What is Divisibility ?

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What is Divisibility ? - PowerPoint PPT Presentation

What is Divisibility ?. Lets say, we have 15 books to be given out to 5 pupils. Can we evenly distribute all the books? 15 ÷ 5 = 3 Each pupil will gets 3 books and we don’t have any extra left. (No Remainder). YES !. 15 is divisible by 5. What is Divisibility ?.

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What is Divisibility?
• Lets say, we have 15 books to be given out to 5 pupils.
• Can we evenly distribute all the books?
• 15 ÷ 5 = 3
• Each pupil will gets 3 books and
• we don’t have any extra left. (No Remainder)

YES!

15 is divisible by 5

What is Divisibility?
• How if, we have 17 books to be given out to 5 pupils.
• Can we evenly distribute all the books

(without any extra left)?

• 17 ÷ 5 = 3 (R)2
• Each pupil will gets 3 books and
• We HAVE2 extra left. (Remainder is 2)

NO!

17 is NOT divisible by 5

25

Divisible

By 5

NO remainder

• 25 toys given out

to 5 children?

• 45 toys given out

to 5 children?

• 75 toys given out

to 5 children?

• 68 toys given out

to 5 children?

45

Divisible

By 5

NO remainder

75

Divisible

By 5

NO remainder

68 NOT

Divisible

By 5

Remainder

3

Divisible by 2, 5; 4, 25; 8, 125
• Numbers that are divisible by 2 or 5:
• Check their last digit
• Numbers that are divisible by 4 or 25:
• Check their last two digits
• Numbers that are divisible by 8 or 125:
• Check their last three digits
Divisible by 3 or 9
• Sum up the all the digits of the number.
• E.g.: 123456,
• The sum of its digits is: 1+2+3+4+5+6 = 21
• 21 is divisible by 3. So, 123456 is divisible by 3
• 21 is NOT divisible by 9. So, 123456 is NOT divisible by 9
• E.g.: 234567,
• The sum of its digits is: 2+3+4+5+6+7 = 27
• 27 is divisible by both 3 and 9 respectively.
• So, 234567 is also divisible by both 3 and 9.
Divisible by 7, 11 or 13
• Find the difference between:
• “the number formed by last three digits”with
• “the number formed by the digits of the rest”
• E.g.: 420321:
• Number formed by last three digits: 321
• Number formed by the digit of the rest: 420
• Their difference: 420 – 321 = 99 (divisible by 11)
• So, 420321 is also divisible by 11.
Divisible by 7 or 21
• Subtract 2 times the last digit from the rest.
• E.g.: 714
• 2X4 = 8; 71-8=63 (divisible by 7 and 21)
• 714 is also divisible by 7 and 21 respectively
• E.g.: 476
• 2X6 = 12; 47-12=35 (divisible by 7 but NOT 21)
• So, 476 is divisible by 7 but NOT by 21
Divisible by 11
• The alternating sum*1 of its digits is divisible by 11
• E.g.: 781
• 7-8+1 =0 or
• (7+1) – (8) =0
• So, 781 is divisible by 11
• E.g.: 34859
• 3-4+8-5+9= ?
• (3+8+9) – (4+5) = 11
• So, 34859 is divisible by 11
Summary of divisibility facts

Fact 1

8 is divisible by 4

12 is divisible by 4

12 + 8 = 20

is divisible by 4

12 - 8 = 4

is divisible by 4

Fact 2

36 is divisible by 3

36 is divisible by 4

(3 and 4 are coprime*2)

36 is divisible

by 12 (= 3X4)

Example 1
• Given a 4-digit number 83Y2, which is divisible by 9. What is the possible digit for Y?
• 83Y2 is divisible by 9 means that:

Sum of digits : 8+3+Y+2 is divisible by 9

• 13+ Y is divisible by 9 （8+3+2=13）
• Numbers that are divisible by 9 after 13 are:

18, 27, 36, ….

• 18= 13 +5; 27 = 13+ 14; 36= 13 + 23;……
• Y is 1-digit number. So, Y = 5.
• 8352 is divisible by 9
Example 2
• Given two numbers, M and N, where both of them are divisible by 7. Their sum, M+N is a 5-digit number, 3802_. What is the last digit of the sum M+N?
• Both M and N are divisible by 7, means that

their sum, (M+N= )3802_ also divisible by 7

• difference between 38 and 02_ is divisible by 7
• 38 – 2_ = multiple of 7 (list out some possible ans)
• 38 – 24 = 14; the last digit is 4
• 38024 is divisible by 7
Example 3
• Find the smallest 4-digit number which is divisible by 2, 5, and 11.
• An integer which is divisible by 2, 5 and 11 is also

divisible by 2x5x11 = 110 (since 2, 5 and 11 are coprime)

• 9x110=990; 10x110=1100; 11x110=1210
• Hence, the smallest 4-digit integer required is 1100.
Exercise 1
• Which of the following numbers are divisible by 8?
• A)34 342
• B)23 624
• C)120 458
• D)348 624
Solution 1
• A)1 234 342
• B)3 273 624
• C)5 120 458
• D)7 348 000
• (B) and (D) since 342 and 000 is divisible by 8 .
• (A) and (C) is not divisible by 8 since 342 and 458 is not divisible by 8
Exercise 2
• Is 10-digit number, 987654321Y a multiple of 45? If ‘yes’, what is the digit for ‘Y’?
Solution 2
• 45=5×9,
• If “987654321Y” is a multiple of 45, then it is divisible by 5 and 9 respectively.
• For integer that is divisible by 5, the last digit must be either 0 or 5. So, Y is either 0 or 5.
• For integer that is divisible by 9: the sum of its digits, that is 9+8+7+6+5+4+3+2+1+Y, must be 0 or multiple of 9.
• 9+8+7+6+5+4+3+2+1=45, a multiple of 9. So, Y must be 0 or multiple of 9.
• Combine the 2 conditions, Y must be 0.
Exercise 3
• Given five numbers: 0, 1, 3, 7, 9. Without repetition, pick any 4 numbers from the list and form a 4-digit number that is divisible by 55.
Solution 3
• 55 = 5 X 11
• Hence, the number formed must be divisible by both 5 and 11.
• Numbers that divisible by 5 ends with digit 0 or 5.
• AXB0 ; the last digit must be 0.
• (A+B) – (X+0) is 0 or multiple of 11 (11, 22, 33….)
• Take X as 1, then A and B could be 3 or 9.
• Ans: 3190 or 9130.
*1) Alternating Sum
• The alternating sum of the digits for a number, let’s say 72853 is as followed:
• -+ -+ or
• The difference between:
• The sum of digits in odd places and
• The sum of digits in even places.
• (++) – (+)

7

7

2

2

8

8

5

5

3

3

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*2) Coprime
• Two numbers are coprime, if
• They are in the simplest form when we put them into fraction form.
• E.g.: 6 and 9 are NOTcoprime, as is NOT in the simplest fraction form.
• E.g.: 2 and 5 are coprime, as is in the simplest fraction form.
• No integers can divide them simultaneously, besides 1.

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