Multiples – For getting multiples of a number we recite the multiplication table of that number .e.g – multiples of 5 are – 5,10,15,20…….

Factors : A factor of a number divides the number exactly ( remainder = 0) e.g factors of 20 are 1,2 ,4,5,10 ,20 Prime number : A number which has only two different factors 1 and the number itself e.g 2,3,5,7,11,13…. Composite number : a number which has more than two different factors . e.g 4,6,8,9..

1.

1 is the factor of every number

2. Every number is the factor of itself

3. Twin prime numbers – two prime numbers whose difference

is two E,g – 5 and 3 , 41 and 43

4. Co prime numbers – Two numbers are said to be co –prime

numbers when they have only1 as common factor .e,g 3 and 5 , 19 and 20

Q.1) Done in book

Q.2)Are the following numbers prime or composite .Show by finding the factors. c) 89 1 X 89 = 89 Factors of 89 are 1 and 89 ,so 89 is a prime number

d) 96 1 X 96 = 96 2 X 48 = 96 3 X 32= 96

4 X 24=96 6 X 16 = 96 8 X 12 = 96 Stop .

Factors of 96 are = 1,2,3,4,6,8,12,16,24,32,48,96. ,so 96 is composite number

f) 101 1 X 101= 101 Factors are 1,101 .so 101 is a prime number . Q.3) Write down the first 10 prime numbers Answer – 2,3,5,7,11,13,17,19,23 and 29 are first 10 prime numbers . Q.4A number lies between 2000 and 2070 and has 5 in its ones place .Is it prime or composite number ? Give reasons A.4) The number that has five in its ones place .The number is 2005 Since the number is divisible by 5 so it has more than 2 factors ,hence 2005 is a composite number. Q.8) Between which multiples of 10 does 3486 lie ? A.8) 10 X 348 = 3480 10 X 349=3490 3480 and 3490 are multiples of 10

Hence 3480 <3486 <3490

Q.9)Write any four pair of twin primes. A.9) Twin prime numbers – two prime numbers whose difference is two . 3and 5 11 and 13 17 and 19

29 and 31 41 and 43

•

9 is divisible by 3 means --------9 ÷3 gives remainder = 0

•

Similarly 45 is divisible by 5 means -----45 ÷ 5 gives remainder = 0

There are certain test which can confirm ( without actually dividing )whether a number is divisible by some other number . 1) Divisibility by 2 346 is divisible by 2 ( Yes /no ) 79 isdivisible by 2 ( Yes /no) A number is divisible by 2 if the digit at ones place is divisible by 2 that is digit at ones place is 0,2,4,6,8 Eg 2) Divisibility by 5 340 is divisible by 5 (Yes /no) 715 is divisible by 5 ( Yes /no ) 506 is divisible by 5 ( Yes /no) A number is divisible by 5 if the digits at ones place is 0 or 5

3) Divisibility by 10 396 is divisible by 1 0 ( Yes /no ) 450 is divisible by 10 ( Yes /no) A number is divisible by 10 if the digit at ones place is 0 4) Divisibility by 4 69432 , is divisible by 4 ? ( check 32 ÷4 ) ( Yes /No) 56728 , is divisible by 4? ( ( check 28 ÷4) (Yes/No) A number is divisible by 4 if the number formed by the digits at tens and ones place is divisible by 4 That is : if number formed by last 2 digits ÷ 4 gives remainder = 0 then we say that complete number is divisible by 4 5) Divisibility by 8 Eg 69432 , is divisible by 8 ?( check432 ÷8 ) ( Yes /no ) 56728 ,is divisible by 8 ? ( check 728 ÷8 ) ( Yes /no ) A number is divisible by 8 if the number formed by the digits at hundreds , tens and ones place is divisible by 8 That is : if number formed by last 3 digits ÷ 8 gives remainder = 0 then we say that complete number is divisible by 8

6) Numbers with trailing zeroes 2300 ÷ 4 -----is it divisible by 4 (Yes/no)

78000 ÷ 8----- is it divisible by 8 ( Yes / no) •

If a number has zeroes at tens and ones place ,it is divisible by 4

•

If a number has zeroes at hundred ,tens and ones place ,it is divisible by 8 .

Q1) e) 789984,365832 ,10098 ,395529 The numbers that are divisible by 8 are _______,___________,___________ Solution 789984 -----consider 984 ÷8 , here remainder = 0 Hence 789984 is divisible by 8 365832 -----consider 832 ÷ 8, here remainder = 0 Hence 365832 is divisible by 8 10098 -----consider 098 ÷ 8, here remainder = 0 Hence 10098 is not divisible by 8 395529 ------ consider 529 ÷ 8 , here remainder = 0 Hence 395529 is not divisible by 8

Q.2) Apply the divisibility rule and show that – a) 432566 is divisible by 2 Divisibility rule of 2 = A number is divisible by 2

Digit at ones place is 0,2,4,6,8 432556 digit at ones place ……. 6 . Therefore 432566 is divisible by 2 b) 352115 is divisible by 5

Divisibility rule of 5 = A number is divisible by 5 digits at ones place is 0 or 5 352115 digit at ones place……. 5 .Therefore 352115 is divisible by 5

d) 3496 is divisible by 4 Divisibility rule of 4 =A number is divisible by 4 if the number formed by the digits at tens and ones place is divisible by 4 Step 1 Step 2

3496 96 ÷ 4 2 4 4 ) 9 6 - 8 1 6 - 1 6 0 0

R= 0 , therefore 96 is divisible by 4 ,hence 3496 is divisible by 4

e) 117904 is divisible by 8 Divisibility rule of 8 =A number is divisible by 8 if the number formed by the digits at hundreds , tens and ones place is divisible by 8

Step 1

117904

Step 2

904 ÷ 8 1 8 ) 9 - 8 1 - 0 0

1 3 0 4 0 8 2 4 -2 4 0 0

R = 0 ,therefore 908 is divisible by 8 ,hence 117904 is divisible by 8

Step 1 – add all the digits of the given number Step 2 – sum of the digits ÷ 3

R=0 no. divisible by 3

R =0 no. not divisible by 3

Step 1) Add the digits 4 + 3 + 6 + 9 = 22 Step 2) 22 ÷ 3 7 3 ) 22

-

21 01

R = 0 So, 4369 is not divisible by 3

Step 1 – add all the digits of the given number Step 2 – sum of the digits ÷ 9

R=0 no. divisible by 9

R =0 no. not divisible by 9

Step 1 – Add the digits 6 + 7 + 5 + 4 + 2 = 24 Step 2 – 24 ÷9 2 9) 24 - 18 06 Remainder = 0 So, 67542 is not divisible by 9

Q.3) )Test the following for divisibility by 3 and 9 b) 145404 Test for divisibility by 3 Step 1 – Add the digits 1 + 4 + 5 + 4 + 0+4 = 18

Step 2 – 18 ÷3 6 3) 18 -

18 00

Remainder = 0

So, 145404 is divisible by 3

Test for divisibility by 9 Step 1 – Add the digits 1 + 4 + 5 + 4 + 0+4 = 18 Step 2 – 18 ÷9 2 9) 18 18 00 Remainder = 0 So, 145404 is divisible by 9

C) 99999 Test for divisibility by 3 Step 1 – Add the digits 9 + 9 + 9 + 9+ 9= 45 Step 2 – 45÷3 15 3) 45 -

45 00

Remainder = 0

So, 99999 is divisible by 3

Test for divisibility by 9 Step 1 – Add the digits 9 + 9 + 9 + 9+ 9= 45 Step 2 – 45÷9 05 9) 45 - 45 00 Remainder = 0 So, 99999 is divisible by 9

Q.6)Replace

by a digit so that the number is divisible by 9 .

b) 1____ 80498 Sum of the digits = 1 +____ + 8+0 +4 +9 +8 = 30 +_____

The given number will be divisible by 9 if sum of all the digits is divisible by 9 30 +_____ = Multiple of 9 that comes after 30 i.e --- 36 30 + ___=36 ______ = 36 – 30 = 6 Hence ____ = 6

d) 4 6 _____ 2 1 Sum of the digits = 4 + 6 + ____ + 2 + 1 = 13 +_____

The given number will be divisible by 9 if sum of all the digits is divisible by 9 13 +_____ = Multiple of 9 that comes after 13 i.e --- 18 13 + ___=18 ______ = 18 – 13 = 5 Hence ____ = 5

Step 1) Starting from ones place add alternate digits .We get Sum 1 and Sum 2 Step 2) Find difference between Sum 1 and Sum 2

0 or multiple of 11

neither 0 nor multiple of 11

no. is divisible by 11

no. is not divisible by 11

Step 1 )

1 3 8 7 5 4 9 2

Sum 1 = 2 + 4+ 7 +3= 16 Sum 2 =9+5+8+1= 23 Step 2 ) 23 – 16 = 7 ----- difference is neither 0 nor multiple of 11

so, 13875492 is not divisible by 11

d) 7 0 2 3 6 4 3 ▪ Step 1 )

7 0 2 3 6 4 3

▪

Sum 1 = 3+ 6+ 2 +7= 18

▪

Sum 2 =4+3+0= 7

▪ Step 2 ) 18 – 7 = 11 ----- difference is multiple of 11

▪

so, 7 0 2 3 6 4 3 is divisible by 11

e) 5 8 3 3 4 6 6 1 ▪ Step 1 ) 5 8 3 3 4 6 6 1 ▪ Sum 1 = 1 +6 + 3 + 8= 18 ▪ Sum 2 =6 + 4 + 3 + 5 = 18 ▪ Step 2 ) 18 – 18 = 10 ----- difference is 0 ▪ so, 5 8 3 3 4 6 6 1 is divisible by 11

f) 6 0 2 1 1 1 2 1 3 Step 1 ) 6 0 2 1 1 1 2 1 3 Sum 1 = 3 +2 + 1 + 2 + 6= 14 ▪ Sum 2 =1 + 1+ 1 + 0 = 3 ▪ Step 2 ) 14 – 3 = 11----- difference is a multiple of 11 so, 6 0 2 1 1 1 2 1 3 is divisible by 11

1) Factors of 6 are ----- 1 ,2,3,6 2 and 3 are coprime factors of 6 So , a number is divisible by 6

if it is divisible by its co prime factors ( i.e 2 and 3 )

Similarly ❖ a number is divisible by 12

if it s divisible by its co prime factors

( i.e 4 and 3 ) ❖ a number is divisible by 15

if it s divisible by its co prime factors ( i.e 5 and 3 )

2) If a number is divisible by another number , factors Example -- 18 is divisible by 6 then 18 is also divisible by 1,2,3

then it is divisible by each of its

factors of 6

3) If

a number is divisible by 2 co prime numbers

Then it is divisible

by their product also .

e.g 5 and 4 are two co prime numbers

40 is divisible by 5 and 4 5 X 4 = 20 40 is divisible by 20 also

4) If two numbers are divisible by a number The their sum is also divisible by that number e.g 10 and 15 are divisible by 5 sum = 10 + 15 = 25 is also divisible by 5 5) ) If two numbers are divisible by a number The their difference is also divisible by that number e.g 10 and 15 are divisible by 5 difference = 15 - 10 = 5 is also divisible by 5

Q.5) Fill ups 2 a) A number is divisible by 6 if it is divisible by its two co prime factors ___and ________. 3 5

3 b) 43185 is divisible by 15 as it is divisible by ___________and _______. 3 c) The number 8625 is not divisible by 6 as it is divisible by _________ but not by 2 _________. 4

3

d) The number 54420 is divisible by 12 as it is divisible by ________ and _______. e) The number 781022 is divisible by 11 as the difference of the sum of the digits at odd places and the sum of the digits at even places is ________. 0 2+0+8 = 10 …sum of digits at odd places = Sum 1 2 + 1 + 7 =10 … sum of the digits at even places = Sum 2 Difference = 0

False a)If a number is divisible by 3,it must be divisible by 9 =_ ________

e.g. – 6 is divisible by 3 but not by 9 b) If a number is divisible by 18 , it must be divisible by 6 and 3 ._ True Because – 6 and 3 are factors of 18 True c) If a number is divisible by both 9 and 10 ,the it must be divisible by 90 ._____ Because – 9 and 10 are co prime numbers and 9 X 10 =90 True d) All numbers which are divisible by 8 are divisible by 4 .________ Because 4 is the factor of 8 e) If a number is exactly divisible by two numbers separately then it must be divisible by their sum False__ .________ e.G .. 6 is divisible by 2 and 3 but not by 5 ( 2 + 3)

PRIME FACTORISATION Representating a number in the form of product of only prime numbers EXAMPLE 12 = 2 x 2 x 3 Methods to find prime factorisation Division method

Factor tree method

Q.1) Find the prime factorisation of the following number : ( use 2,3,5,7,……)

e) 441 3

441

3

147

7

49

7

7

1 prime factorization of 441 = 3 X 3 X 7 X 7

f) 240 2

240

2

120

2

60

2

30

3

15

5

5 1

Prime factorization of 240 = 2 x 2 x 2 x 2 x 3 x 5

g) 2304

2 2 2 2 2 2 2 2 3 3

2304 1152 576 288 144 72 36 18 9 3 1

Prime factorization of 2304 is = 2x2x2x2x2 x2x2x2x3x3

AS A PRODUCT OF PRIME NUMBERS Solution – The greatest 4 – digit number = 9999 prime factorization is the product of prime numbers 3

9999

3

3333

11

1111

101

101 1

Prime factorization of 9999 = 3 x 3 x 11 x 101

Solution : Smallest 4 – digit number is 1000 2

1000

2

500

2

250

5

125

5

25

5

5 1

Prime factorization of 1000 = 2 x 2 x 2 X 5 X 5 X 5

( Start from smallest odd prime number that is 3 then 5 ….) a) 18 = 5 + 13 b) 32 = 3 + 29 c) 66 = 5 + 61 d) 90 = 7 + 83

a) 41 = 3 + 7 + 31 b) 23 = 3 + 3 + 17 c) 75 = 3 + 5 + 67 d) 59 = 3 + 3 + 53

Methods to find HCF

Factor method

Prime factorisation method

Continued division method

▪ 1) Factor method

Factors of 12 = 1 , 2 , 3 , 4 , 6 , 12

Factors of 8 = 1 , 2, 4,8 Factors of 16 = 1,2,4 ,8 ,16 Common factors = 1,2,4

HCF= 4

2 12

2 8

2

16

2 6

2 4

2

8

3 3

2 2

2

4

1

1

2

2 1

12 = 2 x 2 x 3 8=2x2x2 16 = 2 X 2 X 2 X 2 Common factors = 2 ,2 HCF= 2 X 2 = 4

▪ The given numbers are 12 , 8 ,16 ▪ Take any two number out of these

▪ Lets take 12 and 8 ▪

12 > 8 , so 12 ÷ 8 8 ) 12 ( 1

▪

- 8

▪ ▪ ▪

R is not zero so Division continous

4) 8( 2 - 8 0

▪

▪ Last divisor is the HCF ▪

so ,HCF of 12 and 8 = 4

R = 0 so stop

▪ Now consider 4 and 16 ▪

16 > 4 so

▪

16 ÷ 4

▪

4 ) 16 ( 4

▪

- 16

▪

0

▪ ▪ Hence, HCF of 12 ,8 and 16 = 4

Remainder is = 0 so stop .

Q.3)Find HCF by continued division method

d) 208,494 ,949 208 , 494, 949 Consider 1st two number , 494 > 208 494 ÷ 208

208 ) 494 (2 - 416 78 ) 208 (2 - 156 052) 78 ( 1 - 52 26 ) 52 ( 2 - 52 00 = remainder , division stops

Remainder = 0 so 26 is the HCF of 208 and 494 .

now we will find HCF of 26 and 949 949 ÷ 26 26 ) 9 4 9 ( 36 -78 169 -156 0 1 3 ) 2 6 (2

- 26 0 0 = remainder , division stops HCF = 13 So, HCF of 208 , 494 , 949 = 13

e) 1212,6868,1111 Step 1 – 1212 ,6868 ,1111

6868> 1212 6868 ÷ 1212 1212) 6868 ( 5 - 6060 808 )1212 ( 1 - 0808 404 ) 808 (2 - 808 HCF R = 0 STOP

STEP 2 : Consider 404 and 1111 1111 > 404 1111 ÷ 404 404 ) 1111( 2 -

0808 0303 ) 404 ( 1 - 303 101) 303( 3

- 303 HCF

0 = R , STOP

Hence HCF of 1212,6868 and 1111 = 101

G) 70, 105,175

Step 1 – 70, 105 ,175 105 > 70

105 ÷70 70 ) 105 ( 1 - 70

35 ) 70 ( 2 - 70 HCF

0 = R , stop

Step 2 - Consider 35 and 175 175 > 35 175 ÷ 35

35 ) 175 ( 5 - 175 HCF

0 = R , stop

Hence , HCF of 70 , 105 , 175 is 3 5

HCF of 4 and 8 = 4 5 ÷ 4 , remainder = 1 9 ÷4 , remainder = 1 so the required answer is 4 and that is the HCF of 4 ( 5-1 ) , 8 ( 9 – 1) So to start the solution of the above question we will first subtract 1 from 5 and 9 and then find the HCF of 4 and 8

Q.5) FIND THE GREATEST NUMBER WHICH DIVIDES 203 , 434 LEAVING REMAINDER 5 IN EACH CASE Ans ) We have to find a greatest number which exactly divides (203 -5) , (434 -5) The required number is the HCF of 198 , 429. Here 429 >198

429 ÷ 198 198 ) 429 ( 2 - 396 033) 198 ( 6 -

198 HCF 0 = R , stop Hence ,the required number is 33

Q.7) The length , breadth and height of a room are 8.25 m ,6.75 m and 4.50 m respectively, Determine the longest tape which can measure the three dimensions of the room exactly . Ans) 8.25 m = 8.25 X 100 = 825 cm 6.75 m = 6.75 X 100 = 675 cm 4.50 m = 4.50 X 100 = 450 cm The longest tape which can measure the three dimensions of the room exactly is the HCF of 825 ,675 ,450 .( the largest number that divides the given 3 numbers exactly) step 1 – 825 ,675 ,450 825 > 675 ,

825 ÷675 675 ) 825 ( 1 - 675 150 ) 675 ( 4 -

600 075 ) 150 ( 2 - 150

HCF

0=R

Step 2 – consider 75 and 450

450 > 75 450 ÷ 75 75 ) 450 ( 6

- 450 HCF

0 = R ,STOP

HCF of 8.25 m , 6.75 m and 4.50 m is 75 m Hence the longest tape should be of length 75 m so that it can measure the 3 dimensions of the room exactly ( no part of the tape shall be left unused) .

312 MANGO BITES ,260 ECLAIRS AND 156 COFFEE BITES IN A

Ans) To find the maximum number of toffees in each box we have to find the largest number that divides 312 , 260 and 156 exactly . That is we have to find HCF of 312 ,260 and 156

Step 1 – Consider 312 and 260 260 ) 312 ( 1 -

260

052 ) 260 ( 5 - 260 HCF

Step 2 - Consider 52 and 156

0 = R , STOP

52 ) 156 ( 3 - 156

HCF

0 = R,STOP

So HCF of 312 , 260 and 156 is 52. So there are 52 toffees in each packet .

Methods to find LCM

Listing multiples

Prime Factorisation method

Common division method

▪ 1) By listing method ▪ Multiples of 12 = 12,24,36,48,60 , 72 ,84 ,96,108 ,120 …. ▪ Multiples of 8 = 8,16,24,32,40,48,56,64,72,80,88,96,104 ,….. ▪ Multiples of 16 = 16 , 32,48 , 64 80,96,… ▪ Common multiples – 48 ,96 … ▪ L C M = 48

2 12

2

8

2

16

2

6

2 4

2

8

3

3

2 2

2

4

1

1

2

2 1

12 = 2 x 2 x 3 8=2x2x2 16 = 2 x 2 x 2 x 2

( in prime factorisation of 12 , 8 and 16 ,2 occur maximum 4 times in 16 and 3 occur maximum 1 time in 12 )

L C M = 2 X 2 x 2 X 2 x 3 = 48

Find L C M of 12 , 8 , 16 using common division method. ( Use 2,3,5,7,11,…) 2 12 , 8 , 16

2 6 , 4 , 8 2

3 , 2 , 4

2

3 , 1 , 2

3

3 , 1 , 1 1

, 1 , 1

L C M = 2 X 2 X 2 X 2 X 3 = 48

Q.3) Find the L C M by common division method C) 30, 24, 36, 16

2 30, 24, 36, 16 2 15, 12, 18, 8

2 15, 6 , 9 , 4 2 15, 3 , 9 , 2 3 15, 3 , 9 , 1

3 5, 1 , 3 , 1 5 5, 1 , 1 , 1 1,1 , 1 , 1

5

35 , 49 , 91

7

7 , 49, 91

7

1 , 7,

13

1, 1, 13 1 , 1, 1

L.C.M = 5 x 7 X 7 X 13 = 3185

13

2

12 , 16 , 24 , 36

2

6,

8 , 12 , 18

2

3,

4, 6, 9

2

3,

2, 3 , 9

3

3 , 1 , 3 , 9

3

1 , 1, 1, 3 1 , 1, 1, 1

L.C.M = 2 X 2 X 2 X 2 X 3 X 3 = 144

Sol- The required number = 5 + L.C.M of 40,50,60 First find L.C.M of 40, 50 , 60 2

40 , 50 , 60

2

20 , 25 , 30

2

10 , 25 , 15

3

5 , 25 , 15

5

5 , 25 ,5

5

1 , 5 , 1 1 , 1 , 1

L.C.M = 2 X 2 X 2 X 3 X 5 X 5 =600

The required number = 5 + 600 = 605

Sol: We will find L.C.M of 50 , 100 , 125 2

50 , 100 , 125

2

25 , 50 , 125

5

25 , 25 , 125

5

5 , 5 , 25

5

1 , 1, 1

5

,1 , 1

L.C.M = 2 X 2 X 5 X 5 X 5= 500 So, the 3 buses will stop together again after 500 Km .

Sol.) 4 bells toll together at 3 p.m and then they toll at the intervals of 8 , 9 , 12 and 15 minutes respectively . To find when they will toll together next we have to find their L.C.M 2

8, 9, 12, 15

2

4 , 9 , 6 , 15

2

2 , 9 , 3 ,15

3

1 , 9 , 3 ,15

3

1 ,3, 1, 5

5

1, 1, 1, 5 1, 1, 1 , 1

▪ So the three bells will toll together after 360 minutes that is 360/ 60 = 6 hours

So the three bells will toll together at 9 pm Vedio links for understanding concept of L.C.M AND H.C.F

https://www.youtube.com/watch?v=x8oq1u8ytWc

▪ https://www.youtube.com/watch?v=eHxUsLZGEaM