Number Systems

Number system problems test your understanding of divisibility, HCF/LCM, remainders, and number properties. These are foundational for quantitative aptitude.

Types of Numbers

Prime Numbers

First 25 Primes

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

Checking if a Number is Prime

To check if N is prime:

1. Find √N
2. Check divisibility by all primes ≤ √N
3. If none divide N, it’s prime

Example: Is 97 prime?

√97 ≈ 9.85
Primes ≤ 9: 2, 3, 5, 7
97/2 → not divisible (odd)
97/3 → 9+7=16, not divisible by 3
97/5 → doesn't end in 0 or 5
97/7 → 97/7 = 13.86, not divisible
97 is prime ✓

Prime Factorization

Every number > 1 can be uniquely expressed as a product of primes.

360 = 2³ × 3² × 5
120 = 2³ × 3 × 5

Divisibility Rules

Divisibility by 7 (Detailed)

Take 2058:
Step 1: 205 - (2×8) = 205 - 16 = 189
Step 2: 18 - (2×9) = 18 - 18 = 0
2058 is divisible by 7 ✓

Divisibility by 11 (Detailed)

Take 1331:
(1+3) - (3+1) = 4 - 4 = 0 → divisible by 11 ✓

Take 9152:
(9+5) - (1+2) = 14 - 3 = 11 → divisible by 11 ✓

HCF (Highest Common Factor)

Method 1: Prime Factorization

HCF(36, 48):
36 = 2² × 3²
48 = 2⁴ × 3
HCF = 2² × 3 = 12

Method 2: Euclidean Algorithm

HCF(48, 36):
48 = 36 × 1 + 12
36 = 12 × 3 + 0
HCF = 12

Properties of HCF

• HCF divides both numbers
• HCF of co-prime numbers = 1
• HCF(a, b) × LCM(a, b) = a × b

LCM (Least Common Multiple)

Method 1: Prime Factorization

LCM(12, 18):
12 = 2² × 3
18 = 2 × 3²
LCM = 2² × 3² = 36

Method 2: Using HCF

LCM(a, b) = (a × b) / HCF(a, b)

Properties of LCM

• LCM is divisible by both numbers
• LCM ≥ max(a, b)
• LCM of co-prime numbers = a × b

HCF and LCM of Fractions

HCF of fractions = HCF of numerators / LCM of denominators
LCM of fractions = LCM of numerators / HCF of denominators

Example: HCF(2/3, 4/5):

HCF(2, 4) / LCM(3, 5) = 2/15

Remainders

Basic Remainder

When a is divided by b:

a = b × q + r (where 0 ≤ r < b)

Remainder of a Sum

Rem[(a + b) / m] = Rem[a/m] + Rem[b/m]
(If this exceeds m, take remainder again)

Remainder of a Product

Rem[(a × b) / m] = Rem[a/m] × Rem[b/m]
(Again, take remainder if needed)

Example: Remainder of 47 × 53 divided by 7:

47 mod 7 = 5 (47 = 6×7 + 5)
53 mod 7 = 4 (53 = 7×7 + 4)
5 × 4 = 20
20 mod 7 = 6

Remainder of Powers

Wilson’s Theorem (for prime p):

(p-1)! ≡ -1 (mod p)
(p-1)! mod p = p - 1

Fermat’s Little Theorem: If p is prime and gcd(a, p) = 1:

a^(p-1) ≡ 1 (mod p)

Example: 2¹⁰⁰ mod 7:

By Fermat: 2⁶ ≡ 1 (mod 7)
100 = 16×6 + 4
2¹⁰⁰ = (2⁶)¹⁶ × 2⁴ ≡ 1¹⁶ × 16 ≡ 16 ≡ 2 (mod 7)

Cyclicity of Remainders

Find the pattern of remainders for powers:

2¹ mod 7 = 2
2² mod 7 = 4
2³ mod 7 = 1
2⁴ mod 7 = 2
Cycle length = 3

So 2¹⁰⁰ mod 7: 100 mod 3 = 1 → 2¹ mod 7 = 2

Number of Factors

Formula

If N = p₁^a₁ × p₂^a₂ × … × pₖ^aₖ:

Number of factors = (a₁+1)(a₂+1)...(aₖ+1)

Example: 360 = 2³ × 3² × 5¹

Factors = (3+1)(2+1)(1+1) = 4×3×2 = 24

Sum of Factors

Sum = [(p₁^(a₁+1)-1)/(p₁-1)] × [(p₂^(a₂+1)-1)/(p₂-1)] × ...

Number of Odd Factors

Remove the factor of 2 and apply the formula to the remaining part.

Example: 360 = 2³ × 45 = 2³ × 3² × 5

Odd factors of 360 = factors of 45 = (2+1)(1+1) = 6

Number of Even Factors

Even factors = Total factors - Odd factors

Number Properties

Sum of First N Natural Numbers

S = n(n+1)/2

Sum of First N Even Numbers

S = n(n+1)

Sum of First N Odd Numbers

S = n²

Sum of Squares

S = n(n+1)(2n+1)/6

Sum of Cubes

S = [n(n+1)/2]²

Arithmetic Progression (AP)

nth term: aₙ = a + (n-1)d
Sum: S = n/2 × (2a + (n-1)d) = n/2 × (first + last)

Geometric Progression (GP)

nth term: aₙ = ar^(n-1)
Sum: S = a(rⁿ-1)/(r-1) for r ≠ 1
Infinite sum: S = a/(1-r) for |r| < 1

Tricks & Shortcuts

Trick 1: Quick Sum of Consecutive Numbers

Sum from a to b = (a+b) × (b-a+1) / 2

Example: Sum from 51 to 100:

= (51+100) × 50 / 2 = 151 × 25 = 3775

Trick 2: Number of Digits

Number of digits in N = floor(log₁₀N) + 1

Trick 3: Perfect Squares

A perfect square’s last digit can only be: 0, 1, 4, 5, 6, 9 A perfect square ends with even number of zeros. Digital root of a perfect square is 1, 4, 7, or 9.

Trick 4: Quick HCF by Observation

If one number divides the other, the HCF is the smaller number. HCF(a, 0) = a.

Trick 5: Product and HCF/LCM

a × b = HCF(a,b) × LCM(a,b)

Practice Questions

Q1: Divisibility

Is 345678 divisible by 9?

Solution:

Sum of digits: 3+4+5+6+7+8 = 33
33/9 = 3.67 → Not divisible by 9

Q2: HCF and LCM

Find HCF and LCM of 84 and 120.

Solution:

84 = 2² × 3 × 7
120 = 2³ × 3 × 5
HCF = 2² × 3 = 12
LCM = 2³ × 3 × 5 × 7 = 840
Check: 84 × 120 = 10080 = 12 × 840 ✓

Q3: Number of Factors

Find the number of factors of 720.

Solution:

720 = 2⁴ × 3² × 5
Factors = (4+1)(2+1)(1+1) = 5×3×2 = 30

Q4: Remainder

Find the remainder when 2²⁰⁰ is divided by 7.

Solution:

2³ = 8 ≡ 1 (mod 7)
200 = 66×3 + 2
2²⁰⁰ = (2³)⁶⁶ × 2² ≡ 1⁶⁶ × 4 ≡ 4 (mod 7)
Remainder = 4

Q5: HCF and LCM Relationship

If HCF of two numbers is 12 and their LCM is 720, and one number is 72, find the other.

Solution:

a × b = HCF × LCM
72 × b = 12 × 720
b = 8640/72 = 120

Q6: Sum Problem

Find the sum of all 3-digit numbers divisible by 7.

Solution:

First: 105 (7×15), Last: 994 (7×142)
Count: 142 - 15 + 1 = 128
Sum = 128 × (105+994)/2 = 128 × 549.5 = 70,336

Q7: Co-prime Check

Are 17 and 23 co-prime?

Solution:

Both are prime numbers ≠ each other
HCF(17, 23) = 1
Yes, they are co-prime

Q8: Power Remainder

Find the last digit of 7²⁰²³.

Solution:

Cyclicity of 7: 7, 9, 3, 1 (cycle length 4)
2023 mod 4 = 3
3rd in cycle: 3
Last digit = 3

Summary Table