#numbersystem

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A Comprehensive Guide to Computer Number Systems: Binary, Decimal, Octal, and Hexadecimal Explained

Learn everything about computer number systems, including binary, decimal, octal, and hexadecimal, with simple explanations and examples.

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Number System Convertion :- Shortcut methods to convert Binary, Octal and Hexadecimal Number System

1) To convert Binary to Octal we need to make binary numbers in group of Three (3) bits starting from LSB, and convert that binary groups to its equivalent. 2) To convert Binary to Hexadecimal we need to make binary numbers in group of Four (4) bits starting from LSB, and convert that binary groups to its equivalent. 3) To convert Octal to binary write binary equivalent of each octal number using Three (3) bits. 4) To convert Hexadecimal to binary write binary equivalent of each Hexadecimal number using Four (4) bits.

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How to find the number of divisors

In this article, we will explore the divisor concepts, including the number of factors, prime factors, and the fascinating properties of the sum and product of factors. By understanding these concepts, you will gain valuable insights into the relationships between numbers and unlock the secrets hidden within their divisors.

The divisor or factor of any number divides it without leaving any remainder. For example, let’s consider the number 12. The divisors or factors of 12 are 1,2,3,4,6 and 12.

To find the total number of divisors of any number, we use the prime factorization method given below.

For example, the prime factorization of any number (N) is p^a×q^b×r^c, Here p,q, and r are the prime numbers (2,3,5,7,11,…etc.) and a,b, and c are their respective exponents.

The total number of divisors of N= (a+1)×(b+1)×(c+1)

Example: Find the number of divisors of 360

360=2³×3²×5¹

The number of factors= (3+1)×(2+1)×(1+1)=24

24 factors= {1,2,3,4,5,6,8,9,10,12,15,18,20,24,30,36,40,45,60,72,90,120,180,360}

Que 1: Find the number of prime factors of 360.

Solution: Three Prime factors={2,3,5}

Que 2: Find the number of odd factors of 360.

Solution: Remove 2³ from the prime factorization

Remaining Part= 3²×5¹

Number of odd factors= (2+1)×(1+1)=6

Que 3: Find the number of even factors of 360.

Solution: 360=2³×3²×5¹=2×[2²×3²×5¹]

Even Factors= 2K

K=2²×3²×5¹

Number of even factors= (2+1)×(2+1)×(1+1)=18

Que 4: Find the number of factors of 360 that are divisible by 12.

Solution: 360=2³×3²×5¹=2²×3×[2¹×3¹×5¹]

Required factor=12K

K=[2¹×3¹×5¹]

Number of required factors= (1+1)×(1+1)×(1+1)=8

Que 5: Find the number of factors of 360 which are perfect squares.

Solution: 360=2³×3²×5¹

Possible pairs={(2⁰, 2²)×(3⁰, 3²)×(5⁰)}

Number of required factors= 2×2×1=4

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In this article, we will explore the divisor concepts, including the number of factors, prime factors, and the fascinating properties of the

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Number System | Class 9 Maths | Chapter 1 | Introduction Part 1 | NCERT

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Divisibility Rules and Remainders

This video provides an in-depth discussion of the concepts of remainder and divisibility rules for various numbers, including 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, and 37. The aim of this video is to provide viewers with a solid foundation in the techniques required to solve quantitative aptitude questions based on these concepts, which are commonly found in competitive exams.

The video covers theoretical concepts and short tricks that can be used to solve problems quickly. It includes examples of problems that can be solved using these techniques and provides step-by-step explanations of how to arrive at the correct answer.

The concepts of remainder and divisibility are crucial to solving many quantitative aptitude problems, and mastering them can significantly improve one’s chances of success in competitive exams. This video is therefore a valuable resource for anyone looking to improve their skills in this area, whether they are preparing for a specific exam or simply looking to enhance their mathematical abilities.

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Problem on Remainders and Divisors

Que: Find the remainder when the sum of all the divisors of 6480 which are divisible by 18 is divided by 11.

(a) 3

(b) 4

(c) 5

(d) 6

Solution: 6480=2⁴×3⁴×5

6480=18×(2³×3²×5)

The sum of divisors that are divisible by 18= 18×(2⁰+2¹+2²+2³)×(3⁰+3¹+3²)×(5⁰+5¹)

=18×15×13×6

Remainder Calculation:

⇒(18×15×13×6) mod 11

⇒(7×4×2×6) mod 11

⇒(28×12) mod 11

⇒(6×1) mod 11

⇒6

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SI.No. Advance Maths Links 1. Routh's Theorem Click Here 2. Menelaus Theorem Click Here 3. The Pedal and the Orthic Triangle Click Here 4. T

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