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Transcript
```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.
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.
• 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=30
is divisible by 3
Problem 2
• In the number
371a175
try to replace a by a digit so that
the result is divisible by 3.
•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
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
•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,
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
11a1b
so that the answer is divisible by 45.
•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.
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:
56937142251
is 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 number
ABC,
Jean added the same digits in the
reverse order
ABCCBA
Prove that Jean’s number is
divisible by 11.
• 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!
• 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?
• 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.
•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?
• 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.