Think of numbers as a family with a hierarchy. At the top level, you have the real number line — everything from negative infinity to positive infinity. Within that, the subsets you need for RRB Group D are specific and manageable.
Natural Numbers (प्राकृत संख्या): 1, 2, 3, 4, ... — the counting numbers. Some books include 0, but for exam purposes, treat natural numbers as starting from 1 unless the question says otherwise.
Whole Numbers: 0, 1, 2, 3, ... — natural numbers plus zero.
Integers (पूर्णांक): ..., -3, -2, -1, 0, 1, 2, 3, ... — the complete number line without fractions.
Even and Odd Numbers: Even numbers are perfectly divisible by 2 (2, 4, 6, ...). Odd numbers leave a remainder of 1 when divided by 2 (1, 3, 5, ...).
Prime Numbers: Numbers with exactly two factors — 1 and the number itself. The smallest prime is 2. It is also the only even prime — every other even number is divisible by 2, giving it at least three factors.
Composite Numbers: Numbers with more than two factors. Note that 1 is neither prime nor composite — it has exactly one factor (itself).
Here is an analogy that sticks: think of prime numbers as atoms. Just as every molecule is built from atoms, every composite number is built from primes. The process of breaking a composite number into its prime building blocks is called prime factorization, and it is the single most powerful tool in number system problems.
Perfect Squares: Numbers like 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 — results of squaring integers. Recognizing these on sight saves seconds every time a sequence or factor question appears.
| Type | Definition | Examples | |------|-----------|---------| | Natural | Counting numbers from 1 | 1, 2, 3, 100 | | Whole | Natural + 0 | 0, 1, 2 | | Integer | Whole + negatives | -5, 0, 7 | | Prime | Exactly 2 factors | 2, 3, 5, 7, 11, 13 | | Composite | More than 2 factors | 4, 6, 8, 9, 12 |
Primes up to 50 — memorize this list cold: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. That is 15 primes. Any RRB Group D number-theory question can be solved with this list.
To factorize a number, divide repeatedly by the smallest prime that divides it.
Example: Factorize 72.
72 ÷ 2 = 3636 ÷ 2 = 1818 ÷ 2 = 99 ÷ 3 = 33 ÷ 3 = 1So 72 = 2³ × 3².
This compact form is the foundation for HCF, LCM, and factor counting.
Method — Prime Factorization:
Example: HCF of 24 and 36.
24 = 2³ × 3¹36 = 2² × 3²2² and 3¹.HCF = 2² × 3 = 4 × 3 = 12Alternative — Euclidean Algorithm (faster for large numbers): Divide the larger number by the smaller; then divide the divisor by the remainder; repeat until remainder is 0. The last non-zero divisor is the HCF.
HCF(36, 24): 36 = 1 × 24 + 12 → 24 = 2 × 12 + 0 → HCF = 12.
Method — Prime Factorization:
Example: LCM of 24 and 36.
24 = 2³ × 3¹, 36 = 2² × 3²2³ and 3²LCM = 8 × 9 = 72Key relationship: HCF × LCM = Product of the two numbers. So 12 × 72 = 864 = 24 × 36. Always verify with this.
If N = p^a × q^b × r^c, then the total number of factors = (a+1)(b+1)(c+1).
Example: How many factors does 36 have?
36 = 2² × 3²(2+1)(2+1) = 3 × 3 = 9This formula saves you from listing every factor under exam pressure.
| Divisor | Rule | |---------|------| | 2 | Last digit is even | | 3 | Sum of digits divisible by 3 | | 4 | Last two digits divisible by 4 | | 5 | Last digit is 0 or 5 | | 6 | Divisible by both 2 and 3 | | 8 | Last three digits divisible by 8 | | 9 | Sum of digits divisible by 9 | | 11 | (Sum of digits at odd positions) − (Sum of digits at even positions) = 0 or multiple of 11 |
RRB Group D loves simple sequences. Look for three patterns first:
n × (n+1) gives 2, 6, 12, 20, 30 — recognizing this pattern cold is worth marks.The sequence is 2, 4, 6, ..., 48. Use the formula for number of terms in an arithmetic sequence:
n = (Last term − First term) / Common difference + 1 = (48 − 2)/2 + 1 = 23 + 1 = 24.
Note: the question says "between 1 and 50" — this means exclusive of 50, so 50 itself is not counted.
Whenever you need the number of factors, write the number as p^a × q^b and apply (a+1)(b+1). No listing needed.
Example: Factors of 72 = 2³ × 3² → (3+1)(2+1) = 4 × 3 = 12 factors.
Standard method (listing all factors): ~40 seconds. This formula: ~8 seconds. Use it every time.
After computing HCF and LCM, verify instantly: HCF × LCM = a × b (product of the original two numbers).
For HCF = 12, LCM = 72, original numbers 24 and 36: 12 × 72 = 864 and 24 × 36 = 864. They match — your answer is right.
This check takes 5 seconds and catches calculation errors before you mark the wrong option. No standard exam teaches this verification reflex — most candidates skip it and lose marks to careless errors.
The sequence 2, 6, 12, 20, 30, 42... follows n(n+1) for n = 1, 2, 3, 4, 5, 6...
When you see differences between terms increasing by 2 each time (4, 6, 8, 10...), that is the n(n+1) pattern. Next term after 20 (which is 4×5): just compute 5×6 = 30.
Spotting the pattern: 3 seconds. Trying to extend by adding differences: 15 seconds.
For any number, add all digits. If the sum is divisible by 3, the number is divisible by 3. If the sum is divisible by 9, the number is divisible by 9.
Example: Is 4,527 divisible by 9? Sum = 4+5+2+7 = 18. 18 is divisible by 9, so yes.
This works recursively — you can sum the digits of 18 again: 1+8 = 9. Divisible by 9. Confirmed in 4 steps vs long division which takes 6+ steps.
Even numbers from a to b (both inclusive): (b − a)/2 + 1 when a and b are both even.
Between 1 and 50 (exclusive): smallest even = 2, largest = 48. Count = (48−2)/2 + 1 = 24.
Standard approach (listing or guessing 25 by intuition): risky. Formula: 6 seconds, zero guesswork.
When a number system question appears in the exam, run this decision tree in your head:
Step 1 — Identify the question type:
(a+1)(b+1)(c+1).n(n+1) pattern.Step 2 — Verify:
HCF × LCM = product.Step 3 — Eliminate before computing: If two options are obviously too large or too small (e.g., HCF of 24 and 36 cannot be 18 because 18 does not divide 24), cross them out before doing any calculation. This alone eliminates at least one option in most questions.
Total time target per number system question: under 45 seconds.
Why this question: HCF by prime factorization — the most direct application of the core skill.
Solving path: Write 24 = 2³ × 3 and 36 = 2² × 3². Common primes are 2 and 3. Take lower exponents: 2² and 3¹. Multiply: 4 × 3 = 12. Eliminate 18 immediately — 18 does not divide 24. Eliminate 8 — 8 does not divide 36. Between 6 and 12, the HCF is the larger common factor, so 12 wins.
Why this question: Tests understanding of zero's multiplicative property — a conceptual trap for candidates who second-guess themselves.
Solving path: Any number multiplied by 0 equals 0. This is not a trick — do not overthink it. The answer is 0. Candidates sometimes confuse "multiplied by 0" with "raised to the power 0" (which gives 1). These are different operations.
Why this question: Factor counting formula applied directly to a common exam number.
Solving path: 36 = 2² × 3². Apply formula: (2+1)(2+1) = 9. If you want to verify, list: 1, 2, 3, 4, 6, 9, 12, 18, 36 — count them, you get 9. Option (b) is correct.
Why this question: Tests the n(n+1) pattern — one of the most frequently recycled sequence formats in RRB exams.
Solving path: Look at differences: 6−2=4, 12−6=6, 20−12=8. Differences increase by 2 each time. Next difference = 10, so next term = 20+10 = 30. Alternatively, recognize 2=1×2, 6=2×3, 12=3×4, 20=4×5 → next = 5×6=30.
Why this question: Perfect squares sequence — the most basic pattern RRB uses. You should answer this in under 10 seconds.
Solving path: 1=1², 4=2², 9=3², 16=4². Next is 5²=25. No calculation needed if you have squares 1-15 memorized. This is a 5-second question — do not spend more on it.
Why this question: Powers of integers — tests whether you can quickly compute 2³ × 3² without a calculator.
Solving path: 2³ = 2 × 2 × 2 = 8. 3² = 3 × 3 = 9. 8 × 9 = 72. Eliminate 36 (too small for 8×9), 54 (= 6×9, not matching), 64 (= 2^6, not the right expression). Answer is 72.
Why this question: Counting even numbers using the arithmetic sequence formula — tests conceptual clarity on "between" vs "from...to".
Solving path: "Between 1 and 50" means 1 and 50 are excluded. First even number = 2, last = 48. Count = (48−2)/2 + 1 = 46/2 + 1 = 23 + 1 = 24. Watch out for option (b) 25 — that is the count if you include 50 itself.
Confusing HCF and LCM selection rule. For HCF, pick the lower exponent; for LCM, pick the higher exponent. Many candidates reverse these under pressure. Anchor this: HCF is the "smallest" common factor — so you take the smaller power.
Including endpoints in "between" questions. "Even numbers between 1 and 50" means 50 is excluded, giving 24 terms, not 25. The word "between" is exclusive in standard exam usage.
Treating 1 as prime. 1 has only one factor (itself), not two. It is neither prime nor composite. This distinction appears in questions asking you to count primes in a range.
Forgetting that 2 is the only even prime. Every even number greater than 2 is divisible by 2, giving it at least three factors (1, 2, itself). Only 2 satisfies the prime condition among even numbers.
Multiplied by 0 vs raised to the power 0. Any number × 0 = 0. Any non-zero number raised to the power 0 = 1. These are completely different — read the question carefully.
Not verifying HCF/LCM with the cross-product rule. Skipping the verification HCF × LCM = product of numbers means a calculation slip goes undetected. It takes 5 seconds and can save you from choosing a wrong answer you were confident about.