## Math 6 Chapter 1 Lesson 14: Prime numbers, composite numbers and prime numbers table

## 1. Summary of theory Tóm

### 1.1. Prime Numbers – Composite Numbers

**Example 1:**

The number 7 has only two divisors, 1 and 7, then we say 7 is a prime number.

The number 6 has divisors 1, 2, 3, 6, then we say 6 is composite.

Thus, we have **define:**

Given a natural number a > 1

a is said to be prime if U(a) = {1, a} (no divisors other than 1 and itself)

a is said to be composite if U(a) = {1,…,a) (more than 2 divisors)

**Attention: **We should note that:

– The numbers 0 and 1 are neither prime nor composite.

– The smallest prime number is 2 and is the only even prime number.

To prove that a is a prime number, we only need to show that it is not divisible by all primes whose square is less than a.

**Generality: **Tooso integers other than 2 and 3 are of the form: \(6n \pm 1\) with \(n \in {N^*}\)

### 1.2. Factorize a number into prime factors

We have the job definition:

To factor a natural number greater than 1 into prime factors is to write the number as a product of prime factors.

All composite numbers can be factored into primes, and this analysis is unique.

**Example 2: **Factor the number 30 into prime factors:

60 2

30 2

15 3

5 5

first

Thus, the number 30 has been factored into primes. Derive \(60{\rm{ }} = {\rm{ }}2.3.5{\rm{ }} = {\rm{ }}{ 2^2}{.3^1}{.5^1}\)

From the above example we have the following observations:

When written, the prime factors are ordered from smallest to largest.

U(60)={2,3,5,6 =2.3.10=2.5.12=23 .3.15 = 3.5.20=22.5.30=2.3.5.60=22.3.5}

Number 60 has:

(2+1)(1+1)(1+1)=3.2.2=12 (divisor)

**Comment:**

1. The prime factorization form of each prime number is the number itself.

2. If the number A is analyzed as:

\(A = {a^m}. {b^n}. {c^p}…\)

Where a, b, c are prime numbers, then A has all:

(m+1)(n+1)(p+1)…

**Divisor**

**Example 3: **For number 420

a. Factor 420 into prime factors.

b. How many divisors does the number 420 have in all?

c. List all of those divisors.

Solution

a. We have:

\(420 = {2^2}.3.5.7\)

b. The number of divisors of 420 is:

(1+2)(1+1)(1+1)(1+1)=24 (digest)

c. We list them in order of** 4 steps** after:

B1: 420 has divisors: \(1,2,{2^2}\) (1)

B2: Multiplying the terms of the sequence (1) by 3, we get the sequence: 3, 6, 12 (2)

B3: Multiplying the terms of the sequence (1) (2) by 5, we get the sequence: 5, 10, 20, 15, 30, 60 (3)

B4: Multiplying the terms of the sequence (1) (2) (3) by 7, we get the sequence:

7, 14, 28, 21, 42, 84, 53, 70, 170, 105, 210, 420 (4)

So we have 24 divisors of 420:

1 2 3 4 5 6 7 10

12 14 15 20 21 28 30 42

53 60 70 84 105 140 210 420

## 2. Illustrated exercise

**Question 1: **Prove that the following numbers are composite:

a. \({12^{11}} + {13^{17}} + {17^{19}}\)

b. \(1 + {23^{23}} + {29^{29}} + {25^{125}}\)

c. \({45^{25}} + {37^{15}}\)

d. \({95^{354}} + {51^{25}}\)

**Solution guide:**

Prove that the last digit in the power is divisible by 2.

a. Then \({12^{11}} + {13^{17}} + {17^{19}}\) ends with 8

b. Then \(1 + {23^{23}} + {29^{29}} + {25^{125}}\) ends with 4

c. Then \({45^{25}} + {37^{15}}\) ends with 2

d. Then \({95^{354}} + {51^{25}}\) has a digit ending in 6.

**Verse 2: **In a division, the divisor is 99, the remainder is 8. Find the divisor and quotient.

**Solution guide:**

Suppose

99 = a . x + 8 (where a is the divisor, x is the quotient, a > 8)

\( \Rightarrow \) a . x = 91.

So, a must be a divisor of 91 and a > 8

Factoring in primes, we get:

91 = 13 . 7

So we have two answers

* Divisor is 13, quotient is 7

99 = 13 . 7 + 8

* Divisor is 91, quotient is 1

99 = 91 . 1 + 8.

**Question 3: **Find the natural number n that satisfies: n, n + 2, n + 6 are all prime numbers.

**Solution guide:**

From the assumption: n is a prime number

Derived from:

n = 3 or n = 5

For n = 3 deduce n + 6 = 3 + 6 = 9 (not prime)

With n = 5 we get:

n = 5 so n + 2 = 7, n + 6 = 11 (all prime)

So n = 5 is satisfied.

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Prove that the following numbers are composite:

a. \({6^{13}} + {13^{\frac{1}{57}}} + {15^{17}}\)

b. \(1 + {19^{19}} + {21^{21}} + {23^{23}}\)

c. \({25^{15}} + {37^{17}}\)

d. \({75^{342}} + {67^{28}}\)

**Verse 2:** In a division, the divisor is 155, the remainder is 12. Find the divisor and quotient.

**Question 3: **Find the natural number n that satisfies: n, n + 1, n + 3 are all prime numbers.

### 3.2. Multiple choice exercises Bài

**Question 1: **Which of the following numbers is prime: 2, 4, 13, 19, 25, 31

A. 2, 4, 13, 19, 31

B. 4, 13, 19, 25, 31

C. 2, 13, 19, 31

D. 2, 4, 13, 19

**Verse 2: **Which of these following statements is wrong?

A. The number 2 is the smallest prime number

B. All prime numbers are odd

C. A composite number is a natural number greater than 1 that has more than 2 divisors

D. There are 2 consecutive natural numbers that are prime

**Question 3: **Find a natural number a such that \(\overline {6{\rm{a}}} \) is prime?

A. a = 1, a = 3

B. a = 1; a = 5

C. a = 3, a = 7

D. a = 1, a = 7

**Question 4: **What are three consecutive odd natural numbers that are prime?

A. 1, 3, 5

B. 3, 5, 7

C. 5, 7, 9

D. 7, 9, 11

**Question 5: **Choose the incorrect statement:

A. Prime numbers less than 10 are 2, 3, 5, 7

B. 2 is the only even prime number

C. The number 0 is neither prime nor composite

D. The number 1 is the smallest prime number

## 4. Conclusion

Through this prime, composite and prime number lesson, students need to complete some of the objectives given by the lesson, such as:

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