In primary schools, prime numbers often used are less than 50: 2, 3, 5, 7, 1 1 3, 17, 19, 23, 29, 3 1.
How to judge whether a natural number is a prime number or a composite number?
The common methods are: (1) look up the prime number table; (2) use the divisibility of numbers; (3) try to be divisible by prime numbers.
In this lecture, we will further study the problems of prime number, composite number and prime factor decomposition.
Typical example
Example 1. 109 and 437 are prime numbers or composite numbers respectively?
Solution: For natural number n (n >; 1 is an incomplete square number, and it can be judged whether it is a prime number or a composite number by trial and division of prime numbers. )
Find the minimum complete square number k2 greater than n first, and then write all the prime numbers within k. If these prime numbers are not divisible by n, then n is a prime number. If a prime number is divisible by n, then n is a composite number.
Such as: 109
The prime numbers within 437<2 12 and 2 1 are: 2, 3, 5, 7,1,13, 17, 19 respectively. Try dividing by each prime number, 65438.
Example 2. Write five consecutive natural numbers, all of which are composite numbers. What are these five numbers?
Solution: The unit digits are two digits (or more than two digits) of 0, 2, 4, 5, 6 and 8 respectively, and must be a composite number. If you find a 3-digit composite number, you can quickly find that five consecutive natural numbers with 2, 3, 4, 5 and 6 digits must be composite numbers.
For example, 33 is a composite number, so 32, 33, 34, 35 and 36 are five consecutive composite numbers. 63 is a composite number, so 62, 63, 64, 65 and 66 are five consecutive composite numbers. In this way, you can find countless groups of numbers that meet the requirements.
Example 3. Divide 33 into the sum of several different prime numbers. If the product of these prime numbers is to be maximized, what are these prime numbers?
Solution: The prime numbers within 33 are: 2, 3, 5, 7, 1 1 3, 17, 19, 23, 29, 3 1. Divide 33 by the sum of two prime numbers, the two prime numbers must be odd and even, and only two of the prime numbers are even, then the other number must be 3 1. 2×3 1=62。
Divide 33 by the sum of three prime numbers, and all three prime numbers must be odd numbers. The closer these three prime numbers are, the greater their product. Obviously, the product of the three prime numbers 5, 1 1 and 17 is the largest.
5× 1 1× 17=935
Divide 33 into the sum of four prime numbers. 33 is an odd number, there must be an even number 2 among the four prime numbers, the other three are odd numbers, and the sum of the three prime numbers is 3 1. Of the three prime numbers whose sum is 3 1, 7,1and 13 are the closest and the product is the largest.
2×7× 1 1× 13=2002
Is 33 divisible by the sum of five or more prime numbers? It is out of the question. Because the sum of the smallest five consecutive odd prime numbers 3,5,7, 1 1 and 13 is 39, it is known that it exceeds 33.
So when 33 is divided into four prime numbers: 2, 7, 1 1, 13, their products are the largest.
Example 4. Li Ming, a sixth-grade student, took part in the city's math competition. He said, "The product of my ranking, score and my age is 3204". How many points did Li Ming get? Where did you get it?
Solution: Because the product of Li Ming's ranking, score and age is 3204, ranking, score and age are all divisors of 3204. If we decompose 3204 into prime factors, it is
3204=2×2×3×3×89
Obviously, 89 is his test score, and 2×2×3×3 is converted into the form of multiplying two numbers, that is, ranking times age, which has the following situations: 2× 18, 3× 12, 4×9.
As the title shows, Li Ming is a sixth-grade student and should choose 3× 12, so Li Ming won the third place with 89 points.
Simulated test questions (answer time: 15 minutes)
1. Determine which of the following numbers are prime numbers and which are composite numbers?
1 19、 13 1、 143、 157、 16 1、 179
There are five consecutive odd numbers, and their product is 135 135. Find these five consecutive odd numbers.
3. The product of three prime numbers is exactly equal to 1 1 times of their sum. What are the three prime numbers?
The head teacher led the students to plant trees. Students are divided into three groups on average, and the number of trees planted by teachers and students is equal. It is understood that teachers and students have planted 572 trees. A * * *, how many students are there? How many trees are there in each race?
Test answer
1. Determine which of the following numbers are prime numbers and which are composite numbers?
1 19、 13 1、 143、 157、 16 1、 179
Solution: 1 19
13 1 & lt； The prime numbers within 122 and 12 are 2,3,5,7, 1 1. None of these prime numbers can be divisible by 13 1, so 13 1 is a prime number.
The prime numbers within 143< 122, 12 are 2,3,5,7, 1 1. Only 1 1 of these prime numbers is divisible by 143, so 143 is a composite number.
The prime numbers within 157< 132, 13 are 2,3,5,7, 1 1. None of these prime numbers can be divisible by 157, so 157 is a prime number.
16 1 & lt； The prime numbers within 132 and 13 are 2,3,5,7, 1 1. Only seven of these prime numbers are divisible by 16 1, so 16 1 is a composite number.
The prime numbers within 179< 142 and 14 are 2,3,5,7, 1 1, 13. None of these prime numbers can be divisible by 179, so 179 is a prime number.
There are five consecutive odd numbers, and their product is 135 135. Find these five consecutive odd numbers.
Solution:135135 =135×1001= 5× 3× 7×1×/kloc-0
=7×9× 1 1× 13× 15
3. The product of three prime numbers is exactly equal to 1 1 times of their sum. What are the three prime numbers?
Solution: Let three different prime numbers be A, B and C.
Because, in A, B and C, there must be a prime number 1 1. Then let's set A = 1 1
When b = 2, c is 13.
When b = 3, c is 7.
When b = 5, c is 4, and 4 is a composite number, which does not meet the requirements.
When b = 7, c is 3, which is the same as the previous answer.
So these three prime numbers are 2, 1 1, 13 or 3, 7, 1 1 respectively.
The head teacher led the students to plant trees. Students are divided into three groups on average, and the number of trees planted by teachers and students is equal. It is understood that teachers and students have planted 572 trees. A * * *, how many students are there? How many trees are there in each race?
Solution: According to the meaning of the question, the number of students and the number of trees planted per capita are both divisors of 572, the number of students is a multiple of 3, plus 1 teacher, and the number of teachers and students is divided by 3.
572=2×2× 1 1× 13
According to the meaning of the question, we can get: 572 =1/kloc-0 /× (13× 2× 2) =1× (51+1).
A: There are 5 1 students in this class, each race 1 1 tree.