Number System

TYPES OF NUMBERS:

• Natural numbers (N): 1, 2, 3, ...
• Whole numbers (W): 0, 1, 2, 3, ...
• Integers (Z): ..., −2, −1, 0, 1, 2, ...
• Rational numbers (Q): p/q form, q ≠ 0. Decimals terminating or repeating.
• Irrational numbers: π, √2, e. Non-repeating non-terminating decimals.
• Real numbers (R): rational + irrational.
• Complex numbers (C): a + ib form.

Special subsets:

• Prime: divisible only by 1 and itself (2, 3, 5, 7, 11, ...).
• Composite: > 1 and not prime.
• Even: divisible by 2.
• Odd: not divisible by 2.

KEY PROPERTIES:

Closure:

• Addition, multiplication closed on N, W, Z, Q, R, C.
• Subtraction closed on Z, Q, R, C (not N, W in general).
• Division closed on Q{0}, R{0}, C{0}.

Associativity: (a + b) + c = a + (b + c).
Commutativity: a + b = b + a; a × b = b × a.
Distributivity: a × (b + c) = ab + ac.
Identity: 0 (for +); 1 (for ×).
Inverse: −a (for +); 1/a (for ×, a ≠ 0).

DIVISIBILITY RULES (Pack 16 detailed):

By 2: last digit even.
By 3: digit sum div by 3.
By 4: last two digits div by 4.
By 5: last digit 0 or 5.
By 6: div by 2 and 3.
By 7: double last, subtract from rest.
By 8: last three digits div by 8.
By 9: digit sum div by 9.
By 10: last digit 0.
By 11: alternating sum div by 11.

LCM & HCF:

LCM (Least Common Multiple): smallest number divisible by both.
HCF / GCD (Highest Common Factor): largest number that divides both.

Method 1 — Prime factorization:

• 12 = 2² × 3.
• 18 = 2 × 3².
• HCF: take min powers → 2 × 3 = 6.
• LCM: take max powers → 2² × 3² = 36.

Method 2 — Division:

• HCF via Euclidean algorithm: gcd(a, b) = gcd(b, a mod b).

Property: a × b = LCM(a, b) × HCF(a, b).

Co-prime numbers: HCF = 1 (e.g., 7 and 12).

REMAINDER THEOREM:

• a^n mod m: find pattern.
• E.g., 2^10 mod 7? Powers of 2: 2, 4, 1, 2, 4, 1, ... (cycle 3).
  • 10 mod 3 = 1. So 2^10 mod 7 = 2.

Fermat's little theorem: a^p ≡ a (mod p) if p prime.

Wilson's theorem: (p−1)! ≡ −1 (mod p) if p prime.

UNIT DIGIT (LAST DIGIT) PROBLEMS:

Cycle of unit digits of powers:

• 2: 2, 4, 8, 6, 2, 4, 8, 6, ... (cycle 4).
• 3: 3, 9, 7, 1, ... (cycle 4).
• 4: 4, 6, 4, 6, ... (cycle 2).
• 7: 7, 9, 3, 1, ... (cycle 4).
• 8: 8, 4, 2, 6, ... (cycle 4).
• 9: 9, 1, 9, 1, ... (cycle 2).
• 0, 1, 5, 6: cycle 1 (always same).

Q. Unit digit of 7^100?

• 100 mod 4 = 0 → 4th in cycle of 7 (7, 9, 3, 1) → 1.
• (Or 4th position in 0-indexed: position 0 = first... be careful; 100 mod 4 = 0 means we want the LAST in cycle = 1.)

FACTORIAL:

• n! = n × (n−1) × (n−2) × ... × 1.
• 0! = 1.
• Trailing zeros in n!: count of factors of 5.
  • In 100!: 100/5 + 100/25 + 100/125 = 20 + 4 + 0 = 24 trailing zeros.

SURDS & INDICES:

Surd: irrational expression with root, e.g., √2, √(2+√3).

Properties of indices:

• a^m × a^n = a^(m+n).
• a^m / a^n = a^(m−n).
• (a^m)^n = a^(mn).
• a^0 = 1 (a ≠ 0).
• a^(-n) = 1/a^n.
• a^(1/n) = ⁿ√a.

Properties of surds:

• √a × √b = √(ab).
• √a / √b = √(a/b).
• (√a)² = a.
• Rationalizing denominator: × (conjugate).

EXAMPLES:

Q1. Find LCM and HCF of 18 and 24.

• 18 = 2 × 3². 24 = 2³ × 3.
• HCF = 2 × 3 = 6. LCM = 2³ × 3² = 72.

Q2. Number of zeros at end of 100!?

• 24 (as calculated above).

Q3. Unit digit of 3^25?

• Cycle of 3 (3,9,7,1) of length 4. 25 mod 4 = 1. → 1st in cycle = 3.

Q4. Sum of all 2-digit numbers divisible by 7.

• 14, 21, 28, ..., 98. AP with a=14, l=98, d=7.
• n = (98-14)/7 + 1 = 13.
• S = 13/2 × (14+98) = 728.

EXAM HOOKS:

• Prime check: trial division up to √n.
• Trailing zeros in n!: count factors of 5 (Legendre's formula).
• Unit digit: cycle pattern.
• Co-prime: HCF = 1.
• Product of LCM and HCF = product of numbers.
• Memorize first 20 primes.

Numbers are classified into several types essential for SSC CGL. Natural Numbers (N): 1, 2, 3, ... (counting numbers, no zero). Whole Numbers (W): 0, 1, 2, 3, ... (natural + zero). Integers (Z): ...-2, -1, 0, 1, 2, ... (all positive and negative). Rational Numbers: numbers of form p/q where q≠0 (e.g., 3/4, 0.5). Irrational Numbers: cannot be expressed as p/q (e.g., √2, π). Real Numbers: all rational + irrational numbers. Prime Numbers: divisible only by 1 and itself (2 is the only even prime). Composite Numbers: more than 2 factors. Co-prime Numbers: two numbers with HCF = 1 (e.g., 8 and 15). Trick: 1 is neither prime nor composite. There are 25 primes below 100.

Primes up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. To check if a number n is prime, test divisibility by all primes up to √n. Perfect Numbers: sum of all factors (including 1, excluding itself) equals the number. Example: 6 = 1+2+3 (factors 1,2,3). Next: 28. Twin Primes: primes differing by 2 (3,5), (5,7), (11,13), (17,19), (29,31). For SSC CGL: know that the product of two even numbers is even, product of two odd numbers is odd, and sum of two odd numbers is always even.

Q: Which of the following is a prime number — 91, 97, 87, 93? Step 1: Check 91. √91 ≈ 9.5. Test primes 2, 3, 5, 7: 91 = 7 × 13. Not prime. Step 2: Check 87 = 3 × 29. Not prime. Step 3: Check 93 = 3 × 31. Not prime. Step 4: Check 97. √97 ≈ 9.8. Not divisible by 2, 3, 5, 7. So 97 is prime. Answer: 97. Memory trick: For any two-digit number, only check divisibility up to its square root. If none of the primes up to that point divide it, the number is prime.

Divisibility rules (memorize all):

HCF (Highest Common Factor) = GCD:

• Method 1 (prime factorization): write each as product of primes, take minimum power of each common prime.
• Method 2 (Euclidean algorithm — much faster for large numbers):
  HCF(a, b) = HCF(b, a mod b), with HCF(a, 0) = a.

Example: HCF(48, 18) = HCF(18, 48 mod 18) = HCF(18, 12) = HCF(12, 6) = HCF(6, 0) = 6.

LCM (Least Common Multiple):

• Method 1 (prime factorization): take maximum power of each prime appearing.
• Method 2 (using HCF): LCM(a, b) = (a × b) / HCF(a, b).

Example: LCM(48, 18) = (48 × 18) / 6 = 864 / 6 = 144.

Useful identities:

• HCF(a, b) × LCM(a, b) = a × b. (Only for TWO numbers; not three+.)
• HCF divides every linear combination ax + by.
• HCF of fractions: HCF(numerators) / LCM(denominators).
• LCM of fractions: LCM(numerators) / HCF(denominators).

Common SSC question types:

"Find smallest number that when divided by a, b, c leaves remainder r each time."
Answer: LCM(a, b, c) + r.

"Find largest number that divides a, b, c leaving same remainder."
Answer: HCF of differences |a−b|, |b−c|, |a−c|.

"Find the number of trailing zeros in n!"
Answer: floor(n/5) + floor(n/25) + floor(n/125) + ... (each 5 in prime factorization contributes one trailing zero; 2s are always abundant).

Example: trailing zeros in 100! = 20 + 4 + 0 = 24.

"Number of factors of n."
If n = p₁^a · p₂^b · p₃^c · ..., then number of factors = (a+1)(b+1)(c+1)...

Example: 72 = 2³ × 3². Factors = (3+1)(2+1) = 12.

Sum of factors: = (p₁^(a+1) − 1)/(p₁ − 1) × similar for p₂, etc.

TYPES OF NUMBERS:

• Natural numbers (N): 1, 2, 3, ...
• Whole numbers (W): 0, 1, 2, 3, ...
• Integers (Z): ..., −2, −1, 0, 1, 2, ...
• Rational numbers (Q): p/q form, q ≠ 0. Decimals terminating or repeating.
• Irrational numbers: π, √2, e. Non-repeating non-terminating decimals.
• Real numbers (R): rational + irrational.
• Complex numbers (C): a + ib form.

Special subsets:

• Prime: divisible only by 1 and itself (2, 3, 5, 7, 11, ...).
• Composite: > 1 and not prime.
• Even: divisible by 2.
• Odd: not divisible by 2.

KEY PROPERTIES:

Closure:

• Addition, multiplication closed on N, W, Z, Q, R, C.
• Subtraction closed on Z, Q, R, C (not N, W in general).
• Division closed on Q{0}, R{0}, C{0}.

Associativity: (a + b) + c = a + (b + c).
Commutativity: a + b = b + a; a × b = b × a.
Distributivity: a × (b + c) = ab + ac.
Identity: 0 (for +); 1 (for ×).
Inverse: −a (for +); 1/a (for ×, a ≠ 0).

DIVISIBILITY RULES (Pack 16 detailed):

By 2: last digit even.
By 3: digit sum div by 3.
By 4: last two digits div by 4.
By 5: last digit 0 or 5.
By 6: div by 2 and 3.
By 7: double last, subtract from rest.
By 8: last three digits div by 8.
By 9: digit sum div by 9.
By 10: last digit 0.
By 11: alternating sum div by 11.

LCM & HCF:

LCM (Least Common Multiple): smallest number divisible by both.
HCF / GCD (Highest Common Factor): largest number that divides both.

Method 1 — Prime factorization:

• 12 = 2² × 3.
• 18 = 2 × 3².
• HCF: take min powers → 2 × 3 = 6.
• LCM: take max powers → 2² × 3² = 36.

Method 2 — Division:

• HCF via Euclidean algorithm: gcd(a, b) = gcd(b, a mod b).

Property: a × b = LCM(a, b) × HCF(a, b).

Co-prime numbers: HCF = 1 (e.g., 7 and 12).

REMAINDER THEOREM:

• a^n mod m: find pattern.
• E.g., 2^10 mod 7? Powers of 2: 2, 4, 1, 2, 4, 1, ... (cycle 3).
  • 10 mod 3 = 1. So 2^10 mod 7 = 2.

Fermat's little theorem: a^p ≡ a (mod p) if p prime.

Wilson's theorem: (p−1)! ≡ −1 (mod p) if p prime.

UNIT DIGIT (LAST DIGIT) PROBLEMS:

Cycle of unit digits of powers:

• 2: 2, 4, 8, 6, 2, 4, 8, 6, ... (cycle 4).
• 3: 3, 9, 7, 1, ... (cycle 4).
• 4: 4, 6, 4, 6, ... (cycle 2).
• 7: 7, 9, 3, 1, ... (cycle 4).
• 8: 8, 4, 2, 6, ... (cycle 4).
• 9: 9, 1, 9, 1, ... (cycle 2).
• 0, 1, 5, 6: cycle 1 (always same).

Q. Unit digit of 7^100?

• 100 mod 4 = 0 → 4th in cycle of 7 (7, 9, 3, 1) → 1.
• (Or 4th position in 0-indexed: position 0 = first... be careful; 100 mod 4 = 0 means we want the LAST in cycle = 1.)

FACTORIAL:

• n! = n × (n−1) × (n−2) × ... × 1.
• 0! = 1.
• Trailing zeros in n!: count of factors of 5.
  • In 100!: 100/5 + 100/25 + 100/125 = 20 + 4 + 0 = 24 trailing zeros.

SURDS & INDICES:

Surd: irrational expression with root, e.g., √2, √(2+√3).

Properties of indices:

• a^m × a^n = a^(m+n).
• a^m / a^n = a^(m−n).
• (a^m)^n = a^(mn).
• a^0 = 1 (a ≠ 0).
• a^(-n) = 1/a^n.
• a^(1/n) = ⁿ√a.

Properties of surds:

• √a × √b = √(ab).
• √a / √b = √(a/b).
• (√a)² = a.
• Rationalizing denominator: × (conjugate).

EXAMPLES:

Q1. Find LCM and HCF of 18 and 24.

• 18 = 2 × 3². 24 = 2³ × 3.
• HCF = 2 × 3 = 6. LCM = 2³ × 3² = 72.

Q2. Number of zeros at end of 100!?

• 24 (as calculated above).

Q3. Unit digit of 3^25?

• Cycle of 3 (3,9,7,1) of length 4. 25 mod 4 = 1. → 1st in cycle = 3.

Q4. Sum of all 2-digit numbers divisible by 7.

• 14, 21, 28, ..., 98. AP with a=14, l=98, d=7.
• n = (98-14)/7 + 1 = 13.
• S = 13/2 × (14+98) = 728.

EXAM HOOKS:

• Prime check: trial division up to √n.
• Trailing zeros in n!: count factors of 5 (Legendre's formula).
• Unit digit: cycle pattern.
• Co-prime: HCF = 1.
• Product of LCM and HCF = product of numbers.
• Memorize first 20 primes.

The remainder when a number N is divided by d is written as N mod d or N % d. Key rules: If N = d×q + r, then r is the remainder (0 ≤ r < d). Important SSC CGL shortcuts: (a+b) mod d = [(a mod d) + (b mod d)] mod d. (a×b) mod d = [(a mod d) × (b mod d)] mod d. Euler's Theorem for SSC: aᵠ⁽ⁿ⁾ ≡ 1 (mod n) when gcd(a,n)=1. Fermat's Little Theorem: If p is prime and p does not divide a, then aᵖ⁻¹ ≡ 1 (mod p). Practical trick: To find remainder of large powers (like 7⁵⁰ ÷ 5), first find pattern of remainders in cycle.

Cyclicity Method: Find the remainder pattern of successive powers. Example for 7ⁿ ÷ 5: 7¹ = 7 → remainder 2; 7² = 49 → remainder 4; 7³ = 343 → remainder 3; 7⁴ = 2401 → remainder 1; 7⁵ → remainder 2 again. Cycle = 4. So for 7⁵⁰ ÷ 5: 50 mod 4 = 2, so answer = same remainder as 7² = remainder 4. Wilson's Theorem: For prime p, (p−1)! ≡ −1 (mod p). Chinese Remainder Theorem: If N leaves remainder r₁ when divided by d₁ and r₂ when divided by d₂, find N using the LCM-based approach.

Find the remainder when 17³⁵ is divided by 16. Step 1: Note that 17 = 16 + 1. Step 2: 17³⁵ = (16+1)³⁵. Using binomial theorem, all terms except the last contain 16 as a factor. Step 3: The last term = 1³⁵ = 1. Step 4: So 17³⁵ ≡ 1 (mod 16). Remainder = 1. General trick: (kd + 1)ⁿ always leaves remainder 1 when divided by d. Similarly, (kd − 1)ⁿ leaves remainder 1 if n is even, and (d−1) if n is odd.

BODMAS stands for: Brackets → Orders (powers/roots) → Division → Multiplication → Addition → Subtraction. Always follow this left-to-right within same priority. Types of brackets in order: (i) Vinculum (bar over numbers), (ii) ( ) Round brackets, (iii) { } Curly brackets, (iv) [ ] Square brackets. Solve innermost bracket first. Key mistakes to avoid: 20 ÷ 4 × 5 = 5 × 5 = 25 (NOT 20 ÷ 20 = 1). Division and multiplication have equal priority — go left to right. Similarly, addition and subtraction — go left to right. In SSC CGL, 3-4 questions per year come from simplification.

Algebraic identities used in simplification: (a+b)² = a²+2ab+b²; (a−b)² = a²−2ab+b²; a²−b² = (a+b)(a−b); (a+b)³ = a³+3a²b+3ab²+b³; a³+b³ = (a+b)(a²−ab+b²); a³−b³ = (a−b)(a²+ab+b²). Fraction shortcuts: a/b ÷ c/d = a/b × d/c. Mixed fractions: 3½ = 7/2. Surds: √a × √b = √(ab); √a / √b = √(a/b); (√a + √b)(√a − √b) = a−b. For quick calculation: 1/8 = 0.125, 1/6 ≈ 0.1667, 1/7 ≈ 0.1429, 1/9 ≈ 0.1111.

Simplify: 18 − [6 − {4 + (8 − 3)}]. Step 1: Innermost ( ): 8 − 3 = 5. Expression: 18 − [6 − {4 + 5}]. Step 2: { }: 4 + 5 = 9. Expression: 18 − [6 − 9]. Step 3: [ ]: 6 − 9 = −3. Expression: 18 − (−3). Step 4: 18 + 3 = 21. Answer: 21. Common mistake: forgetting that subtracting a negative number is addition. Another example: 5 + 3 × 4 − 2 = 5 + 12 − 2 = 15 (NOT 8 × 4 − 2 = 30).

Key series formulas tested in SSC CGL: Sum of first n natural numbers: n(n+1)/2. Sum of squares of first n natural numbers: n(n+1)(2n+1)/6. Sum of cubes of first n natural numbers: [n(n+1)/2]². Sum of first n even numbers: n(n+1). Sum of first n odd numbers: n². For arithmetic series: Sum = n/2 × (first term + last term). For finding missing term in series: identify the pattern (difference, ratio, or mixed). Common patterns: +2, +4, +6... (arithmetic); ×2, ×3... (geometric); prime numbers; perfect squares/cubes.

The unit digit of powers follows a cycle — crucial for SSC CGL. Digit 0,1,5,6: unit digit never changes (always 0,1,5,6). Digit 4: cycle of 2 (4,6,4,6,...). Digit 9: cycle of 2 (9,1,9,1,...). Digits 2,3,7,8: cycle of 4. For 2ⁿ: unit digits are 2,4,8,6,2,4,8,6,... For 3ⁿ: 3,9,7,1,3,9,7,1,... For 7ⁿ: 7,9,3,1,7,9,3,1,... For 8ⁿ: 8,4,2,6,8,4,2,6,... To find unit digit of aⁿ: find (n mod 4), look up position in cycle. If n mod 4 = 0, use position 4.

Find the unit digit of 7⁹⁵. Step 1: Cycle of 7's unit digit has length 4 (7,9,3,1). Step 2: Find 95 mod 4 = 3 (since 95 = 4×23+3). Step 3: Position 3 in cycle (7,9,3,1) → unit digit = 3. Answer: 3. For the series question: Find the sum 1+3+5+7+...+99. This is sum of first n odd numbers. Number of terms = (99+1)/2 = 50. Sum = 50² = 2500. Formula shortcut: sum of first n odd numbers = n².