Topic Included: | Formulas, Definitions & Exmaples. |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Number System Aptitude Notes & Questions. |

Questions for practice: | 10 Questions & Answers with Solutions. |

All-natural numbers that are greater than 1 but not the prime numbers are known as composite numbers, also a composite number must have more than two factors. 1 is not a composite number. The numbers {4, 6, 8, 9, 10, 12, 14, 15,.........} are the composite numbers. A Composite number can be factorized into more than two prime factors.

The number \(9\) which is a composite number, can be factorized into two prime factors \((3 \times 3)\). Here the composite number \(9\) have \(3\) factors 1, 3, and 9.

The composite number \(18\) can be factorized into three prime factors \((2 \times 3 \times 3)\). Here the composite number \(18\) have \(6\) factors 1, 2, 3, 6, 9, and 18.

now we are going to discuss how to find out factors of a composite number by using a formula.

$$Composite \ Numbers \ = A_{1}^{x_1} \times A_{2}^{x_2} \times A_{3}^{x_3} \times A_{4}^{x_4} \times ........A_{n}^{x_n} $$

Composite Numbers = $$ A_{1}^{x_1} \times A_{2}^{x_2} \times A_{3}^{x_3} \times A_{4}^{x_4} \times ........A_{n}^{x_n} $$

Where, \(A_1, A_2, A_3, A_4,..............A_n\) are prime factors, and \(x_1, x_2, x_3, x_4,..............x_n\) are their respective powers.

now to find the factors of a composite number, add 1 in every respective power of prime numbers.

$$Factors \ = (x_1 + 1) \times (x_2 + 1) \times (x_3 + 1) \times (x_4 + 1) \times...............(x_n + 1)$$

\((x_1 + 1) \times (x_2 + 1) \times (x_3 + 1) \\ \times (x_4 + 1) \times............(x_n + 1)\)

**Example(1):** Find the factors of composite number \(250\)?

**Solution:** Factors of composite number \(250 = 2 \times 5 \times 5 \times 5 = 2^1 \times 5^3\)

Where, \(A_1 = 2\), \(A_2 = 5\) are prime numbers, and \(x_1 = 1\), \(x_2 = 3\) are their respective powers.

Now factors of composite number \(= (x_1 + 1) \times (x_2 + 1)\) \(= (1 + 1) \times (3 + 1) = 2 \times 4 = 8\), here we found the factors of composite number \(250\) are \(8\).

**Example(2):** Find the factors of composite number \(180\)?

**Solution:** Factors of composite number \(180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5^1\)

Where, \(A_1 = 2\), \(A_2 = 3\) and \(A_3 = 5\) are prime numbers, and \(x_1 = 2\), \(x_2 = 2\) and \(x_3 = 1\) are their respective powers.

Now factors of composite number \(= (x_1 + 1) \times (x_2 + 1) \times (x_3 + 1)\) \(= (2 + 1) \times (2 + 1) \times (1 + 1)\) \(= 3 \times 3 \times 2 = 18\), here we found the factors of composite number \(180\) are \(18\).

Any single prime number has only two factors \(1\), and the number itself, like \(5\) is a prime number and has only two factors \(1\), and itself. A prime number can not be a composite number.

**Example(3):** Find the factors of composite number \(47\)?

**Solution:** Factors of composite number \(47 = 47^1\)

Where, \(A_1 = 2\) is a prime number, and \(x_1 = 1\) is their respective power.

Hence factors of composite number \(= (x_1 + 1)\) \(= (1 + 1) = 2\), here we found the factors of composite number \(47\) are \(2\), The number 47 itself and 1.

Lec 1: Introduction to Number System
Lec 2: Factors of Composite Number
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Lec 3: Basic Remainder Theorem
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Lec 4: Polynomial Remainder Theorem
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Lec 5: LCM of Numbers
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Lec 6: HCF of Numbers
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Lec 7: Divisibility Rules of Numbers
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