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#### Page No 2:

#### Question 1:

Choose the correct answer from the given four options

For some integer *m*, every even integer is of the form

(A) *m*

(B) *m *+ 1

(C) 2*m*

(D) 2*m* + 1

#### Answer:

As we know that an integer is said to be even when it is divisible by 2.

Let *m* be any integer.

i.e* m* = ..., -5,-4,-3,-2,-1,0,1,2,3,4,5,...

so* m* can be even or not even.

hence we can say '*m*' cannot always be even.

Now for , *m*+1 =..., -4,-3,-2,-1,0,1,2,3,4,...

here also '*m*+1' cannot always be even.

Now consider 2*m* =..., -10,-8,-6,-4,-2,0,2,4,6,8,10,...

here '2*m*' is always even.

For 2*m*+1 = ..., -9,-7,-5,-3,-1,1,3,5,7,9,11,....

here also '2*m*+1' is always an odd integer.

So, we can say '2*m*' is the answer.

Hence, the correct answer is option C.

#### Page No 2:

#### Question 2:

Choose the correct answer from the given four options

For some integer *q*, every odd integer is of the form

(A) *q*

(B) *q* + 1

(C) 2*q*

(D) 2*q* + 1

#### Answer:

As we know an integer is said to be odd if it is not divisible by 2.

Let *q* be any integer i.e* q *= ...,-2,-1,0,1,2,...

Now multiplying both sides by 2 we get:

2*q* = ...,-4,-2,0,2,4,...

Now adding 1 on both sides we get:

2*q* + 1 = ...,-3,-1,1,3,5,...

Thus for any integer *q* , integer of the form 2*q* +1 is always odd.

Hence,the correct answer is option D.

#### Page No 3:

#### Question 3:

Choose the correct answer from the given four options

*n*^{2} – 1 is divisible by 8, if *n* is

(A) an integer

(B) a natural number

(C) an odd integer

(D) an even integer

#### Answer:

Let *a *= *n*^{2} − 1

Here *n* can be odd or even.

Case 1: *n* = even , let *n* = 4*k*, where *k* is any integer.

$\Rightarrow a={\left(4k\right)}^{2}-1$

$\Rightarrow a=16{k}^{2}-1$

Which is not divisible by 8.

Case 2: *n* = odd, let *n *= 4*k* + 1, where *k* is any integer.

$\Rightarrow a={\left(4k+1\right)}^{2}-1$

$\Rightarrow a=16{k}^{2}+1+8k-1\phantom{\rule{0ex}{0ex}}\Rightarrow a=16{k}^{2}+8k\phantom{\rule{0ex}{0ex}}$

$\Rightarrow a=8\left(2{k}^{2}+k\right)$

Which is always divisible by 8.

Thus we can conclude from above two cases , if n is odd , then ${n}^{2}-1$ is divisible by 8.

Hence, the correct answer is option C.

#### Page No 3:

#### Question 4:

Choose the correct answer from the given four options

If the HCF of 65 and 117 is expressible in the form 65*m* – 117, then the value of *m* is

(A) 4

(B) 2

(C) 1

(D) 3

#### Answer:

We know by Euclid's division algorithm,

*b* = *aq *+ *r*, where $0\le ra$

$\Rightarrow 117=65\times 1+52\phantom{\rule{0ex}{0ex}}\Rightarrow 65=52\times 1+13\phantom{\rule{0ex}{0ex}}\Rightarrow 52=13\times 4+0$

∴ HCF(65,117) = 13 .........(1)

Also, given that HCF(65,117) = 65m − 117 ...........(2)

From (1) and (2) we get:

$65m-117=13\phantom{\rule{0ex}{0ex}}\Rightarrow 65m=130\phantom{\rule{0ex}{0ex}}\Rightarrow m=2$

Hence, the correct answer is option B.

#### Page No 3:

#### Question 5:

Choose the correct answer from the given four options

The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is

(A) 13

(B) 65

(C) 875

(D) 1750

#### Answer:

First, we need to subtract the remainders 5 and 8 from corresponding numbers respectively and then get the HCF of resulting numbers using Euclid's division algorithm, which will be the required number.

$\Rightarrow $ After subtracting these remainders from the numbers we have:

(70 − 5) = 65

(125 − 8) = 117

Now, required number is HCF(65,117).

$\Rightarrow $ Using Euclid's division algorithm:

$117=65\times 1+52\phantom{\rule{0ex}{0ex}}65=52\times 1+13\phantom{\rule{0ex}{0ex}}52=13\times 4+0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

∴ HCF = 13

$\Rightarrow $ 13 is the largest number which divides 70 and 125 leaving remainders 5 and 8.

Hence, the correct answer is option A.

#### Page No 3:

#### Question 6:

Choose the correct answer from the given four options

If two positive integers *a *and *b* are written as *a* = *x*^{3}*y*^{2} and *b* = *xy*^{3}; *x, y* are prime numbers, then HCF (*a, b*) is

(A) *xy*

(B) *xy*^{2}

(C) *x*^{3}*y*^{3}

(D) *x*^{2}*y*^{2}

#### Answer:

Given:

$a={x}^{3}{y}^{2}=x\times x\times x\times y\times y\phantom{\rule{0ex}{0ex}}b=x{y}^{3}=x\times y\times y\times y$

And we know that HCF is the largest common factor of two or more numbers.

$\Rightarrow $HCF of* a* and *b *= HCF(*x*^{3}*y*^{2}, *xy*^{3}) = $x\times y\times y=x{y}^{2}$

Hence, the correct answer is option B.

#### Page No 3:

#### Question 7:

Choose the correct answer from the given four options

If two positive integers *p* and *q *can be expressed as *p* = *ab*^{2} and *q* = *a*^{3}*b*;* a, b* being prime numbers, then LCM (*p, q*) is

(A) *ab*

(B) *a*^{2}*b*^{2}

(C)* a*^{3}*b*^{2}

(D)* a*^{3}*b*^{3}

#### Answer:

Given:

$p=a{b}^{2}=a\times b\times b\phantom{\rule{0ex}{0ex}}q={a}^{3}b=a\times a\times a\times b$

And we know that LCM is smallest positive integer that is divisible by both given numbers.

$\Rightarrow $LCM of *p* and *q* = LCM(*ab*^{2}, *a*^{3}*b*) = $a\times b\times b\times a\times a={a}^{3}{b}^{2}$

Hence, the correct answer is option C.

#### Page No 3:

#### Question 8:

Choose the correct answer from the given four options

The product of a non-zero rational and an irrational number is

(A) always irrational

(B) always rational

(C) rational or irrational

(D) one

#### Answer:

Product of a rational $\frac{7}{2}$ and an irrational $\frac{\sqrt{5}}{2}=\frac{7}{2}\times \frac{\sqrt{5}}{2}=\frac{7\sqrt{5}}{4}$,which is irrational.

Hence, the correct answer is option A.

#### Page No 3:

#### Question 9:

Choose the correct answer from the given four options

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(A) 10

(B) 100

(C) 504

(D) 2520

#### Answer:

We know that LCM is the smallest positive integer that is divisible by given numbers.

$\Rightarrow $We need to find LCM of the numbers from 1 to 10 (both inclusive).

$\Rightarrow $Factors of 1 to 10 numbers:

$\begin{array}{rcl}1& =& 1\\ 2& =& 1\times 2\\ 3& =& 1\times 3\\ 4& =& 1\times 2\times 2\\ 5& =& 1\times 5\\ 6& =& 1\times 2\times 3\\ 7& =& 1\times 7\\ 8& =& 1\times 2\times 2\times 2\\ 9& =& 1\times 3\times 3\\ 10& =& 1\times 2\times 5\end{array}$

∴ LCM of numbers from 1 to 10 = LCM(1,2,3,4,5,6,7,8,9,10).

$\Rightarrow $ LCM = $1\times 2\times 2\times 2\times 3\times 3\times 5\times 7=2520$

Hence, the correct answer is option D.

#### Page No 3:

#### Question 10:

Choose the correct answer from the given four options

The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after:

(A) one decimal place

(B) two decimal places

(C) three decimal places

(D) four decimal places

#### Answer:

As we know, in terminating rational number the denominator always have the form ${2}^{m}\times {5}^{n}$.

$\Rightarrow $ Rational number = $\frac{14587}{1250}=\frac{14587}{{2}^{1}\times {5}^{4}}$

$\Rightarrow \frac{14587}{10\times {5}^{3}}\times \frac{{2}^{3}}{{2}^{3}}=\frac{14587\times 8}{10\times 1000}=\frac{116696}{10000}=11.6696\phantom{\rule{0ex}{0ex}}$

Thus, we can say given rational number will terminate after four decimal places.

Hence, the correct answer is option D.

#### Page No 4:

#### Question 1:

Write whether every positive integer can be of the form 4*q* + 2, where *q* is an integer. Justify your answer.

#### Answer:

By Euclid's division lemma.

*b *= *aq* +* r* , where *b* is any positive integer and 0 ≤ *r *< *a.*

Now, if we divide *b* by 4, we get

*b* = 4*q* +* r* for 0 ≤ *r *< 4.

$\Rightarrow $*r *= 0 ,1 ,2 ,3

So, *b* can be in the form of 4*q*, 4*q *+ 1, 4*q *+ 2, 4*q *+ 3.

Hence, Every positive integer cannot be of the form 4*q *+ 2.

#### Page No 4:

#### Question 2:

“The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Give reasons.

#### Answer:

Let the two consecutive positive integers be *n* and *n *+ 1.

Now, the product of two consecutive positive integers will be *n*(*n *+ 1).

Thus, We will have the following two cases possible:

Case 1:

if *n* is even, then *n *+ 1 will be odd.

i.e. if *n* = 2, then *n *+ 1 = 3.

This implies that *n*(*n *+ 1) will always be even, since one of these two numbers is always divisible by 2.

Case 2:

if *n* is odd, then* n *+ 1 will be even.

i.e. if *n *= 1, then* n *+ 1 = 2.

This implies that *n*(*n *+ 1) will always be even, since one of these two numbers is always divisible by 2.

Hence, we can conclude that product of two consecutive positive integers is divisible by 2.

#### Page No 4:

#### Question 3:

“The product of three consecutive positive integers is divisible by 6”. Is this statement true or false”? Justify your answer.

#### Answer:

Let the three consecutive positive integers be *n*, *n *+ 1 and *n *+ 2.

Now, the product of three consecutive positive integers will be *n*(*n *+ 1)(*n *+ 2).

Case 1:

if *n* is even, then *n *+ 1 will be odd and *n *+ 2 will be even.

i.e. if *n* = 2, then *n *+ 1 = 3 and n + 2 = 4.

This implies that *n*(*n *+ 1)(*n *+ 2) will always be divisible by 2 and 3.

Case 2:

if *n* is odd, then* n *+ 1 will be even and *n *+ 2 will be odd.

i.e. if *n *= 1, then* n *+ 1 = 2 and *n *+ 2 = 3.

This implies that *n*(*n *+ 1)(*n *+ 2) will always be divisible by 2 and 3.

We know any number divisible by 2 and 3 is also divisible by 6.

Hence, we can conclude that product of three consecutive positive integers is divisible by 6.

#### Page No 4:

#### Question 4:

Write whether the square of any positive integer can be of the form 3*m* + 2, where *m* is a natural number. Justify your answer.

#### Answer:

By Euclid's division lemma:

*b* = *aq* + *r *and 0 ≤ *r *< *a*, where *b* can be any positive integer.

Now, if we divide *b* by 3, we get

*b* = 3*q* +* r* for 0 ≤ *r *< 3

Now, *b* can be written in the form of 3*q*, 3*q* + 1, 3*q* + 2.

Now squaring each possible form of* b*, we have three possible cases:

Case I:

If *b* = 3*q*, squaring both sides, we get

$\Rightarrow {\left(3q\right)}^{2}=9{q}^{2}=3\left(3{q}^{2}\right)=3m$, where *m *= 3*q*^{2}.

Case II:

If *b* = 3*q* + 1, squaring both sides, we get

$\Rightarrow {\left(3k+1\right)}^{2}=9{k}^{2}+6k+1\phantom{\rule{0ex}{0ex}}\Rightarrow 3(3{k}^{2}+2k)+1=3m+1$

where* m* = 3*k*^{2} + 2*k.*

Case III:

If *b* = 3*q* + 2, squaring both sides, we get

$\Rightarrow {(3k+2)}^{2}=9{k}^{2}+12k+4\phantom{\rule{0ex}{0ex}}\Rightarrow 9{k}^{2}+12k+3+1\phantom{\rule{0ex}{0ex}}\Rightarrow 3(3{k}^{2}+4k+1)+1\phantom{\rule{0ex}{0ex}}\Rightarrow 3m+1$

where *m* = 3*k*^{2} + 4*k *+ 1.

So, we can conclude, square of any positive integer can be of the form 3*m* and 3*m *+ 1

Hence, square of any positive number cannot be of the form 3*m *+ 2.

#### Page No 4:

#### Question 5:

A positive integer is of the form 3*q* + 1, *q* being a natural number. Can you write its square in any form other than 3*m* + 1, i.e., 3*m* or 3*m* + 2 for some integer *m*? Justify your answer.

#### Answer:

By Euclid's division lemma:

*b* = *aq* + *r*, 0 ≤ *r *< *a*, where *b* can be any positive integer.

Now, if we divide *b* by 3, we get

*b* = 3*q* +* r* for 0 ≤ *r *< 3

Thus, *b* can be written in the form of 3*q*, 3*q* + 1, 3*q* + 2.

Now, squaring 3*q *+ 1, form of the positive integer, we get

${\left(3q+1\right)}^{2}=9{q}^{2}+6q+1\phantom{\rule{0ex}{0ex}}\Rightarrow 3(3{q}^{2}+2q)+1=3m+1$

where *m* = 3*q*^{2} + 2*q.*

So we can conclude,square of the form 3*q *+ 1 can be written as 3*m *+ 1 always Hence, it cannot be written in the form of 3*m* or 3*m *+ 2.

#### Page No 4:

#### Question 6:

The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is HCF (525, 3000)? Justify your answer.

#### Answer:

By Euclid's division algorithm, we can find the HCF(525, 3000).

$\begin{array}{rcl}& \Rightarrow & 3000=525\times 5+375\\ & \Rightarrow & 525=375\times 1+150\\ & \Rightarrow & 375=150\times 2+75\\ & \Rightarrow & 150=75\times 2+0\end{array}$

Now, the numbers 3, 5, 15, 25 and 75 are the common factors of 525 and 3000, but the highest common factor is 75.

So, we can say that HCF(525, 3000) is 75.

#### Page No 4:

#### Question 7:

Explain why 3 × 5 × 7 + 7 is a composite number.

#### Answer:

A number which has more than two factors is known as composite number.

We have,

$3\times 5\times 7+7=105+7=112$

$\Rightarrow 112=2\times 2\times 2\times 2\times 7={2}^{4}\times 7$

So, it is the product of prime factors 2 and 7. i.e. it has more than two factors.

Hence, it is a composite number.

#### Page No 4:

#### Question 8:

Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.

#### Answer:

We know,

LCM ⨰ HCF = Product of two numbers.

We know that HCF will always be the factor of LCM.

Since, 18 is not factor of 380

So, we can say that two numbers cannot have 18 as their HCF and 380 as their LCM.

#### Page No 4:

#### Question 9:

Without actually performing the long division, find if $\frac{987}{10500}$ will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.

#### Answer:

In terminating rational number, the denominator always have the form 2^{m }⨰ 5* ^{n}*.

$\Rightarrow $$\frac{987}{10500}=\frac{47}{500}=\frac{47}{{5}^{3}\times {2}^{2}}$

Since, the denominator is of the form 2

^{m }⨰ 5

*, where m = 2 and n = 3.*

^{n}Hence, this is terminating decimal rational number.

#### Page No 4:

#### Question 10:

A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of *q*, when this number is expressed in the form $\frac{p}{q}$? Give reasons.

#### Answer:

Given that 327.7081 is terminating decimal number. So, it represents a rational number and also its denominator must have the form 2^{m }⨰ 5* ^{n}*.

Thus,

$327.7081=\frac{3277081}{10000}=\frac{p}{q}$

$\begin{array}{rcl}& \Rightarrow & q={10}^{4}=2\times 2\times 2\times 2\times 5\times 5\times 5\times 5\\ & =& {2}^{4}\times {5}^{4}={(2\times 5)}^{4}\end{array}$

Hence, the prime factors of

*q*are 2 and 5.

#### Page No 6:

#### Question 1:

Show that the square of any positive integer is either of the form 4*q* or 4*q* + 1 for some integer *q*.

#### Answer:

By Euclid's division lemma,

*b *= *aq *+ *r* 0 ≤ *r *< *a, *where *b* is any positive integer.

Now, if we divide *b* by 4, then* b* can written in the form of 4*m*, 4*m *+ 1,4*m *+ 2,4*m *+ 3.

This implies that We will have the four possible cases:

Case I:

If *b* = 4*m*, then squaring both sides, we get

${b}^{2}={\left(4m\right)}^{2}=16{m}^{2}=4\left(4{m}^{2}\right)$

$\Rightarrow {b}^{2}=4q$, where *q* is any positive integer.

Case II:

If* b* = 4*m *+ 1, then squaring both sides, we get

${b}^{2}={(4m+1)}^{2}=16{m}^{2}+1+8m$

$\Rightarrow {b}^{2}=4(4{m}^{2}+2m)+1\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=4q+1$

where *q *is any positive integer.

Case III:

If *b* = 4*m *+ 2, then squaring both sides, we get

${b}^{2}=16{m}^{2}+4+16m\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=4(4{m}^{2}+4m+1)\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=4q$

where *q* is any positive integer.

Case IV:

If *b *= 4*m *+ 3, then squaring both sides, we get

${b}^{2}=16{m}^{2}+9+24m=16{m}^{2}+24m+8+1\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=4(4{m}^{2}+6m+2)+1\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {b}^{2}=4q+1$

where *q* is any positive integer.

Hence, we can conclude the square of any positive integer is either of the form 4*q* or 4*q *+ 1 for some integer.

#### Page No 6:

#### Question 2:

Show that cube of any positive integer is of the form 4*m*, 4*m *+ 1 or 4*m* + 3, for some integer *m*.

#### Answer:

By Euclid's division lemma,

*b *= *aq *+ *r * 0 ≤ *r *< *a, *where *b* is any positive integer.

Now, if we divide *b* by 4, then* b* can written in the form of 4*q*, 4*q *+ 1, 4*q *+ 2, 4*q *+ 3.

This implies that We will have the four possible cases:

Case I:

If *b* = 4*q*, then taking cube both sides, we get

${b}^{3}={\left(4q\right)}^{3}=64{q}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{3}=4\left(16{q}^{3}\right)=4m$

where *m* is any positive integer.

Case II:

If *b* = 4*q *+ 1, then taking cube both sides, we get

${b}^{3}={(4q+1)}^{3}=64{q}^{3}+1+48{q}^{2}+12q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{3}=4(16{q}^{3}+12{q}^{2}+3q)+1\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{3}=4m+1$

where *m* is any positive integer.

Case III:

If *b* = 4q + 2, then taking cube both sides, we get

${b}^{3}={(4q+2)}^{3}=16{q}^{3}+8+96{q}^{2}+48q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{3}=4(4{q}^{3}+2+24{q}^{2}+12q)\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{3}=4m$

where *m* is any positive integer.

Case IV:

If* b* = 4q + 3, then taking cube both sides, we get

${b}^{3}={(4q+3)}^{3}=64{q}^{3}+27+96{q}^{2}+72q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{3}=64{q}^{3}+24+3+96{q}^{2}+72q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{3}=4(16{q}^{3}+8+24{q}^{2}+18q)+3\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{3}=4m+3$

where *m* is any positive integer.

Hence, we can conclude the cube of any positive integer is of the form 4*m*, 4*m *+ 1 or 4*m *+ 3 for some integer *m*.

#### Page No 6:

#### Question 3:

Show that the square of any positive integer cannot be of the form 5*q* + 2 or 5*q* + 3 for any integer *q*.

#### Answer:

By Euclid's division lemma,

*b *= *aq *+ *r* 0 ≤ *r *< *a, *where *b* is any positive integer.

Now, if we divide *b* by 5, then* b* can written in the form of 5*m*, 5*m*+1, 5*m*+2, 5*m*+3, 5*m*+4.

This implies that We will have the five possible cases:

Case I:

If *b *= 5*m*, then squaring both sides, we get

${b}^{2}={\left(5m\right)}^{2}=25{m}^{2}=5\left(5{m}^{2}\right)$

$\Rightarrow $${b}^{2}=5q$

where *q* is any positive integer.

Case II:

If *b *= 5*m *+ 1, the squaring both sides, we get

${b}^{2}={(5m+1)}^{2}=25{m}^{2}+1+10m\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=5(5{m}^{2}+2m)+1\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=5q+1$

where *q* is any positive integer.

Case III:

If *b* = 5*m *+ 2, then squaring both sides, we get

${b}^{2}={(5m+2)}^{2}=25{m}^{2}+4+20m\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=5(5{m}^{2}+8m)+4\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=5q+4$

where *q *is any positive integer.

Case IV:

If* b* = 5*m *+ 3, then squaring both sides, we get

${b}^{2}={(5m+3)}^{2}=25{m}^{2}+9+30m\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=25{m}^{2}+5+4+30m\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=5(5{m}^{2}+1+6m)+4\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=5q+4$

where *q* is any positive integer.

Case V:

If *b* = 5*m *+ 4, then squaring both sides, we get

${b}^{2}={(5m+4)}^{2}=25{m}^{2}+16+40m\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=25{m}^{2}+15+1+40m\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=5(5{m}^{2}+3+8m)+1\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=5q+1$

where *q* is any positive integer.

Hence, we can conclude the square of any positive integer cannot be of the form 5*q *+ 2 or 5*q *+ 3 for any integer.

#### Page No 6:

#### Question 4:

Show that the square of any positive integer cannot be of the form 6*m* + 2 or 6*m* + 5 for any integer *m*.

#### Answer:

By Euclid's division lemma,

*b *= *aq *+ *r* 0 ≤ *r *< *a, *where *b* is any positive integer.

Now, if we divide *b* by 6, then* b* can written in the form of 6*q*, 6*q *+ 1, 6*q *+ 2, 6*q *+ 3, 6*q *+ 4, 6*q *+ 5.

This implies that We will have the six possible cases:

Case I:

If *b* = 6*q*, then squaring both sides, we get

${b}^{2}={\left(6q\right)}^{2}=36{q}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=6\left(6{q}^{2}\right)=6m$

where *m* is any positive integer.

Case II:

If *b* = 6*q *+ 1, then squaring both sides, we get

${b}^{2}={(6q+1)}^{2}=36{q}^{2}+1+12q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=6(6{q}^{2}+2q)+1=6m+1$

where *m* is any positive integer.

Case III:

If *b* = 6*q *+ 2, then squaring both sides, we get

${b}^{2}={(6q+2)}^{2}=36{q}^{2}+4+24q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=6(6{q}^{2}+4q)+4\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=6m+4$

where *m* is any positive integer.

Case IV:

If *b* = 6*q *+ 3, then squaring both sides, we get

${b}^{2}={(6q+3)}^{2}=36{q}^{2}+9+36q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=36{q}^{2}+6+3+36q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=6(6{q}^{2}+1+6q)+3=6m+3$

where *m* is any positive integer.

Case V:

If *b* = 6*q *+ 4, then squaring both sides, we get

${b}^{2}={(6q+4)}^{2}=36{q}^{2}+16+48q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=36{q}^{2}+12+4+48q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=6(6{q}^{2}+2+8q)+4=6m+4$

where *m* is any positive integer.

Case VI:

If *b* = 6*q *+ 5, then squaring both sides, we get

${b}^{2}={(6q+5)}^{2}=36{q}^{2}+25+60q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=36{q}^{2}+24+1+60q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=6(6{q}^{2}+4+10q)+1=6m+1$

where *m* is any positive integer.

Hence, we conclude the square of any positive integer cannot be of the form 6*m *+ 2 or 6*m *+ 5 for any integer *m*.

#### Page No 6:

#### Question 5:

Show that the square of any odd integer is of the form 4*q *+ 1, for some integer* q*.

#### Answer:

By Euclid's division lemma,

*b *= *aq *+ *r* 0 ≤ *r *< *a, *where *b* is any positive integer.

Now, if we divide *b* by 4, then* b* can written in the form of 4*q*, 4*q *+ 1, 4*q *+ 2, 4*q *+ 3.

Now, 4*q* and 4*q *+ 2 are even integers, since they are divisible by 2.

Hence, we need to ignore 4*q* and 4*q *+ 2 form, since we need to show for odd integers only.

This implies that We will have the two possible cases:

Case I:

If *b *= 4*q *+ 1, the squaring both sides, we get

${b}^{2}={(4q+1)}^{2}=16{q}^{2}+1+8q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=4(4{q}^{2}+2q)+1=4m+1$

where m is any positive integer.

Case II:

If *b *= 4*q *+ 3, then squaring both sides, we get

${b}^{2}={(4q+3)}^{2}=16{q}^{2}+9+24q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=16{q}^{2}+8+1+24q\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=4(4{q}^{2}+2+6q)+1\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=4m+1$

where *m* is any positive integer.

Hence, for some integer *m*, the square of any odd integer is of the form 4*m *+ 1.

#### Page No 6:

#### Question 6:

If *n* is an odd integer, then show that *n*^{2} – 1 is divisible by 8.

#### Answer:

By Euclid's division lemma,

*n *= *aq *+ *r*, 0 ≤ *r *< *a, *where *n* is any positive integer.

Now, if we divide *n* by 4, then* n* can written in the form of 4*q*, 4*q *+ 1, 4*q *+ 2, 4*q *+ 3.

Now, 4*q* and 4*q *+ 2 are even integers, since they are divisible by 2.

Hence, we need to ignore 4*q* and 4*q *+ 2 form, since we need to show for odd integers only.

This implies that We will have the two possible cases:

Case I:

If *n *= 4*q *+ 1, then ${n}^{2}-1={(4q+1)}^{2}-1$

$\Rightarrow {n}^{2}-1=16{q}^{2}+1+8q-1\phantom{\rule{0ex}{0ex}}\Rightarrow 16{q}^{2}+8q=8q(2q+1)$

which is clearly, divisible by 8.

Case II:

If *n* = 4*q *+ 3, then ${n}^{2}-1={(4q+3)}^{2}-1$

$\Rightarrow {n}^{2}-1=16{q}^{2}+9+24q-1\phantom{\rule{0ex}{0ex}}\Rightarrow 16{q}^{2}+24q+8=8(2{q}^{2}+3q+1)\phantom{\rule{0ex}{0ex}}$

which is clearly, divisible by 8.

Hence, ${n}^{2}-1$ is divisble by 8, if *n* is odd integer.

#### Page No 6:

#### Question 7:

Prove that if *x* and *y* are both odd positive integers, then *x*^{2} + *y*^{2} is even but not divisible by 4.

#### Answer:

Let *m* be any positive integer.

Now, 2*m* will always be even integer.

This implies that 2*m *+ 1 and 2*m *+ 3 will always be odd integer.

Let *x* = 2*m *+ 1 and *y* = 2*m *+ 3 are odd positive integers, for every positive integer *m*.

Thus*, x*^{2} + *y*^{2} can be written as:

$={\left(2m+1\right)}^{2}+{\left(2m+3\right)}^{2}\phantom{\rule{0ex}{0ex}}=4{m}^{2}+1+4m+4{m}^{2}+9+12m\phantom{\rule{0ex}{0ex}}=8{m}^{2}+16m+10\phantom{\rule{0ex}{0ex}}=2(4{m}^{2}+8m+5)\phantom{\rule{0ex}{0ex}}$

Which is always even, but not always divisible by 4.

Hence, *x*^{2} + *y*^{2} is even for every positive integer *m* but not divisible by 4.

#### Page No 6:

#### Question 8:

Use Euclid’s division algorithm to find the HCF of 441, 567, 693.

#### Answer:

Let *a *= 693, *b *= 567 and* c *= 441

By Euclid's division algorithm,

*a *= b*q *+ *r*, 0 ≤ *r *< *b*.

where *a* is any positive integer.

First, we take *a* and* b* and find their HCF.

$\Rightarrow 693=567\times 1+126\phantom{\rule{0ex}{0ex}}\Rightarrow 567=126\times 4+63\phantom{\rule{0ex}{0ex}}\Rightarrow 126=63\times 2+0\phantom{\rule{0ex}{0ex}}$

∴ HCF(693, 567) = 63.

Now, we take *c* and HCF(693, 567), then find their HCF.

This implies that using Euclid's division algorithm,

$\Rightarrow 441=63\times 7+0$

∴ HCF(693, 567, 441) = 63.

#### Page No 6:

#### Question 9:

Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.

#### Answer:

Now, we are given 1, 2 and 3 are the remainders of 1251, 9377 and 15628, respectively.

Thus, after subtracting these remainders from the numbers.

We have,

$\begin{array}{rcl}1251-1& =& 1250\\ 9377-2& =& 9375\\ 15628-3& =& 15625\end{array}$

which will be divisible by required number.

Thus, required number will be HCF(1250, 9375, 15625).

First, we take largest numbers, 15625 and 9375 and find their HCF.

Using Euclid's division algorithm,

$\Rightarrow 15625=9375\times 1+6250\phantom{\rule{0ex}{0ex}}\Rightarrow 9375=6250\times 1+3125\phantom{\rule{0ex}{0ex}}\Rightarrow 6250=3125\times 2+0\phantom{\rule{0ex}{0ex}}$

∴ HCF(15625, 9375) = 3125.

Now, we take 1250 and HCF(15625, 9375) and find their HCF, we get

Using Euclid's division algorithm,

$\Rightarrow 3125=1250\times 2+625\phantom{\rule{0ex}{0ex}}\Rightarrow 1250=625\times 2+0$

∴ HCF(1250, 9375, 15625) = 625.

Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3 respectively.

#### Page No 6:

#### Question 10:

Prove that $\sqrt{3}+\sqrt{5}$ is irrational.

#### Answer:

To prove these type of questions, we use contradiction method i.e. we assume given number is rational and at last we have to prove our assumption is wrong, i.e. the number is irrational.

Now, let us suppose that $\sqrt{3}+\sqrt{5}$ is rational.

Let $\sqrt{3}+\sqrt{5}=a$, where *a *is rational number.

Thus, we can write $\sqrt{3}=a-\sqrt{5}$

Now, squaring both sides, we get

${\left(\sqrt{3}\right)}^{2}={\left(a-\sqrt{5}\right)}^{2}$

$\Rightarrow 3={a}^{2}+5-2a\sqrt{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 2a\sqrt{5}={a}^{2}+2$

∴ $\sqrt{5}=\frac{{a}^{2}+2}{2a}$, which is contradiction, since right hand side is rational number while $\sqrt{5}$ is irrational.

Hence, we conclude our assumption is wrong and $\sqrt{3}+\sqrt{5}$ is irrational number.

#### Page No 6:

#### Question 11:

Show that 12^{n} cannot end with the digit 0 or 5 for any natural number *n*.

#### Answer:

As we know, any number ending with 0 or 5 should always be divisible by 5.

This implies that 12* ^{n} *should end with the digit zero, if it must be divisible by 5.

Now, this is possible only if prime factorisation of 12

*contains the prime number 5.*

^{n}$\Rightarrow 12=2\times 2\times 3={2}^{2}\times 3\phantom{\rule{0ex}{0ex}}\Rightarrow {12}^{n}={\left({2}^{2}\times 3\right)}^{n}={2}^{2n}\times {3}^{n}$

Now there is no term that contains 5.

Hence, we conclude there is no value

*n*for which 12

*ends with digit 0 or 5.*

^{n}#### Page No 6:

#### Question 12:

On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

#### Answer:

Now, in this question we need to find the minimum distance each should walk.

This implies that we need to find least possible distance, so that each can cover the same distance in each steps.

Hence, we have to find LCM of 40 cm,42 cm and 45 cm.

Factors of 40 = $2\times 2\times 2\times 5$

Factors of 42 = $2\times 3\times 7$

Factors of 45 = $5\times 3\times 3$

Thus, LCM(40,42,45) is

$=2\times 3\times 5\times 2\times 2\times 3\times 7\phantom{\rule{0ex}{0ex}}=30\times 12\times 7\phantom{\rule{0ex}{0ex}}=210\times 12\phantom{\rule{0ex}{0ex}}=2520$

Hence, minimum distance each should walk 2520 cm, so that each can cover the same distance in compete steps.

#### Page No 7:

#### Question 13:

Write the denominator of the rational number $\frac{257}{5000}$ in the form 2^{m} × 5^{n}, where *m, n *are non-negative integers. Hence, write its decimal expansion, without actual division.

#### Answer:

Now, the denominator of the rational number $\frac{257}{5000}$ is 5000.

Thus, Factors of 5000 = $2\times 2\times 2\times 5\times 5\times 5\times 5={2}^{3}\times {5}^{4}$, which is of the type 2* ^{m} *⨯ 5

*, where*

^{n}*m*= 3 and

*n*= 4 are non-negative integers.

Now, we need to write the decimal expansion of the rational number.

∴ Rational number $\frac{257}{5000}$ can be written as:

$=\frac{257}{{2}^{3}\times {5}^{4}}\times \frac{2}{2}\phantom{\rule{0ex}{0ex}}=\frac{514}{{2}^{4}\times {5}^{4}}=\frac{514}{{10}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{514}{10000}=0.0514$

Hence, the decimal expansion of the rational number $\frac{257}{5000}$ is 0.0514.

#### Page No 7:

#### Question 14:

Prove that $\sqrt{p}+\sqrt{q}$ is irrational, where *p, q* are primes.

#### Answer:

To prove these type of questions, we use contradiction method i.e. we assume given number is rational and at last we have to prove our assumption is wrong, i.e. the number is irrational.

Now, let us suppose $\sqrt{p}+\sqrt{q}$ is rational.

Let $\sqrt{p}+\sqrt{q}=a$, where *a *is rational number.

This implies that we can write $\sqrt{q}=a-\sqrt{p}$

Now, squaring both sides, we get

${\left(\sqrt{q}\right)}^{2}={(a-\sqrt{p})}^{2}$

$\Rightarrow $$q={a}^{2}+p-2a\sqrt{p}$

∴ $\sqrt{p}=\frac{{a}^{2}+p-q}{2a}$, which is contradiction, since right hand side is rational while $\sqrt{p}$ is irrational, where *p* and *q *are prime numbers.

Hence, $\sqrt{p}+\sqrt{q}$ is irrational number.

#### Page No 7:

#### Question 1:

Show that the cube of a positive integer of the form 6*q* + *r, q* is an integer and *r *= 0, 1, 2, 3, 4, 5 is also of the form 6*m* +* r*.

#### Answer:

By Euclid's division lemma,

*b *= *aq *+ *r*, 0 ≤ *r *< *a*.

where *b* is any positive integer.

If we divide *b* by 6, then* b* can written in the form of 6*q*, 6*q *+ 1, 6*q *+ 2, 6*q *+ 3, 6*q *+ 4, 6*q *+ 5.

Thus, We will have the six possible cases:

Case I

If *b* = 6*q*, then taking cube both sides, we get

${b}^{3}={\left(6q\right)}^{3}=216{q}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 216{q}^{3}=6\left(36{q}^{3}\right)=6m$

where *m* is any integer.

Case II

If *b* = 6*q *+ 1, then taking cube both sides, we get

${b}^{3}={\left(6q+1\right)}^{3}=216{q}^{3}+1+108{q}^{2}+18q\phantom{\rule{0ex}{0ex}}\Rightarrow 6\left(36{q}^{3}+18q+3q\right)+1=6m+1$

where *m* is any integer.

Case III

If *b* = 6*q *+ 2, then taking cube both sides, we get

${b}^{3}={\left(6q+2\right)}^{3}=216{q}^{3}+216{q}^{2}+72q+8\phantom{\rule{0ex}{0ex}}\Rightarrow 6\left(36{q}^{3}+36{q}^{2}+12q+1\right)+2=6m+2$

where *m* is any integer.

Case IV

If *b* = 6*q *+ 3, then taking cube both sides, we get

${b}^{3}={\left(6q+3\right)}^{3}=216{q}^{3}+324{q}^{2}+162q+27\phantom{\rule{0ex}{0ex}}\Rightarrow 6\left(36{q}^{3}+54{q}^{2}+27q+4\right)+3=6m+3$

where *m* is any integer.

Case V

If *b* = 6*q *+ 4, then taking cube both sides, we get

${b}^{3}={\left(6q+4\right)}^{3}=216{q}^{3}+432{q}^{2}+288q+64\phantom{\rule{0ex}{0ex}}\Rightarrow 6(36{q}^{3}+72{q}^{2}+48q+10)+4=6m+4$

where *m* is any integer.

Case VI

If *b* = 6*q *+ 5, then taking cube both sides, we get

${b}^{3}={\left(6q+5\right)}^{3}=216{q}^{3}+540{q}^{2}+450q+125\phantom{\rule{0ex}{0ex}}\Rightarrow 6(36{q}^{3}+90{q}^{2}+75q+20)+5=6m+5$

where *m* is any integer.

Hence, we conclude the cube of a positive integer of the form 6*q*+*r*, where *q* is an inetger and *r*=0, 1, 2, 3, 4, 5 is also of the form 6*m*, 6*m *+ 1, 6*m *+ 2, 6*m *+ 3, 6*m *+ 4, 6*m *+ 5 i.e. 6*m *+ *r*.

#### Page No 7:

#### Question 2:

Prove that one and only one out of *n, n* + 2 and *n* + 4 is divisible by 3, where *n* is any positive integer.

#### Answer:

By Euclid's division lemma,

*n*=*aq *+ *r*, 0 ≤ *r *< *a*.

where *n* is any positive integer.

Now, if we divide *n* by 3, then* n* can written in the form of 3*q*, 3*q *+ 1, 3*q *+ 2.

This implies that we will three possible cases:

Case I:

If *n *= 3*q*, then *n* is divisible by 3.

However, *n *+ 2 and *n* + 4 are not divisible by 3.

Case II:

If *n* = 3*q* + 1, then *n* + 2 = 3*q* + 3 = 3(*q* + 1), which is divisible by 3.

However,* n* and *n* + 4 are not divisible by 3.

Case III:

If *n* = 3*q* + 2, then *n* + 4 = 3*q* + 6 = 3(*q* + 2), which is divisible by 3.

However, *n* and* n* + 2 are not divisible by 3.

Hence, we conclude one and only one out of *n*, *n* + 2 and *n* + 4 is divisible by 3.

#### Page No 7:

#### Question 3:

Prove that one of any three consecutive positive integers must be divisible by 3.

#### Answer:

Let *n*, *n *+ 1, *n* + 2 be three consecutive positive integers.

By Euclid's division lemma,

*n* = *aq *+ *r*, 0 ≤ *r *< *a*.

where *n* is any positive integer.

Now, if we divide *n* by 3, then* n* can be written in the form of 3*q*, 3*q*+1, 3*q*+2.

This implies that we will have three possible cases:

Case I:

If *n *= 3*q*, then *n* is only divisible by 3.

However, *n *+ 1 and *n* + 2 are not divisible by 3.

Case II:

If *n* = 3*q* + 1, then *n* + 2 = 3*q* + 3 = 3(*q* + 1), which is only divisible by 3.

However,* n* and *n* + 1 are not divisible by 3.

Case III:

If *n* = 3*q* + 2, then *n* + 1 = 3*q* + 3 = 3(*q* + 1), which is only divisible by 3.

However, *n* and* n* + 2 are not divisible by 3.

Hence, we conclude one and only one out of *n*, *n* + 1 and *n* + 2 is divisible by 3.

#### Page No 7:

#### Question 4:

For any positive integer *n*, prove that *n*^{3} – *n* is divisible by 6.

#### Answer:

Let $a={n}^{3}-n$, where *n *is any positive integer.

$\Rightarrow a=n\left({n}^{2}-1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow a=n(n-1)(n+1)$

We know that,

I. If a number is completely divisible by 2 and 3, then it is also divisible by 6.

II. If the sum of digits of any number is divisible by 3, then it is divisible by 3.

III. If a number is an even number, then it is divisible by 2.

∴ $a=(n-1)n(n+1)$

Now, sum of digits $=n-1+n+n+1=3n$, which is a multiple of 3, and $(n-1)n(n+1)$ will always be even, since they are multiple of consecutive positive integers.

Hence, Condition II and III are completely satisfied.

∴ we conclude by condition I the number ${n}^{3}-n$ is divisible by 6.

#### Page No 7:

#### Question 5:

Show that one and only one out of *n, n* + 4,* n* + 8,* n *+ 12 and *n* + 16 is divisible by 5, where *n* is any positive integer.

#### Answer:

Given numbers are *n*, (*n* + 4), (*n* + 8), (*n *+ 12) and* n* + 16.

By Euclid's division lemma,

*n *= *aq *+ *r* ; 0 ≤ *r *< *a*.

where *n* is any positive integer.

Now, if we divide *n* by 5, then* n* can written in the form of 5*q, *5*q *+ 1,5*q *+ 2,5*q *+ 3, 5*q *+ 4.

This implies that We will have the five possible cases:

Case I:

If *n* = 5*q, *then *n *is only divisible by 5.

However (*n* + 4), (*n* + 8), (*n *+ 12) and (n + 16) are not divisible by 5.

Case II:

If *n* = 5*q *+ 1, then *n* + 4 = 5*q *+ 1 + 4 = 5*q + *5 = 5(*q *+ 1), which is divisible by 5.

However *n,* (*n* + 8), (*n *+ 12) and (n + 16) are not divisible by 5.

Case III:

If *n* = 5*q *+ 2, then *n* + 8 = 5*q *+ 2 + 8 = 5*q *+ 10 = 5(*q *+ 2), which is divisible by 5.

However *n,* (*n* + 4), (*n *+ 12) and (n + 16) are not divisible by 5.

Case IV:

If *n* = 5*q *+ 3, then *n* + 12 = 5*q *+ 3 + 12 = 5*q *+ 15 = 5(*q *+ 3), which is divisible by 5.

However *n,* (*n* + 4), (*n *+ 8) and (n + 16) are not divisible by 5.

Case V:

If *n* = 5*q *+ 4, then *n* + 16 = 5*q *+ 4 + 16 = 5*q *+ 20 = 5(*q *+ 4), which is divisible by 5.

However *n,* (*n* + 4), (*n *+ 8) and (n + 12) are not divisible by 5.

Hence, one and only one out of *n*, (*n* + 4), (*n* + 8), (*n *+ 12) and* *(*n* + 16) is divisible by 5

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