Every number question in SSC GD starts somewhere inside the number system. Before you touch percentages, ratios, or time-and-work, you need a solid mental map of how numbers are classified — because a surprising number of marks are lost not on hard calculations but on basic definitional errors like "Is 1 prime?" or "Is 0 a natural number?"
Here is the cleanest way to picture it. Think of a set of nested boxes:
Natural Numbers — the counting numbers you learned as a child: 1, 2, 3, 4, ... They go on forever. Zero is not in this box.
Whole Numbers — exactly the same set, but now with 0 added: 0, 1, 2, 3, 4, ...
Integers — the whole numbers plus their negative counterparts: ..., −3, −2, −1, 0, 1, 2, 3, ... Think of a number line stretching in both directions.
Rational Numbers — any number that can be written as p/q where p and q are integers and q ≠ 0. Every integer qualifies (just write it over 1). Fractions like 3/4 and terminating/repeating decimals like 0.75 or 0.333... are rational.
Irrational Numbers — numbers that cannot be expressed as p/q. √2, √3, π are the classic examples. Their decimal expansions never terminate and never repeat.
The analogy that sticks: natural numbers are your coins, whole numbers are your coins plus an empty wallet (zero), integers add the concept of debt (negatives), rationals add fractions, and irrationals are the leftover "weird" numbers that do not fit any fraction.
For SSC GD, you mostly operate in natural numbers and integers. But the definitions of prime numbers, factors, LCM, and HCF all live in natural numbers — so anchor yourself there.
| Set | Contains | Example |
|-----|----------|---------|
| Natural (N) | 1, 2, 3, ... | 7 |
| Whole (W) | 0, 1, 2, 3, ... | 0 |
| Integer (Z) | ..., −2, −1, 0, 1, 2, ... | −5 |
| Rational (Q) | p/q, q≠0 | 3/4 |
| Irrational | Non-terminating, non-repeating | √5 |
A prime number has exactly two distinct factors: 1 and itself. That "exactly two" is the gate — it immediately disqualifies 1 (only one factor: itself) and 0.
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.
Primes up to 50 worth memorising: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
A composite number has more than two factors. Example: 12 has factors 1, 2, 3, 4, 6, 12 — six factors.
Special cases exam setters love:
Factor of n: a number that divides n exactly (remainder = 0). Factors of 18: 1, 2, 3, 6, 9, 18.
Multiple of n: n × 1, n × 2, n × 3, ... Multiples of 6: 6, 12, 18, 24, ...
Prime factorisation: write any natural number as a product of primes.
12 = 2² × 3
18 = 2 × 3²
This is the engine behind both HCF and LCM.
The largest number that divides both given numbers without remainder.
Method — Prime factorisation:
Example: HCF(12, 18)
12 = 2² × 3¹18 = 2¹ × 3²2¹ × 3¹ = 6Method — Successive division (Euclid): Divide the larger by the smaller; replace the larger with the remainder; repeat until remainder = 0. The last non-zero divisor is the HCF.
18 ÷ 12 = 1 remainder 6
12 ÷ 6 = 2 remainder 0
HCF = 6
For two-number problems in the exam, Euclid's method is faster when numbers are large.
The smallest number that is a multiple of both given numbers.
Method — Prime factorisation:
Example: LCM(6, 8)
6 = 2¹ × 3¹8 = 2³2³ × 3¹ = 24Key relationship (applies to exactly two numbers):
LCM × HCF = Product of the two numbers
Check: 24 × 2 = 48 = 6 × 8. Use this to verify or to find one when the other is known.
| Divisor | Rule | |---------|------| | 2 | Last digit even | | 3 | Sum of digits divisible by 3 | | 4 | Last two digits divisible by 4 | | 5 | Last digit 0 or 5 | | 6 | Divisible by both 2 and 3 | | 8 | Last three digits divisible by 8 | | 9 | Sum of digits divisible by 9 | | 10 | Last digit 0 | | 11 | (Sum of digits at odd positions) − (Sum at even positions) = 0 or multiple of 11 |
S = n(n+1)/2
For n = 10: 10 × 11 / 2 = 55. This formula comes up repeatedly — in number series, AP, and even age problems in disguise.
aⁿ means a multiplied by itself n times.
2³ = 2 × 2 × 2 = 8
3² = 9
10³ = 1000
The unit digit of powers follows a cycle — useful for large-exponent questions:
For two numbers a and b: LCM = (a × b) / HCF. Once you find HCF by quick inspection or Euclid, divide the product by it to get LCM instantly — no separate factorisation needed. Example: HCF(12,18) = 6 by inspection. LCM = (12 × 18) / 6 = 216 / 6 = 36. Standard factorisation method: 6 steps. This method: 2 steps.
First n natural numbers: n(n+1)/2. To avoid multiplying, notice: if n ends in an even digit, halve n first, then multiply by (n+1). For n=10: halve 10 → 5, then 5 × 11 = 55. For n=20: halve 20 → 10, then 10 × 21 = 210. Standard repeated addition for n=10 takes ~9 additions. This formula: 2 mental steps.
To test if a number N is prime, check divisibility only by primes up to √N. For N = 97: √97 ≈ 9.8, so only check 2, 3, 5, 7. None divide 97 → it is prime. Standard method (check all numbers up to 97): 96 checks. This method: 4 checks.
For unit digit of aⁿ: find (n mod 4). If remainder is 0, use the 4th position of the cycle; if 1, use 1st position, and so on. Example: unit digit of 2⁷. 7 mod 4 = 3, so use 3rd position of 2's cycle (2, 4, 8, 6) → 8. Verify: 2⁷ = 128. Correct. Standard method: compute 128. This method: 3 mental steps.
To count how many numbers between 300 and 400 contain the digit 7 exactly once, split into two slots: tens place and units place. Numbers with 7 in tens place: 370–379 = 10 numbers, but 377 has 7 twice, so exclude it → 9 numbers. Numbers with 7 in units place: 307, 317, 327, 337, 347, 357, 367, 387, 397 = 9 numbers (skip 377 already counted). Total = 9 + 9 = 18. This systematic slot-by-slot count takes ~30 seconds vs. listing all candidates: 2+ minutes.
When you see a number system question in the exam hall, run through this decision tree in under 5 seconds:
Step 1 — What type is it?
n(n+1)/2 immediately.Step 2 — LCM or HCF?
LCM × HCF = a × b after quick Euclid for HCF.Step 3 — Sanity check.
For definition questions (prime, zero-property, integer classification), don't overthink — the answer is the one that matches the textbook definition exactly. Exam setters use these as gift marks.
Why this question: Counting digit occurrences in a range tests your ability to organise counting systematically. A brute-force list will cost you 2+ minutes. The slot method is the exam-hall weapon.
Solving path: Break the range 300–399 (not "between", but we work with the range anyway). The digit 7 must appear exactly once. Two slots can carry a 7: the tens place and the units place (the hundreds digit is always 3, never 7). Case 1 — 7 in tens place: numbers 370–379. That is 10 numbers. Subtract 377 (has 7 in both tens and units) → 9 valid. Case 2 — 7 in units place: 307, 317, 327, 337, 347, 357, 367, 387, 397. That is 9 numbers (377 is excluded because it already appeared in Case 1 and has two 7s). Total = 9 + 9 = 18.
Why this question: The smallest prime is a classic definitional trap — SSC setters always include 0 and 1 as distractors.
Solving path: Apply the definition — a prime has exactly two factors: 1 and itself. Check each option: 0 → infinite factors (every non-zero number divides 0); 1 → only one factor (itself); 2 → factors are 1 and 2 only — exactly two, so prime; 3 → also prime but larger. Answer: 2.
Why this question: LCM via prime factorisation is the most commonly tested HCF-LCM format in SSC GD.
Solving path: Factorise: 6 = 2¹ × 3¹, 8 = 2³. LCM takes the maximum power of each prime: 2³ × 3¹ = 8 × 3 = 24. Cross-check: 24 ÷ 6 = 4 (exact), 24 ÷ 8 = 3 (exact). Answer: 24.
Why this question: Sum of natural numbers is a formula question that appears repeatedly in different disguises (sum of first n odd numbers, AP sums). Knowing the formula cold is non-negotiable.
Solving path: Formula: S = n(n+1)/2. For n = 10: S = 10 × 11 / 2 = 110 / 2 = 55. Answer: 55. Time: under 5 seconds.
Why this question: HCF by listing factors is the beginner method; knowing which to use based on number size is the skill.
Solving path: Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 18: 1, 2, 3, 6, 9, 18. Common factors: 1, 2, 3, 6. Highest = 6. Alternatively, Euclid: 18 ÷ 12 = 1 R 6; 12 ÷ 6 = 2 R 0. HCF = 6.
Calling 1 a prime number. 1 has only one factor (itself). A prime needs exactly two distinct factors. This mistake costs easy marks on definition questions.
Excluding 0 from whole numbers. Natural numbers start at 1. Whole numbers start at 0. If the question asks "Is 0 a natural number?" — the answer is no. If it asks "Is 0 a whole number?" — yes.
Taking LCM as the product of numbers. LCM(6, 8) ≠ 48. LCM is 24. The product equals LCM × HCF, not LCM itself (unless HCF = 1).
Forgetting that LCM ≥ larger number. If your calculated LCM is smaller than the larger of the two numbers, you have made an error. This is the fastest self-check.
Double-counting in digit-occurrence problems. When counting numbers where a specific digit appears exactly once, the number 377 (for digit 7) has 7 in both tens and units — it must be excluded from both cases, not counted in one or both. Missing this gives the wrong total.
Confusing 2³ with 2 × 3 = 6. The exponent means repeated multiplication, not multiplication of base and exponent. 2³ = 2 × 2 × 2 = 8, not 6. This confusion shows up on power-based MCQs where 6 is always planted as a distractor.