Number System

Divisibility, factors, prime numbers, unit digit

The Number System is a fundamental topic in quantitative aptitude that deals with the properties and operations of numbers. It encompasses natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers. Mastery of divisibility rules, factors, multiples, prime factorisation, and unit digit patterns is essential for solving competitive exam questions efficiently.

Key Formulas

\text{Sum of first n natural numbers = n(n+1)/2}

\text{Sum of first n odd numbers = n\textasciicircum{}2}

\text{Sum of first n even numbers = n(n+1)}

\text{Sum of squares of first n natural numbers = n(n+1)(2n+1)/6}

\text{Sum of cubes of first n natural numbers = [n(n+1)/2]\textasciicircum{}2}

\text{Number of factors of N = (a+1)(b+1)(c+1)... where N = p\textasciicircum{}a x q\textasciicircum{}b x r\textasciicircum{}c...}

\text{Product of all factors of N = N\textasciicircum{}(number of factors/2)}

\text{LCM x HCF = Product of two numbers}

\text{Unit digit patterns cycle every 4 powers (for digits 2,3,7,8), every 2 powers (for 4,9), always 1 for 1, always 5 for 5, always 6 for 6}

\text{Remainder when divided by 9 = sum of digits mod 9 (digital root)}

\text{Last two digits of a number = remainder when divided by 100}

Key Concepts

Divisibility Rules

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. 7: Double last digit, subtract from rest, check divisibility. 8: Last three digits divisible by 8. 9: Sum of digits divisible by 9. 10: Last digit is 0. 11: Alternating sum of digits divisible by 11. 12: Divisible by both 3 and 4.

Prime Factorisation

Breaking down a number into its prime components. Every composite number can be expressed uniquely as a product of prime numbers (Fundamental Theorem of Arithmetic). For example, 72 = 2^3 x 3^2. This is crucial for finding HCF, LCM, number of factors, and sum of factors.

Factors and Multiples

A factor divides a number completely. A multiple is the product of a number with an integer. To find the number of factors: add 1 to each exponent in prime factorisation and multiply. To find sum of factors: for N = p^a x q^b, sum = [(p^(a+1)-1)/(p-1)] x [(q^(b+1)-1)/(q-1)].

Unit Digit Patterns

Cyclicity determines the unit digit of powers. Digits 0,1,5,6 have unit digit same as base. Digits 4,9 cycle every 2 powers. Digits 2,3,7,8 cycle every 4 powers. To find unit digit of a^b: find b mod cycle length, then compute. For example, 3^47: 47 mod 4 = 3, so unit digit is same as 3^3 = 7.

Remainder Theorems

Euler's theorem: a^ (n) 1 (mod n) when gcd(a,n)=1. Fermat's little theorem: a^(p-1) 1 (mod p) for prime p. Chinese Remainder Theorem solves systems of congruences. Remainder when dividing by 5 depends only on last digit. Remainder when dividing by 4 depends on last two digits.

Types of Numbers

Natural numbers: counting numbers 1,2,3... Whole numbers: 0,1,2,3... Integers: ...-2,-1,0,1,2... Rational numbers: p/q where q!=0. Irrational numbers: cannot be expressed as p/q (e.g., sqrt2, pi). Real numbers: rational + irrational. Even numbers: divisible by 2. Odd numbers: not divisible by 2. Prime numbers: exactly two factors. Composite numbers: more than two factors.

Solved Examples

Problem 1:

Find the number of factors of 360 and the sum of all its factors.

Solution:

Step 1: Prime factorise 360.
360 = 2^3 x 3^2 x 5^1

Step 2: Number of factors = (3+1)(2+1)(1+1) = 4x3x2 = 24

Step 3: Sum of factors = [(2 -1)/(2-1)] x [(3^3-1)/(3-1)] x [(5^2-1)/(5-1)]
= [(16-1)/1] x [(27-1)/2] x [(25-1)/4]
= 15 x 13 x 6 = 1170

Answer: 24 factors, sum = 1170

Problem 2:

What is the unit digit of 7^2023 + 8^2024 + 9^2025?

Solution:

Step 1: Find unit digit of 7^2023.
Cyclicity of 7 is 4 (7,9,3,1).
2023 mod 4 = 3, so unit digit is same as 7^3 = 3.

Step 2: Find unit digit of 8^2024.
Cyclicity of 8 is 4 (8,4,2,6).
2024 mod 4 = 0, so unit digit is same as 8 = 6.

Step 3: Find unit digit of 9^2025.
Cyclicity of 9 is 2 (9,1).
2025 mod 2 = 1, so unit digit is 9.

Step 4: Sum unit digits: 3 + 6 + 9 = 18.
Unit digit of 18 is 8.

Answer: 8

Problem 3:

Find the remainder when 2^100 is divided by 5.

Solution:

Step 1: Observe pattern of 2^n mod 5.
2^1 mod 5 = 2
2^2 mod 5 = 4
2^3 mod 5 = 3
2 mod 5 = 1
The pattern cycles every 4.

Step 2: Find 100 mod 4 = 0.

Step 3: When remainder is 0, it corresponds to the last position in cycle.
So 2^100 mod 5 = 2 mod 5 = 1.

Answer: 1

Problem 4:

A number when divided by 5 leaves remainder 3, when divided by 7 leaves remainder 2. What is the smallest such number?

Solution:

Step 1: Let the number be N.
N 3 (mod 5), so N = 5k + 3
Possible values: 3, 8, 13, 18, 23, 28, 33...

Step 2: Check which gives remainder 2 when divided by 7.
3 mod 7 = 3 != 2
8 mod 7 = 1 != 2
13 mod 7 = 6 != 2
18 mod 7 = 4 != 2
23 mod 7 = 2

Step 3: The smallest number is 23.

Answer: 23

Tips & Tricks

Memorise divisibility rules up to 12 thoroughly - they save significant calculation time
For unit digit problems, remember cyclicity patterns: 2,3,7,8 cycle every 4; 4,9 cycle every 2; 0,1,5,6 remain constant
When finding number of factors, always start with prime factorisation - it's the foundation for multiple concepts
For remainder problems with large exponents, look for patterns or use Euler's theorem when applicable
In competitive exams, estimation and digit analysis often help eliminate wrong options faster than full calculation
Practice mental calculation of digital roots (sum of digits mod 9) for quick divisibility checks and error detection

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