Download Exercise 1.1 Q1. Use Euclid`s division algorithm to find the HCF of: I

Exercise1.1
Q1.UseEuclid’sdivisionalgorithmtofindtheHCFof:
I.
135and225
II.
196and38220
III.
867and255
Ans.(i)135and225
Since225>135,weapplythedivisionlemmato225and135toobtain
225=135×1+90
Sinceremainder90≠0,weapplythedivisionlemmato135and90to
obtain
135=90×1+45
Weconsiderthenewdivisor90andnewremainder45,andapplythe
divisionlemmatoobtain
90=2×45+0
Sincetheremainderiszero,theprocessstops.
Sincethedivisoratthisstageis45,
Therefore,theHCFof135and225is45.
(ii)196and38220
Since38220>196,weapplythedivisionlemmato38220and196to
obtain
38220=196×195+0
Sincetheremainderiszero,theprocessstops.
Sincethedivisoratthisstageis196,
Therefore,HCFof196and38220is196.
(iii)867and255
Since867>255,weapplythedivisionlemmato867and255toobtain
867=255×3+102
Sinceremainder102≠0,weapplythedivisionlemmato255and102to
obtain
255=102×2+51
Weconsiderthenewdivisor102andnewremainder51,andapplythe
divisionlemmatoobtain
102=51×2+0
Sincetheremainderiszero,theprocessstops.
Sincethedivisoratthisstageis51,Therefore,
HCFof867and255is51.
Q2. Showthatanypositiveoddintegerisoftheform6q+1,or6q+3,or6q+
5,whereqissomeinteger.
Ans. Letabeanypositiveintegerandb=6.
Then,byEuclid’salgorithm,a=6q+rforsomeintegerq≥0,and
r=0,1,2,3,4,5because0≤r<6.
Therefore,a=6qor6q+1or6q+2or6q+3or6q+4or6q+5Also,6q
+1=2×3q+1=2k1+1,wherek1isapositiveinteger
6q+3=(6q+2)+1=2(3q+1)+1=2k2+1,wherek2isaninteger
6q+5=(6q+4)+1=2(3q+2)+1=2k3+1,wherek3isaninteger
Clearly,
6q+1,6q+3,6q+5areoftheform2k+1,wherekisaninteger.
Therefore,6q+1,6q+3,6q+5arenotexactlydivisibleby2.
Hence,theseexpressionsofnumbersareoddnumbers.
Andtherefore,anyoddintegercanbeexpressedintheform6q+1,or6q
+3,or6q+5
Q3
Anarmycontingentof616membersistomarchbehindanarmybandof
32membersinaparade.Thetwogroupsaretomarchinthesame
numberofcolumns.Whatisthemaximumnumberofcolumnsinwhich
theycanmarch?
Ans. HCF(616,32)willgivethemaximumnumberofcolumnsinwhichthey
canmarch.
WecanuseEuclid’salgorithmtofindtheHCF.
616=32×19+8
32=8×4+0
TheHCF(616,32)is8.
Therefore,theycanmarchin8columnseach.
Q4
UseEuclid’sdivisionlemmatoshowthatthesquareofanypositive
integeriseitherofform3mor3m+1forsomeintegerm.
[Hint:Letxbeanypositiveintegerthenitisoftheform3q,3q+1or3q+
2.Nowsquareeachoftheseandshowthattheycanberewritteninthe
form3mor3m+1.]
Ans. Letabeanypositiveintegerandb=3.
Thena=3q+rforsomeintegerq≥0
Andr=0,1,2because0≤r<3
Therefore,a=3qor3q+1or3q+2Or,
a2=(3q)2or(3q+1)2or(3q+2)2
=(3q)2or9q2+6q+1or9q2+12q+4
=3x(3q2)or3×(32+2)+13×(32+4+1)+1
=3k1or3k2+1or3k3+1
Wherek1,k2,andk3aresomepositiveintegers
Hence,itcanbesaidthatthesquareofanypositiveintegeriseitherofthe
form3mor3m+1.
Q5
UseEuclid’sdivisionlemmatoshowthatthecubeofanypositiveinteger
isoftheform9m,9m+1or9m+8.
Ans Letabeanypositiveintegerandb=3
a=3q+r,whereq≥0and0≤r<3
a=3qor3q+1or3q+2
Therefore,everynumbercanberepresentedasthesethreeforms.There
arethreecases.
Case1:Whena=3q,
a3=(3q)3=27q3=9(3q3)=9m
Wheremisanintegersuchthatm=3q3
Case2:Whena=3q+1,
a3=(3q+1)3
a3=27q3+27q2+9q+1
a3=9(3q3+3q2+q)+1
a3=9m+1
Wheremisanintegersuchthatm=(3q3+3q2+q)
Case3:Whena=3q+2,
a3=(3q+2)3
a3=27q3+54q2+36q+8
a3=9(3q3+6q2+4q)+8
a3=9m+8
Wheremisanintegersuchthatm=(3q3+6q2+4q)
Therefore,thecubeofanypositiveintegerisoftheform9m,9m+1,or
9m+8.