fadors and multiples

Ifa given number is exactly divisible by anotler number, that nunber is called afactor of the given number, while the glven number is called a multiple of theother number.E&unples (t) 10=2x5.So,2andSarefactorsof 10. Also, 10 is amultiple of

2 and 5.

III)7O-2xbx7. So, 2, 5 and 7 are factors ol 70. Also, 70 is amul t ip le o l2 , 5 and 7 . Fur ther ,2x5 ,2x7,5v7 and 2x5 x7 arefactors of 70.

Note . Every number is a factor of itself and 1 ls a factor of every number.. Every number is a multlpl€ of itself and every number is a multlpb of L. Every numb€r has countless multlples.

Classificatioh of numbers

Even nunlberc

Numbers that leave no remairder on b€lng dt\,'tded by 2 a.re called evenDunrbers. In other words, numbers that are multiples of 2 are even numbers.For exarnple, O, 4,78, 126, 5132 a]rd 23450 are even numbers.Note The even dlglts are 0, 2, 4, 6 and 8. Any number har.lng an even dlglt ln the units

place ls a.n even number,

Odd numbers

Numbers not exactly di!'tsible by 2 are called odd numbers. In other words,numbers that are not multlple of2 are called odd numbers. For example, l, 13,125. 4137 and 18249 are odd numbers.Note The odd digits ar€ l, 3, 5, 7 and L Any nunber haling an odd digit in the unlts

place ls an odd number

Prime numbet.

A number is called a pdme number or prime if it has no other factors except Iand itself. 2, 3, 5, 7, 11, 13, 17, 19 and 23 are some prime numbers. I is notconsidered a prime number. 2 ts the only even prime nurnber and it is thesmallest prime number.

ComDosite numberc

A number that has two or more prtne factors is called a composite number.4,6,8, 21, 136 are some composite numbers.

N.6

Faclors and Muttlples

lbo numbers are called co_prime orGumon factor except l. For example.co-prime numbers.

N-7

reldl ivel j prime if they do not have aJ and 4: I 7 dnd 2 t: and 23 and 25 a-ce

itted aof the

ple of

l i sa7 are

Fairs of prime numbers that diJTer by-2.are called twin primes. 3 and 5; S and 7;l r and l3: and I 7 and | 9 arF some sxanptes ol twjn primes.

o.ization

wriung a number as a product of prime numbers is caled prime factorizatron.Exanrples ( i) 48 =2 x2 x2 x2 xS.

l ir) 1260 =2 x2 xS x3 xS x7.

ot pirrc factorization

a4O=2x42O= 2 x2 x2 tO=2x2x2 \ tOE=2x2x2x3x \ s=2x2x2xBxSxz

35

Divide the number repeatedly by pdlne numbers 2. S, 5, Z, ... till the quotient is 1.Altcr rativety, breal{ the number tnto smaller factors till all the fa",;.";;;;;r-".Expr€ss 84O as a product of prlmes.

)ers.

rnrLs

-ds.

13 ,

nits

r l

he

2l a40214n-2121os fios-57

: . 84O=2x2x2x3x ' x? .

Highest Common Factor (HCF)The greatest number whiclhighest connron fac,o. (t;-t::-TTTol

factor of two or more numbers is theirLxdmp,e rn. racro,.s or ztl 1,1;T:.::1";:il:;visor rccD.

The fac to rs o f80 _ 1 ,2 ,4 .5 ,8 , lO , 16 ,20 .40 ,80 .... the common factors of 24-and 80 = 1, 2, 4, 8 and of these, g istJre greatest number. So, the HCF of 2;;;,;*= s.

\ lethods of f inding HCF

There€re two methods of finding the HCF of two or more numbers.(i) By prime factorizaflon (ii) By di\,,ts.on

Express each Dumber as a product oI prime numbers.Fhd the common factors of the numbers.

1 .

2 .

time factotization

Steps

N-8 ICSE Mathemattcs for Clss a

3. Find the product of the common factors This product is

requlred HCF

Ftnd the HcF of 252, 315 aDd 441.

252 = 2x r26 =2x2x6g = 2x2 x3x2 \ =2x2x3 x3 x7 '

315 -3 . I 05=3 3 ,35 .3 3 ' 5 7

441 =31147 = 3 x3 x 49 = 3xgx7 x7

The common factors of 252, 315 alird 441 = 3' 3 ' 7

.'. the HCF = 3 x 3 x 7 = 63.

TXAMPI.E

To fin.l HCr bv division

Steps 1. Divlde the bigger numb€r by the smaller number'

2, ll ttrere ls no remainder then the smaller number is the HCF Ifthere

a remainder, take tt as the nertr divisor aDd take the pre!'loutdifisor

*re new divldend

EXAMPLE

Solution

3. Contilirue Step 2 til there is no remainder' The last dtvisor is the

required HCF

To find the HCF of three o! more numbers. take the following steps

Steps l. Ftnd the HCF of any two of the numbers'

2. Find the HCF of the third number and th€ HcF obtained in Step 1'

This ts the HCF of the three numbers'

Continue similarly for more numbers'

ftnd the HcF of 594, 81O atrd 1260.

Ffst we flnd the HcF of 594 and 810,

594 < 810, so, 594ls the dit'isorand 810 ts the div'tdend.

594) 810 (1- 594--zl6l5e4 (2

- 432--Idt2r6 0

- 162---541162 (3- 162

the HcF of594 and 810 = 54.

Lowest Common Multiple (tcM)

The lowest common multiple (LCM) of two or more numbers is the smallest of

the common multiples of those numbers'

Example The multiples of 4 ar e 4' a' 12, 16, 20, @ 28' elc'

The mulliples c,f 6 are 6 12 I& @ 30 36 elc'

The muttiples of8 are 8' 16' E 32' 40' etc

Now we flnd the HcF of 54 and 1 26054) 1260 (23

- 108---l€o, - 762-l{1I54 (3.+

. . the HCF of 54 and 1260 = 18.

Hence, the HcF ofthe glven numbers = 18'

}Iere isisor as

rs tne

; t c p l .

0.

Factors dd MuiuPles

CleaJly, 2+ is the smallest or the lowest common multiple of 4, 6and 8.

So, the LCM of 4, 6 and 8 = 24.

r of f indinS LCM

There are two methods of finding the LCM of two or more numbers(a) By prtne factorization {b) By division

LCM ime lacto zation

Steps l. Express each number as a product ofpdme lactors

2. Take each prime factor the €Featest number of times it appears ln anyof the prlme factorizations ofthe numbers.

3. The product of the pdme factors in Step 2 is the LCM of the numbers

rind the LcM of 75, 125 and 15O,

75 = 3 x 25 = 3 x 5 x 5. 125 = 5 x 25 = 5 x 5 x 5,150 = 2 x 75 = 2 x 3 x 25 = 2, 3 x 5 x 5.

The greatest Dumber of times that 2 appears as a factor ofany of the numbers ls

one, 3 appears once, 5 appears thrice.

.. LcM = 2 x 3 x 5 x 5 x 5 = 750.

find LCM by division

Steps 1. Di!'ide the numbers by a prime number which ls a facto! of at least two^ f l L a . r l v p r h , ! h h p r .

2. Write the quotients and carry lorward the numbers lvhich are notdt!'tslble.

3. Repeat Steps 1 and 2 ttll no 1wo ofthe numbers have a common factor'

4. The product of the dil'lsors of all the sleps and the remaining numbersis the LCM of the given numbers.

Ftnd the LcM of 24, 28, 35 and 75.

. . L C M = 2 x 2 x 3 x 5 x 7 x 2 x 5 = 4 2 O O .

bme important prcperties of HCF and LCM

22357

1. The product of two numbers is equal to the product oftheir HcF and LCM.

24, 2a, 35,75t 2 , t 4 , 35 ,756 ,7 ,35 ,752, 7 ,35 ,25

HCF x LCM : flrst number: second.number

lest of

EXAMPLE The product of two numbers is looS and their LcM is 168, Find their Hcr.

. Drodurl of llre Nvo numbers | 008 ^tlLI OI IWO nUmDerS ' = b.

LCM of the numbers r68The HCF of two numbers cannot exceed e'lther of the numbers.

The LCM of two numbers cannot be less than either of the numbers.

The HCF of two co-prime numbers = 1 , and tf tlle HCF oi two numbers is 1 , thetwo numbers are co-pdme,

The LCM of two co-prime numbers = the product of the two nunbers.

Flnd the IlcF a[d LcM of 14 and 25,

14 and 25 are co'prlme numbers... HcF = I and LcM = 14 x 25 = 350.

Solved [xamples

Find the greatest rlumt er that wiU dtvide 326, 895 and 1790 leavtng thercmainders 11, 13 arrd 5 respecttvely.

326 - l l =315 , 895 -13 =882 , 1790-5 =1785 .The required number ls the HCF of 315, 882 a.nd 1785.

2.

3 .

4 .

5 .

EXAMPTf

EXAMPLT I

EXAMPTE 2

315) 882 (2- 630--252)

3 r5 0- 252---6q282 (4

- 252' . X

. . the HcF of315 and 882 = 63.

.. the HcF of315, 882 and 1785Hence, the required number = 21.

Now, 63) 1785 (28- 126----w

- 504---tlt63 (B

-63'

. . the HCF of63 and. l7a5 = 21.

(i) nind the LcM of 24, 36, 60, 72 and 90.(ii) Find tne smallest nlrnber which when divid€d by 24,36', 60,72 and 90

leaves the remalnder 5 {n €ach case.

tiiil ritrd th€ smallest number which when increased by 8 is exactly dlvisible by24. 36. 60, 72 and 90.

22335

(ii) The required number = LCM + 5 : 360 + 5 = 365.

(i i i ) The required number : LCM-8:360-8 = 352.

12 , 18 ,30 ,36 ,456 . 9 , 15 . 18 ,453 , 9 , 15 , 9 ,45 . . L C M = 2 x 2 x 2 x 3 x 3 x 5 = 3 6 0 .r , 3 , 5 , 3 , 15l , 1 . 5 , 1 , 5l . I , t , l , I

N, l l

Find the greatest sumber of six digits $,hich is exacuy dtvisible by 4, a, L2, li, 2lanal25.

First. we Iind the LCM of 4. 8, 12, 15,2I and 25.2 4 , A , 12 . 15 .2 t .252 | 2 ,4 ,6 , t 5 ,2 ' 1 ,253 r , 2 , 3 , 15 . 2 t . 25 . . LCM 2 .2 . 3 . 5 . 2 . 7 . 5 = 42005 | t , 2 , l , 5 , 7 , 25

1 ,2 , 1 . t , 7 . 5

The greatest number of six digits = 999999.L€t us dtvide this by the LCM.

4200) 9e9999 (2388400rE50t

- t?990 .. 999999 + LCM leaves the rematnder 399.33999

_ 33600

Hence, the required number = 999999 - remalnder = 999999 _ 399 = 999600.

Find the small€st number of ftve digits which is exactly divistbte by B, 12, t4, zaand 56.

First, we find the LcM of 8, 12, 14, 28 and 56.

Factors dd Muluples

l . the

90

) b y

. . LcM = 2 x2 x2 x7 xA = t68 .

The smallest number of ftve dtgits = IOOOO.Let us dlvide this by the LCM.

168) 10000 (59- 840---I60d- I 5 l 2

So. the 59th multiple of 168 is less than IOOOO,.'. the required number = 60th multlple of l6a = 60 x 168 = IOO8O.ALt'L'",)rtitl!The requlred number = IOOOO-remainder + LCM = IOOO0_88+ 168 = IO0SO.

Five bells start rlnging together. Ifth€ bells rtng at intervals of 12, la, 2b, 40 and54 s€conds respectively, after what interval of time will they rlng togeth€r agaid?

The required time (tn seconds) = the LCM of t2, 18,25,40 and 54.2 12 , 1A ,25 ,40 ,54

2 1.. - 3, L?, 142q,562 4 ,6 .7 . t4 .28z ). z,-s,-z,:, ty7 l r ,3 .7 .7 .7

f r , s. l , - r . I

fIAMPIE 5

2335

. . the required LCM = 2 x 2 x 3 x 3 x 5 x 5 x 2 x S= 5400.

r , 1 .5 ,2 ,3

l , 3 . 25 . lO , 91 . 1 ,25 , lO . 3

Hence, the required dme = 54OO s = 90 mtn = t h 30 min.

N-12

EXAMPTE 6

ICSE Maihcmatl.s fo. Class 8

The LCM of two numbers i6 12 times their HCr and the diff€rence of the LCM aldthe HCr is I IO. If one of the numbers is 30, lind the other,

Let the HcF of the two numbers = r.. . the LCM oI the two numbers = 12i.Given, the LCM the HCF = IIO+ 12x r=1 IO e l l x= I lO ... the HCF : 10 and the LCM = 120.'. the product of two numbers = HCF x LCM

.. 30 x the other number = HCF x LCM.. H c F L c M l o 1 2 0 - ^

Lne olner numoer ' - , lo 30

x=10 .

1.

2 ,

3 .

Remember These

Expreqsing a number ac a product ot prlme lactors is called prime factorlza(lon,

The HCF of two o! more numbers is the greatest number whlch exactly divides a-ithc numbers,

Thcre arc two methods of finding the HCF| (i) by prime lacto zatio! and (it) by division.

The LCM oI two or more numbers is the smallest number whtch can be dil'ided o{actly byall lhe given numbers.

5. There are lwo methods ofllnding lhe LcM: (i) by prtne factortzatlon and (ti) by dt!'ision.

6. The producl oI Lrvo flumbers = the LcM of the numbers x the HcF of the numbers.

7. The HCF of two co-prlme numbers = l.

E. Thc LCM of two co-pdme numbers = the product of the two co-prlme numbers.

l . Find lhe HCF of drc fo owing by pr imr fuclorrzai lon.( i) 48, 12O and 216

2. Find the HCF of the followtng.

t i) 216, 756 and 1260(i i i) 528. 1584 and 2I78

(i i) 21O. 280 and 525

{i i) 108, 288 and 720(lv) 90. 216, 720 and 1908

3. Find the gfeatest number that \a'lll divide(tl 399 and 828 learing the remalnder 9 in each case,

lli) 445,721 and 942 leaving the remainders 5, 6 and 7 respectively.

Find the LCM of the following by p me factorization.+,(t 16. 28, 84 lrj.) 132, 165,22O ( j i i l 36 ,64 ,96 , 100

- . -.-pl6

r r -u ^ f rh- a^r^- i - -

2r ,3030: 24, 54

sallest number

t i t j 48 ,56 , 105,225 ( i j i ) 12 , 1a , s0 .48 ,60l,r) 12, Ia,24,30,86, 42

which rrhen divided by f8, 28, fO8 and 105 leaves the

90 and 108they change

N-t3

7ln each case,

amallest number wbtch .lvhen increased by 5 is exacdy dMsible by 12, 16, la,2Z

least number whtch when decreased by 2 is exac y divisible by 6, g, 12, lg and

greatest number of six digits which is exac y dtvisible by 8, tZ,16,20 a:Ild,24_sma.llest number of six dlgtts which is exacdy di\,,isible by 12, 16, 30, 48 and €jO.start rtngtng together. If the bells ring at intervats of 2, 4, 6, 8 and lO seconctsLy, after what hterval of Ume q,ill they-ring together €ain?

l: ltgh,: ,., four. dlfferenr crosstngs change ar Inrervals of 48: 72,respecuvely. I these lighrs change togetler at 7rO5 p.m.. when fillagain?

I,cM and HCF of two numbers are I 44 and 24 respecflvely. Find the product of the two

p.oduct of two numbers is 22176. If thelr HCF ts 12, flnd thetr LCM.rcM and HCF ol two numbers are 960 and 16 respecuvely. If one of the numbers lsfind the other number.

- ll|Ll. or rwo numbers ls _: of thef LcM, and the sum of the LcM and the HcF is 116. Ifof the numbers is 16, llnd the other,

ANSWIRS4 { i i ) 35

(,lJ 660 (iu) 14400

2. ( l ) 36 (u)36 ( i i i ) 66 0vJ l86. (i) 840 (lt 25200 UIr) Z2O (iv) IO8O7 , 4 2 7 a . 3 6 2

3. (i) 39 0l) 55(vl 2520

9. 99984013. s456

3:]6

r1. 2 mlnui€s15. 20

t 2 , 7 t 4 L p . m .1 6 . 2 8

.|.