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Criterions for divisibility The ancient Greeks knew criterions for divisibility by 2, 3, 5 and 9 in the third century B.C. In this presentation we will.

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Presentation on theme: "Criterions for divisibility The ancient Greeks knew criterions for divisibility by 2, 3, 5 and 9 in the third century B.C. In this presentation we will."— Presentation transcript:

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2 Criterions for divisibility The ancient Greeks knew criterions for divisibility by 2, 3, 5 and 9 in the third century B.C. In this presentation we will learn criterions for divisibility by 2, 3, 5, 9 and 11. Created by Inna Shapiro ©2007

3 The criterion for divisibility by 2 If the last digit of a number is divisible by two, then the number is also divisible by two.

4 Problem 1 Kathy wrote down three natural numbers. Prove that she can always choose two of them so that their sum is divisible by two.

5 Answer If Kathy wrote down three numbers, then one can choose either two even numbers, or two odd ones, from written numbers. The sum of the chosen numbers will be even.

6 The criterion for divisibility by 3 A natural number is divisible by 3 if and only if the sum of its digits is divisible by 3 Example: the number is divisible by 3, because the sum =30 is divisible by 3

7 Problem 2 In the number 371a175 try to replace a by a digit so that the result is divisible by 3. Write all possible answers.

8 Answer Let us calculate the sum of written digits: =24. That means that a could be equal to 0, or 3, or 6, or 9, so that the total sum a is divisible by 3 Answers: , , , and

9 The criterion for divisibility by 5 If the last digit of a number is 0 or 5, then the number is divisible by five

10 Problem 3 In the number 72a3b Replace a and b so that the answer is divisible by 15.

11 Answer The last digit could be either 0 or 5, so we have to add one digit to 72a30 or 72a35. Using to criterion of divisibility by 3, we get seven answers: 72030, 72330, 72630,72930, 72135, 72435,

12 The criterion for divisibility by 9 A natural number is divisible by 9 if and only if the sum of its digits is divisible by 9 Example: the number is divisible by 9 because the sum =36 is divisible by 9.

13 Problem 4 Replace a and b in the number 11a1b so that the answer is divisible by 45.

14 Answer The answer is divisible by 45 if it is divisible both by 5 and by 9 That means that the last digit is 5 or 0. We have to replace a with a digit in 11a10 or in 11a15 so that 1+1+a+1+0 or 1+1+a+1+5 is divisible by 9. Answer: and

15 The criterion for divisibility by 11 A natural number is divisible by 11 if the sum of its digits in odd positions minus the sum of its digits in even positions is either equal to zero or divisible by eleven. Example: is divisible by 11, because the sum ( ) – ( ) = 11, which is divisible by 11.

16 Problem 5 Peter wrote down some number ABC, Jean added the same digits in the reverse order ABCCBA Prove that Jeans number is divisible by 11.

17 Answer In the number ABCCBA A sum of digits on even positions A+B+C is the same as a sum of digits on odd ones => this number is divisible by 11. (Try to divide by 11).

18 Problem 6 Prove that 2002 is divisible by 11. Dont try to divide!

19 Answer In the number 2002 sum of odd digits 2+0=2 and sum of even digits 0+2=2 so 2002 is divisible by 11.

20 Problem 7 Mary has to pack 1001 apples into equal boxes. How many boxes can she use and how many apples does she have to put in each box if one box can contain no more than 20 and no less than 10 apples?

21 Answer 1001 is divisible by 11, because 1+0= /11=91=7*13, where 11, 7 and 13 are prime numbers. Each box can contain no more than 20 and no less than 10 apples. That means that Mary can put 11 or 13 apples into each box. She has to pack 91 or 77 boxes.

22 Problem 8 Prove that the difference between a three-digit number and the sum of its digits is divisible by 9.

23 Answer Let us consider a number ABC. It consists of A hundreds, B tens and C ones. We have ABC=(A*100)+(B*10)+C For example, 456=4*100+5*10+6 The difference between ABC and the sum of its digits is (A*100+B*10+C)-(A+B+C)=99*A+9*B, which is always divisible by 9.

24 Problem 9 A librarian ordered several books for a school library. All books had the same price, and the total cost was $187. What was the price of a book if the librarian ordered more than fifteen books?

25 Answer You can see that 187 is divisible by 11, because 1+7=8 So 187=11*17, where 11 and 17 are prime numbers. That means that the librarian ordered 17 books and the price of each was $11.

26 Problem 10 Replace a and b in the number 399a68b so that a result is divisible by 55.

27 Answer If the number is divisible by 55, it is divisible by 5 and 11. That means that b could either be 5 or 0. We have to determine the digit a in 399a680, so that =9+a+8, that is, a=1; or in 399a685 so that =9+a+8, that is, a=6. Answer: or


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