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 learncriterions for divisibility by 2, 3, 5, 9 and 11.

Created by Inna Shapiro ©2007

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.

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.

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.

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 1234569 is divisible by 3, because the sum

1+2+3+4+5+6+9=30is divisible by 3

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.

Answer

•Let us calculate the sum of written digits: 3+7+1+1+7+5=24. That means that a could be equal to 0, or 3, or 6, or 9, so that the total sum 3+7+1+a+1+7+5 is divisible by 3•Answers: 3713175, 3716175, 3719175, and 3710175

The criterion for divisibility by 5

• If the last digit of a number is 0 or 5, then the number is divisible by five

Problem 3

• In the number

72a3b

Replace a and b so that the

answer is divisible by 15.

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, 72735.

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 5274567 is divisible by 9 because the sum

5+2+7+4+5+6+7=36

is divisible by 9.

Problem 4

• Replace a and b in the number

11a1bso that the answer is divisible by 45.

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: 11610 and 11115.

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:

56937142251is divisible by 11, because the sum (5+9+7+4+2+1) – (6+3+1+2+5) = 11, which is divisible by 11.

Problem 5

• Peter wrote down some numberABC,

Jean added the same digits in the reverse order

ABCCBAProve that Jean’s number is divisible by 11.

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 378873 by 11).

Problem 6

• Prove that 2002 is divisible by 11.

• Don’t try to divide!

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.

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?

Answer

• 1001 is divisible by 11, because 1+0=0+1.

• 1001/11=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.

Problem 8

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

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.

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?

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.

Problem 10

• Replace a and b in the number 399a68b

so that a result is divisible by 55.

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 3+9+6+0=9+a+8, that is, a=1;

or

in 399a685 so that 3+9+6+5=9+a+8, that is, a=6.

Answer: 3991680 or 3996685.