789x531y$ is divisible by $72$.
Solution:
Step 1: Break $72$ into co-prime factors: $72 = 8 \times 9$. The number must be divisible by both $8$ and $9$.
Step 2: Apply divisibility by $8$ on the last three digits ($31y$). For $31y$ to be divisible by $8$, $310 + y$ divided by $8$ gives remainder $6 + y$. So, $6 + y = 8 ightarrow y = 2$.
Step 3: Substitute $y = 2$ into the number: $789x5312$.
Step 4: Apply divisibility by $9$ on the sum of digits: $7 + 8 + 9 + x + 5 + 3 + 1 + 2 = 35 + x$.
Step 5: The next multiple of $9$ after $35$ is $36$. Therefore, $35 + x = 36 ightarrow x = 1$.
Step 6: Compute $(x + y) = 1 + 2 = 3$.
Answer: The value of $(x + y)$ is 3.
Example 3: Division Algorithm Problem
Question: In a division sum, the divisor is $10$ times the quotient and $5$ times the remainder. If the remainder is $46$, find the dividend.
Solution:
Step 1: Given $ \text{Remainder} = 46$.
Step 2: Divisor is $5$ times remainder: $ \text{Divisor} = 5 \times 46 = 230$.
Step 3: Divisor is $10$ times quotient: $230 = 10 \times \text{Quotient} ightarrow \text{Quotient} = 23$.
Step 4: Use Division Algorithm: $ \text{Dividend} = ( \text{Divisor} \times \text{Quotient}) + \text{Remainder}$.
$$ \text{Dividend} = (230 \times 23) + 46 = 5290 + 46 = 5336$$
Answer: The dividend is 5336.
Common Mistakes to Avoid
- Confusing Divisibility of 3 and 9 with 6: Do not check digit sums for divisibility by 6. For 6, test parity (even number) and sum of digits (divisible by 3).
- Forgetting Position Counting in Divisibility by 11: Positions should be counted from left to right or right to left consistently (1st, 2nd, 3rd, etc.), not based on digit values.
- Assuming 1 is a Prime Number: Remember that $1$ is neither prime nor composite; $2$ is the smallest and only even prime number.
- Ignoring Co-prime Conditions: When checking composite divisibility like 12 or 72, factors must be co-prime ($ \text{HCF}=1$). For 12 use 3 and 4 (not 2 and 6).
Practice Questions with Solutions
- Q1: What is the remainder when $7^{84}$ is divided by $342$?
- Q2: Find the smallest 5-digit number which is exactly divisible by $12, 18,$ and $21$.
- Q3: How many prime numbers exist between $1$ and $50$?
- Q4: If $48327*8$ is divisible by $11$, find the digit in place of $*$.
- Q5: Find the total number of factors of $360$.
Solutions:
Solution 1: Rewrite $7^{84}$ as $(7^3)^{28} = 343^{28}$. Notice $343 = 342 + 1$. By remainder theorem, $(342 + 1)^{28} \bmod 342 = 1^{28} = 1$. The remainder is 1.
Solution 2: $ \text{LCM}(12, 18, 21) = 252$. Smallest 5-digit number is $10000$. Divide $10000$ by $252$: $10000 = 252 \times 39 + 172$. Required number $= 10000 + (252 - 172) = 10080$. The answer is 10080.
Solution 3: The prime numbers under 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. There are 15 prime numbers.
Solution 4: Let the unknown digit be $x$. Sum of digits at odd positions: $8 + 7 + 3 + 4 = 22$. Sum of digits at even positions: $x + 2 + 8 = 10 + x$. Difference: $22 - (10 + x) = 12 - x$. For divisibility by 11, $12 - x = 11 ightarrow x = 1$. The answer is 1.
Solution 5: Prime factorization of $360 = 2^3 \times 3^2 \times 5^1$. Total factors $= (3 + 1)(2 + 1)(1 + 1) = 4 \times 3 \times 2 = 24$. The answer is 24.
Frequently Asked Questions (FAQs)
Q1: What is the fastest way to check divisibility by 11?
Add alternate digits from left to right to create two sums (odd-positioned digits and even-positioned digits). Subtract one sum from the other. If the result is 0, 11, -11, 22, etc., the number is divisible by 11.
Q2: Why are co-prime numbers important in divisibility rules?
To check divisibility for a composite divisor $N = a \times b$, $a$ and $b$ must be co-prime ($ \text{HCF}(a, b) = 1$). If they share factors, checking divisibility by both will produce false positives.
Q3: What is the unit digit formula for exponents?
Unit digits repeat in cyclicity of 4 (except 0, 1, 5, 6 which have cyclicity 1). Divide the power by 4 and use the remainder as the new exponent to find the unit digit.
Conclusion and Final Tips
Number System and Divisibility Rules form the backbone of numerical aptitude in RRB exams. Memorize the divisibility rules up to 13 and prime factor rules for 72 and 88, as these appear repeatedly in RRB NTPC and Group D papers. Practice 10-15 questions daily to build speed and accuracy. Good luck with your preparation!
