# Class 10 NCERT Solutions- Chapter 1 Real Numbers – Exercise 1.1

### Question 1. Use Euclid’s division algorithm to find the HCF of:

### (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

**Solution:**

i)135 and 225We need to apply Euclid’s algo, for finding HCF, now First Step to take Divisor ,we always take Divisor as smaller than Dividend.

Divisor=135

Dividend=225

Quotient=225/135=1(quotient is an integer value)

Euclid’s Div. Algo: 225=135*1+90

Since 90!=0 that means We have to again apply division lemma for remainder:90,Now Remainder Will become Divisor and Divisor

will become Dividend.

Euclid’s Div. Algo: 135=90*1+45

Again we Need to apply Division lemma because Yet remainder is not equal to 0 (45!=0) and just similar to prev. ways we will get

Divisor =45 and Dividend=90

Euclid’s Div. Algo:90=45*2+0

Here Remainder=0 that means We need to stop here,

When Remainder = 0 Then HCF = Divisor => HCF(225,135)=HCF(135,90)

=HCF(90,45)

=45.

ii)196 and 38220Divisor=196

Dividend=38220

Quotient=38220/196=195

Euclid’s Div. Algo: 38220=196*195+0

Remainder=0,We don’t need to apply Division lemma further,

HCF(38220,196)=196

iii) 867 and 255Divisor=255

Dividend=867

Quotient=867/225=3

Euclid’s Div. algo: 867=255*3+102

Remainder = 102 (!=0) That means Again we need to apply Division lemma method, Divisor=102,Dividend=255,Quotient=255/102=2

Euclid’s Div. algo: 255=102*2+51

Again Remainder =51(!=0), We need to apply Division lemma method again, Divisor=51,Dividend=102,Quotient=102/51=2

Euclid’S Div. algo:102=51*2+0

Remainder =0,We need to stop here ,

HCF(867,255)=HCF(255,102)

= HCF(102,51)

= 51

### Question 2.Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

**Solution:**

**We know that any odd no integer is not divisible by 2,

When we divide 6q by 2 then it is perfectly divisible by 2 and give result 3q without any remainder,=>6q is even number

let take any positive integer x and y=6

as Per Euclid’s Div algo x=6q+r, q>=0 & 0<=r<6

if r=0 => x=6q, we earlier concluded that 6q is divisible by 2 =>x=6q is positive even integerif r=2, 4 =>x=6q+2, x=6q+4 also An positive even integer because 6q is even as well as 2,4 also even and even divided by even gives

evenif r=1

Dividing 6q+1 by 2:

Dividend=6q+1,Divisor=2

Quotient = (6q+1)/2 =3q

as per Euclid’s div algo 6q+1=3q*2+1

As We Got Remainder = 1 after dividing by 2 that means 6q+1 is odd number, As q>=0 that means 6q+1 is positive odd integerif r=3

Dividing 6q+3 by 2:

Dividend=6q+3,Divisor=2

Quotient = (6q+3)/2 =3q+1

as per Euclid’s div algo 6q+3=(3q+1)*2+1

As We Got Remainder = 1 after dividing by 2 that means 6q+3 is odd number, As q>=0 that means 6q+3 is positive odd integerif r=5

Dividing 6q+5 by 2:

Dividend=6q+5,Divisor=2

Quotient = (6q+5)/2 =3q+2

as per Euclid’s div algo 6q+5=(3q+2)*2+1

As We Got Remainder = 1 after dividing by 2 that means 6q+5 is odd number, As q>=0 that means 6q+5 is positive odd integer

Overall Conclusion : Any positive integer x can be in form 6q+1,6q+3 or 6q+5 if it is odd otherwise it is in form 6q, 6q+2, 6q+4

### Question 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

**Solution:**

Given,

Total number of army contingent members=616

Total number of army band members=32Given two groups are to march in same number of column then maximum number of columns will be Highest Common Factor between two

groups ie HCF(616,32)Finding HCF(616,32)

———————

Divisor=32,Dividend=616,quotient=616/32=19

Euclid’s Div. Algo: 616=32*19+8

Remainder = 8(!=0), again Apply Division lemmaDivisor=8,Dividend=32,Quotient=32/8=4

Euclid’s Div. Algo: 32=8*4+0

Remainder = 0,We have to stop here

HCF(616,32)=8Therefore, Maximum Number of Columns Will be 8 in which Both groups can march

### Question 4.Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

**Solution:**

Let x be any positive integer, And y=3,

As per Euclid’s Div algo. x=3q+r;q>=0 && 0 <= r < 3 => r=(0,1,2)If r=0: x=3q

Squaring both sides: x^{2 }=3*3*q^{2}

let assume m1=3q^{2 },Here m1 will be any positive integer because q>=0.

=>x^{2}=3m1

if r=1: x=3q+1

Squaring both sides:x^{2 }=(3q+1)^{2}

=>x^{2}=3(3q^{2}+ 2q)+1

let assume m2=3q^{2}+2q

=> x^{2 }= 3m2 + 1 ,Here m2 will be any positive integer because q>=0.

if r=2 : x=3q+2

Squaring both sides:x^{2}=(3q+2)^{2}

=>x^{2}=3(3q^{2}+ 4q+1)+1

let assume m3=3q^{2}+4q+1

=> x^{2}= 3m3 + 1 ,Here m3 will be any positive integer because q>=0.

Since m1,m2,m3 are positive integer ,therefore we can Conclude that x^{2}=3m or 3m+1 ,Where m is some integer.

This proved that the square of any positive integer(x) is either of the form 3m or 3m + 1 for some integer m

### Question 5.Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

**Solution:**

Let x be any positive integer, and y=3

As per Euclid’s Div algo. x=3q+r; q>=0 && 0 <= r < 3 => r=(0,1,2)if r=0: x=3q

Cubing both sides: x^{3}=9*3*q^{3}

let assume m1=3q^{3},Here m1 will be any positive integer because q>=0.

=>x2=9m1

if r=1: x=3q+1

Cubing both sides: x^{3}=(3q+1)^{3}

=>x^{3}=9(3q^{3}+3q^{2}+q)+1

Let assume m2=3q^{3}+3q^{2}+q

=> x^{3}= 9m2 + 1 ,Here m2 will be any positive integer because q>=0.

if r=2 : x=3q+2

Cubing both sides: x^{3}=(3q+2)^{3}

=>x^{3}=9(3q^{3}+ 6q^{2}+4q)+8

let assume m3=3q^{3}+6q^{2}+4q

=> x^{3}= 9m3 + 8 ,Here m3 will be any positive integer because q>=0.

Since m1,m2,m3 are positive integer ,therefore we can Conclude that x^{3}=9m,9m+1 or 3m+8, where m is some integer.

This proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

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