Solution

(c) We know that, even integers are 2, 4, 6,..

So, it can be written in the form of $2 m$.

where,

$ \begin{aligned} & m=\text{ Integer }=Z \quad[\text{ since, integer is represented by } Z] \\ & m=\cdots,-1,0,1,2,3, \ldots \end{aligned} $

$ \begin{matrix} \text{ or } & m=\cdots,-1,0,1,2,3, \ldots \\ \therefore & 2 m=\cdots,-2,0,2,4,6 . . \end{matrix} $

Alternate Method

Let ’ $a$ ’ be a positive integer. On dividing ’ $a$ ’ by 2 , let $m$ be the quotient and $r$ be the remainder. Then, by Euclid’s division algorithm, we have

$ a=2 m+r \text{, where } a \leq r<2 \text{ i.e., } r=0 \text{ and } r=1 \text{. } $

$ \Rightarrow \quad a=2 m \text{ or } a=2 m+1 $

when, $a=2 m$ for some integer $m$, then clearly $a$ is even.

Solution

(d) We know that, odd integers are 1, 3, 5, ,

So, it can be written in the form of $2 q+1$.

where,

or

$ q=\text{ integer }=Z $

Alternate Method

Let ’ $a$ ’ be given positive integer. On dividing ’ $a$ ’ by 2 , let $q$ be the quotient and $r$ be the remainder. Then, by Euclid’s division algorithm, we have

$ \begin{matrix} \Rightarrow & a=2 q+r, \text{ where } 0 \leq r<2 \\ \Rightarrow & a=2 q+r, \text{ where } r=0 \text{ or } r=1 \\ \Rightarrow & a=2 q \text{ or } 2 q+1 \end{matrix} $

when $a=2 q+1$ for some integer $q$, then clearly $a$ is odd.

Solution

(c) Let $a=n^{2}-1$

Here $n$ can be ever or odd.

Case I $n=$ Even i.e., $n=2 k$, where $k$ is an integer.

$ \begin{matrix} \Rightarrow & a=(2 k)^{2}-1 \\ \Rightarrow & a=4 k^{2}-1 \end{matrix} $

At $k=-1,=4(-1)^{2}-1=4-1=3$, which is not divisible by 8 .

At $k=0, a=4(0)^{2}-1=0-1=-1$, which is not divisible by 8 , which is not

Case II $n=$ Odd i.e., $n=2 k+1$, where $k$ is an odd integer.

$$ \begin{array}{ll} \Rightarrow & a=2 k+1 \\ \Rightarrow & a=(2 k+1)^{2}-1 \\ \Rightarrow & a=4 k^{2}+4 k+1-1 \\ \Rightarrow & a=4 k^{2}+4 k \\ \Rightarrow & a=4 k(k+1) \\ \end{array} $$

$ \begin{matrix} \text{ At } k=-1, a=4(-1)(-1+1)=0 \text{ which is divisible by } 8 . \\ \text{ At } k=0, a=4(0)(0+1)=4 \text{ which is divisible by } 8 . \\ \text{ At } k=1, a=4(1)(1+1)=8 \text{ which is divisible by } 8 . \\ \text{ Hence, we can conclude from above two cases, if } n \text{ is odd, then } n^{2}-1 \text{ is } \\ \text{ divisible by } 8 . \end{matrix} $

Thinking Process

Apply Euclid’s division algorithm until the remainder is 0 . Finally we get divisor, which is the required HCF of 65 and 117. Now, put $65 m-117=HCF(65,117)$ and get the value of $m$.

Solution

(b) By Euclid’s division algorithm,

$$ \begin{aligned} & b=a q+r, 0 \leq r<a \quad[\because \text { dividend }=\text { divisor } \times \text { quotient }+ \text { remainder }] \\ & \Rightarrow \quad 117=65 \times 1+52 \\ & \Rightarrow \quad 65=52 \times 1+13 \\ & \Rightarrow \quad 52=13 \times 4+0 \\ & \therefore \quad \operatorname{HCF}(65,117)=13 \\ & \text { Also, given that, } \operatorname{HCF}(65,117)=65 m-117 \\ & 65 m-117=13 \\ & \Rightarrow \quad 65 m=130 \\ & \Rightarrow \quad m=2 \\ & \end{aligned} $$

From Eqs. (i) and (ii),

$$ \begin{array}{ll} & 65 m-117 & =13 \\ \Rightarrow & 65 m & =130 \\ \Rightarrow & m & =2 \end{array} $$

Thinking Process

First, we subtract the remainders 5 and 8 from corresponding numbers respectively and then get HCF of resulting numbers by using Euclid’s division algorithm, which is the required largest number.

Solution

(a) Since, 5 and 8 are the remainders of 70 and 125 , respectively. Thus, after subtracting these remainders from the numbers, we have the numbers $65=(70-5)$, $117=(125-8)$, which is divisible by the required number.

Now, required number $=HCF$ of 65,117

[for the largest number]

For this, $\quad 117=65 \times 1+52 \quad[\because$ dividend $=$ divisor $\times$ quotient + remainder $]$

$\Rightarrow$ $65=52 \times 1+13$

$\Rightarrow \quad 52=13 \times 4+0$

$\therefore \quad HCF=13$

Hence, 13 is the largest number which divides 70 and 125 , leaving remainders 5 and 8 .

Solution

(b) Given that,

$ a=x^{3} y^{2}=x \times x \times x \times y \times y $

and

$ b=x y^{3}=x \times y \times y \times y $

$\therefore$ HCF of $a$ and $b$

$ =HCF(x^{3} y^{2}, x y^{3})=x \times y \times y=x y^{2} $

[since, HCF is the product of the smallest power of each common prime facter involved in the numbers]

Solution

(c) Given that,

$ p=a b^{2}=a \times b \times b $

and

$ q=a^{3} b=a \times a \times a \times b $

$\therefore \quad$ LCM of $p$ and $q=LCM(a b^{2}, a^{3} b)=a \times b \times b \times a \times a=a^{3} b^{2}$

[since, LCM is the product of the greatest power of each prime factor involved in the numbers]

Solution

(a) Product of a non-zero rational and an irrational number is always irrational i.e., $\frac{3}{4} \times \sqrt{2}=\frac{3 \sqrt{2}}{4}$ (irrational).

Solution

(d) Factors of 1 to 10 numbers

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

$$\therefore \quad \text{LCM of number } 1 \text{ to } 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) \\ = 1 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 2520$$

Thinking Process

In terminating rational number the denominator always have the form $2^{m} \times 5^{n}$.

Solution

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

$ \begin{aligned} & =\frac{14587}{10 \times 5^{3}} \times \frac{(2)^{3}}{(2)^{3}} \\ & =\frac{14587 \times 8}{10 \times 1000} \\ & =\frac{116696}{10000}=11.6696 \end{aligned} $

Hence, given rational number will terminate after four decimal places.

Solution

No, by Euclid’s Lemma, $b=a q+r, 0 \leq r<a \quad[\because$ dividend $=$ divisor $\times$ quotient + remainder $]$ Here, $b$ is any positive integer $a=4, b=4 q+r$ for $0 \leq r<4$ i.e., $r=0,1,2,3$

So, this must be in the form $4 q, 4 q+1,4 q+2$ or $4 q+3$.

Thinking Process

The product of two consecutive numbers i.e., $n(n+1)$ will always be even, as one out of $n$ or $(n+1)$ must be even.

Solution

Yes, two consecutive integers can be $n,(n+1)$. So, one number of these two must be divisible by 2 . Hence, product of numbers is divisible by 2 .

Solution

Yes, three consecutive integers can be $n,(n+1)$ and $(n+2)$.

So, one number of these three must be divisible by 2 and another one must be divisible by 3. Hence, product of numbers is divisible by 6 .

Solution

No, by Euclid’s lemma, $b=a q+r, 0 \leq r \leq a$ Here, $b$ is any positive integer, $a=3, b=3 q+r$ for $0 \leq r \leq 2$

So, any positive integer is of the form $3 k, 3 k+1$ or $3 k+2$.

Now, $$ \begin{aligned} (3 k)^2 & =9 k^2=3 m & [\text{where, } m=3K^2]\\ (3 k+1)^2 & =9 k^2+6 k+1 \\ & =3\left(3 k^2+2 k\right)+1=3 m+1 & [where, \left.m=3 k^2+2 k\right] \end{aligned} $$

Also

$$ \begin{aligned} (3 k+2)^2 & =9 k^2+12 k+4\left[\because(a+b)^2=a^2+2 a b+b^2\right] \\ & =9 k^2+12 k+3+1 \\ & =3\left(3 k^2+4 k+1\right)+1 \\ & \left.=3 m+1 \quad \text { [where, } m=3 k^2+4 k+1\right] \end{aligned} $$

which is in the form of $3 m$ and $3 m+1$. Hence, square of any positive number cannot be of the form $3 m+2$.

Solution

No, by Euclid’s Lemma, $b=a q+r, 0 \leq r<a$

Here, $b$ is any positive integer $a=3, b=3 q+r$ for $0 \leq r<3$

So, this must be in the form $3 q, 3 q+1$ or $3 q+2$.

Now, $$ \begin{array}{rlrl} (3 q)^2 & =9 q^2=3 m & [here, m=3q^2]\\ (3 q+1)^2 & =9 q^2+6 q+1 & \\ & =3\left(3 q^2+2 q\right)+1=3 m+1 & [where \left.m=3 q^2+2 q\right] \end{array} $$ Also, $$ \begin{aligned} (3 q+2)^2 & =9 q^2+12 q+4 \\ & =9 q^2+12 q+3+1 \\ & =3\left(3 q^2+4 q+1\right)+1 \\ & =3 m+1 & [here, \left.m=3 q^2+4 q+1\right] \end{aligned} $$

Hence, square of a positive integer is of the form $3 q+1$ is always in the form $3 m+1$ for some integer $m$.

Solution

Since, the $HCF(525,3000)=75$

By Euclid’s Lemma, $\quad 3000=525 \times 5+375[\because$ dividend $=$ divisor $\times$ quotient + remainder $]$

$525=375 \times 1+150$

$375=150 \times 2+75$

$150=75 \times 2+0$

and the numbers $3,5,15,25$ and 75 divides the numbers 525 and 3000 that mean these terms are common in both 525 and 3000 . So, the highest common factor among these is 75 .

Thinking Process

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

Solution

We have, $3 \times 5 \times 7+7=105+7=112$

Now, $\quad 112=2 \times 2 \times 2 \times 2 \times 7=2^{4} \times 7$

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.

Solution

No, because HCF is always a factor of LCM but here 18 is not a factor of 380 .

Solution

Yes, after simplification denominator has factor $5^{3} \cdot 2^{2}$ and which is of the type $2^{m} \cdot 5^{n}$. So, this is terminating decimal.

$ \begin{aligned} \because \quad \frac{987}{10500} & =\frac{47}{500}=\frac{47}{5^{3} \cdot 2^{2}} \times \frac{2}{2} \\ & =\frac{94}{5^{3} \times 2^{3}}=\frac{94}{(10)^{3}}=\frac{94}{1000}=0.094 \end{aligned} $

Solution

327.7081 is terminating decimal number. So, it represents a rational number and also its denominator must have the form $2^{m} \times 5^{n}$.

$ \begin{aligned} & \text{ Thus, } \quad 327.7081=\frac{3277081}{10000}=\frac{p}{q} \\ & \therefore \quad 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{aligned} $

Hence, the prime factors of $q$ is 2 and 5 .

Thinking Process

Use Euclid’s division algorithm, put the value of divisor as 4 and then put the value of remainder from 0 to 3 and get the different form. Now, squaring every different form and get the required form.

Solution

Let a be an arbitrary positive integer. Then, by, Euclid’s division algorithm, corresponding to the positive integers $a$ and 4 , there exist non-negative integers $m$ and $r$, such that

$\Rightarrow$ $\begin{aligned} a & =4 m+r, \text{ where } 0 \leq r<4 \\ a^{2} & =16 m^{2}+r^{2}+8 m r\end{aligned}$

where,

$ 0 \leq r<4 $

Case I When $r=0$, then putting $r=0$ in Eq. (i), we get

$ a^{2}=16 m^{2}=4(4 m^{2})=4 q $

$ \text{ where, } q=4 m^{2} \text{ is an integer. } $

Case II When $r=1$, then putting $r=1$ in Eq. (i), we get

$$ \begin{equation*} [\because(a+b)^{2}=a^{2}+2 a b+b^{2}] \tag{i} \end{equation*} $$

$$ \begin{aligned} a^{2} & =16 m^{2}+1+8 m \\ & =4(4 m^{2}+2 m)+1=4 q+1 \end{aligned} $$

where, $q=(4 m^{2}+2 m)$ is an integer.

Case III When $r=2$, then putting $r=2$ in Eq. (i), we get

$$ \begin{aligned} a^{2} & =16 m^{2}+4+16 m \\ & =4(4 m^{2}+4 m+1)=4 q \end{aligned} $$

where, $q=(4 m^{2}+4 m+1)$ is an integer.

Case IV When $r=3$, then putting $r=3$ in Eq. (i), we get

$$ \begin{aligned} a^{2} & =16 m^{2}+9+24 m=16 m^{2}+24 m+8+1 \\ & =4(4 m^{2}+6 m+2)+1=4 q+1 \end{aligned} $$

where, $q=(4 m^{2}+6 m+2)$ is an integer.

Hence, the square of any positive integer is either of the form $4 q$ or $4 q+1$ for some integer $q$.

Thinking Process

Use Euclid’s division algorithm. Put the value of remainder from 0 to 3, get the different forms of any positive integer. Now cubing to every different form of positive integer and get the required forms.

Solution

Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers $a$ and 4 , there exist non-negative integers $q$ and $r$ such that

$ \begin{array}{ll} & a=4 q+r \text{, where } 0 \leq r<4 \\ \Rightarrow & \quad a^{3}=(4 q+r)^{3}=64 q^{3}+r^{3}+12 q r^{2}+48 q^{2} r \\ & {[\because(a+b)^{3}=a^{3}+b^{3}+3 a b^{2}+3 a^{2} b]} \\ \Rightarrow & \quad a^{3}=(64 q^{2}+48 q^{2} r+12 q r^{2})+r^{3} \\ & \text{ where, } \quad 0 \leq r<4 \end{array} $

Case I When $r=0$,

Putting $r=0$ in Eq. (i), we get

$a^{3}=64 q^{3}=4(16 q^{3})$

$\Rightarrow a^{3}+4 m$ where $m=16 q^{3}$ is an integer.

Case II When $r=1$, then putting $r=1$ in Eq. (i), we get

$ \begin{aligned} a^{3} & =64 q^{3}+48 q^{2}+12 q+1 \\ & =4(16 q^{3}+12 q^{2}+3 q)+1 \\ & =4 m+1 \end{aligned} $

where, $m=(16 q^{2}+12 q^{2}+3 q)$ is an integer.

Case III When $r=2$, then putting $r=2$ in Eq. (i), we get

$ \begin{aligned} a^{3} & =64 q^{3}+144 q^{2}+108 q+27 \\ & =64 q^{3}+144 q^{2}+108 q+24+3 \\ & =4(16 q^{3}+36 q^{2}+27 q+6)+3=4 m+3 \end{aligned} $

where, $m=(16 q^{3}+36 q^{2}+27 q+6)$ is an integer.

Hence, the cube of any positive integer is of the form $4 m, 4 m+1$ or $4 m+3$ for some integer $m$.

Solution

Let a be an arbitrary positive integer.

Then, by Euclid’s divisions Algorithm, corresponding to the positive integers a and 5, there exist non-negative integers $m$ and $r$ such that

$ \begin{array}{ll} \Rightarrow & a^{2}=(5 m+r)^{2}=25 m^{2}+r^{2}+10 m r \\ & {[\because(a+b)^{2}=a^{2}+2 a b+b^{2}]} \\ \Rightarrow & a^{2}=5(5 m^{2}+2 m r)+r^{2} \\ \text{ where, } & 0 \leq r<5 \end{array} $

$ a=5 m+r \text{, where } 0 \leq r<5 $

Case I When $r=0$, then putting $r=0$ in Eq. (i), we get

$ a^{2}=5(5 m^{2})=5 q $

where, $q=5 m^{2}$ is an integer.

Case II When $r=1$, then putting $r=1$ is Eq. (i), we get

$ \begin{matrix} \Rightarrow \quad & a^{2} & =5(5 m^{2}+2 m)+1 \\ & & q & =5 q+1 \end{matrix} $

where, $q=(5 m^{2}+2 m)$ is an integer.

Case III When $r=2$, then putting $r=2$ in Eq. (i), we get

$ a^{2}=5(5 m^{2}+4 m)+4=5 q+4 $

where, $q=(5 m^{2}+4 m)$ is an integer.

Case IV When $r=3$, then putting $r=3$ in Eq. (i), we get

$ \begin{aligned} a^{2} & =5(5 m^{2}+6 m)+9=5(5 m^{2}+6 m)+5+4 \\ & =5(5 m^{2}+6 m+1)+4=5 q+4 \end{aligned} $

where, $q=(5 m^{2}+6 m+1)$ is an integer.

Case V When $r=4$, then putting $r=4$ in Eq. (i), we get

$ \begin{aligned} & \quad a^{2}=5(5 m^{2}+8 m)+16=5(5 m^{2}+8 m)+15+1 \\ & \Rightarrow \quad a^{2}=5(5 m^{2}+8 m+3)+1=5 q+1 \\ & \text{ where, } q=(5 m^{2}+8 m+3) \text{ is an integer. } \end{aligned} $

Hence, the square of any positive integer cannot be of the form $5 q+2$ or $5 q+3$ for any integer $q$.

Solution

Let a be an arbitrary positive integer, then by Euclid’s division algorithm, corresponding to the positive integers $a$ and 6 , there exist non-negative integers $q$ and $r$ such that

$$ \begin{array}{ll} \Rightarrow & a^{2} =(6 q+r)^{2}=36 q^{2}+r^{2}+12 q r \\ & \quad[\because(a+b)^{2}=a^{2}+2 a b+b^{2}] \\ \Rightarrow & a^{2}=6(6 q^{2}+2 q r)+r^{2} \tag{i}\\ \text{ where, } & 0 \leq r<6 \end{array} $$

Case I When $r=0$, then putting $r=0$ in Eq. (i), we get

$ a^{2}=6(6 q^{2})=6 m $

where, $m=6 q^{2}$ is an integer.

Case II When $r=1$, then putting $r=1$ in Eq. (i), we get

$ a^{2}=6(6 q^{2}+2 q)+1=6 m+1 $

where, $m=(6 q^{2}+2 q)$ is an integer.

Case III When $r=2$, then putting $r=2$ in Eq (i), we get

$ a^{2}=6(6 q^{2}+4 q)+4=6 m+4 $

where, $m=(6 q^{2}+4 q)$ is an integer.

Case IV When $r=3$, then putting $r=3$ in Eq. (i), we get

$ \begin{aligned} a^{2} & =6(6 q^{2}+6 q)+9 \\ & =6(6 q^{2}+6 a)+6+3 \\ \Rightarrow \quad a^{2} & =6(6 q^{2}+6 q+1)+3=6 m+3 \end{aligned} $

where, $m=(6 q+6 q+1)$ is an integer.

Case V When $r=4$, then putting $r=4$ in Eq. (i), we get

$ \begin{aligned} a^{2} & =6(6 q^{2}+8 q)+16 \\ & =6(6 q^{2}+8 q)+12+4 \\ \Rightarrow \quad a^{2} & =6(6 q^{2}+8 q+2)+4=6 m+4 \end{aligned} $

where, $m=(6 q^{2}+8 q+2)$ is an integer.

Case VI When $r=5$, then putting $r=5$ in Eq. (i), we get

$ \begin{aligned} a^{2} & =6(6 q^{2}+10 q)+25 \\ & =6(6 q^{2}+10 q)+24+1 \\ \Rightarrow \quad a^{2} & =6(6 q^{2}+10 q+4)+1=6 m+1 \\ \text{ where, } m & =(6 q^{2}+10 q+1) \text{ is an integer. } \end{aligned} $

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

Solution

By Euclid’s division algorithm, we have $a=b q+r$, where $0 \leq r<4$

On putting $b=4$ in Eq. (i), we get

$ a=4 q+r \text{, where } 0 \leq r<4 \text{ i.e., } r=0,1,2,3 $

If $r=0 \Rightarrow a=4 q, 4 q$ is divisible by $2 \Rightarrow 4 q$ is even.

If $r=1 \Rightarrow a=4 q+1,(4 q+1)$ is not divisible by 2 .

If $r=2 \Rightarrow a=4 q+2,2(2 q+1)$ is divisible by $2 \Rightarrow 2(2 q+1)$ is even.

If $r=3 \Rightarrow a=4 q+3,(4 q+3)$ is not divisible by 2 .

So, for any positive integer $q,(4 q+1)$ and $(4 q+3)$ are odd integers.

Now,

$ a^{2}=(4 q+1)^{2}=16 q^{2}+1+8 q=4(4 q^{2}+2 q)+1 $

$ [\because(a+b)^{2}=a^{2}+2 a b+b^{2}] $

is a square which is of the form $4 m+1$, where $m=(4 q^{2}+2 q)$ is an integer.

and $a^{2}=(4 q+3)^{2}=16 q^{2}+9+24 q=4(4 q^{2}+6 q+2)+1$ is a square

$ [\because(a+b)^{2}=a^{2}+2 a b+b^{2}] $

which is of the form $4 m+1$, where $m=(4 q^{2}+6 q+2)$ is an integer.

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

Solution

Let

$$ \begin{equation*} a=n^{2}-1 \tag{i} \end{equation*} $$

Given that, $n$ is an odd integer.

$\therefore \quad n=1,3,5, \ldots$

From Eq. (i), at $n=1, a=(1)^{2}-1=1-1=0$,

which is divisible by 8 .

From Eq. (i), at $n=3, a=(3)^{2}-1=9-1=8$,

which is divisible by 8 .

From Eq. (i), at $n=5, a=(5)^{2}-1=25-1=24=3 \times 8$,

which is divisible by 8 .

From Eq. (i), at $n=7, a=(7)^{2}-1=49-1=48=6 \times 8$,

which is divisible by 8 .

Hence, $(n^{2}-1)$ is divisible by 8 , where $n$ is an odd integer.

Alternate Method

We know that an odd integer $n$ is of the from $(4 q+1)$ or $(4 q+3)$ for some integer $q$.

Case I When $n=4 q+1$

In this case, we have

$ \begin{aligned} (n^{2}-1) & =(4 q+1)^{2}-1 \\ & =16 q^{2}+1+8 q-1 \\ (a+b)^{2}&=[a^{2}+2 a b+b^{2}] \quad & \\ & =16 q^{2}+8 q=8 q(2 q+1) \\ & =16 q^{2}+8 q=8 q(2 q+1) \end{aligned} $

which is clearly, divisible by 8 .

Case II When $n=4 q+3$

In this case, we have

$ \begin{aligned} (n^{2}-1) & =(4 q+3)^{2}-1 \\ & =16 q^{2}+9+24 q-1 \quad[\because(a+b)^{2}=a^{2}+2 a b+b^{2}] \\ & =16 q^{2}+24 q+8 \\ & =8(2 q^{2}+3 q+1) \end{aligned} $

which is clearly divisible by 8 .

Hence, $(n^{2}-1)$ is divisible by 8 .

Thinking Process

Here we have to take any two consecutive odd positive integers. After that squaring and adding both the number and check it is even but not divisible by 4.

Solution

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

$ \begin{aligned} x^{2}+y^{2} & =(2 m+1)^{2}+(2 m+3)^{2} \\ & =4 m^{2}+1+4 m+4 m^{2}+9+12 m \\ &[\because(a+b)^{2}=a^{2}+2 a b+b^{2}] \\ & =8 m^{2}+16 m+10=\text{ even } \\ & =2(4 m^{2}+8 m+5) \text{ or } 4(2 m^{2}+4 m+2)+1 \end{aligned} $

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

Thinking Process

First we use Euclid’s division algorithm between two larger numbers any of three and get the HCF between these two. After that we take the third number and resulting HCF of two numbers and apply again Euclid’s division algorithm and get the required HCF.

Solution

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

By Euclid’s division algorithms,

$$ \begin{equation*} a=b q+r \tag{i} \end{equation*} $$

$[\because$ dividend $=$ divisor $\times$ quotient + remainder $]$

First we take, $a=693$ and $b=567$ and find their HCF.

$ \begin{aligned} & 693=567 \times 1+126 \\ & 567=126 \times 4+63 \\ & 126=63 \times 2+0 \end{aligned} $

$ \therefore \quad HCF(693,567)=63 $

Now, we take $c=441$ and say $d=63$, then find their HCF.

Again, using Euclid’s division algorithm,

$ \begin{aligned} & & c & =d q+r \\ \Rightarrow & & 441 & =63 \times 7+0 \\ \therefore & & HCF(693,567,441) & =63 \end{aligned} $

Solution

Since, 1, 2 and 3 are the remainders of 1251, 9377 and 15628, respectively. Thus, after subtracting these remainders from the numbers.

We have the numbers, $1251-1=1250,9377-2=9375$ and $15628-3=15625$

which is divisible by the required number.

Now, required number $=$ HCF of 1250, 9375 and 15625 [for the largest number]

By Euclid’s division algorithm,

$$ \begin{equation*} a=b q+r \tag{i} \end{equation*} $$

$[\because$ dividend $=$ divisor $\times$ quotient + remainder $]$

For largest number, put $a=15625$ and $b=9375$

$ 15625=9375 \times 1+6250 $

$ \begin{matrix} \Rightarrow & & 9375 & =6250 \times 1+3125 \\ \Rightarrow & & 6250 & =3125 \times 2+0 \\ \therefore & HCF(15625,9375) & =3125 \end{matrix} $

Now, we take $c=1250$ and $d=3125$, then again using Euclid’s division algorithm,

$ \begin{aligned} & \Rightarrow \quad 3125=1250 \times 2+625 \\ & \Rightarrow \quad 1250=625 \times 2+0 \\ & \therefore \quad HCF(1250,9375,15625)=625 \end{aligned} $

[from Eq. (i)]

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

Thinking Process

In this type of question, we use the contradiction method i.e., we assume given number is rational and at last we have to prove our asssumption is wrong, i.e., the number is irrational.

Solution

Let us suppose that $\sqrt{3}+\sqrt{5}$ is rational.

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

Therefore,

$\sqrt{3}=a-\sqrt{5}$

On squaring both sides, we get

$\Rightarrow \quad 3=a^{2}+5-2 a \sqrt{5} \quad[\because(a-b)^{2}=a^{2}+b^{2}-2 a b]$

$\Rightarrow$

$2 a \sqrt{5}=a^{2}+2$

Therefore,

$ \sqrt{5}=\frac{a^{2}+2}{2 a} \text{ which is contradiction. } $

As the right hand side is rational number while $\sqrt{5}$ is irrational. Since, 3 and 5 are prime numbers. Hence, $\sqrt{3}+\sqrt{5}$ is irrational.

Solution

If any number ends with the digit 0 or 5 , it is always divisible by 5 .

If $12^{n}$ ends with the digit zero it must be divisible by 5 .

This is possible only if prime factorisation of $12^{n}$ contains the prime number 5 .

Now,

$ 12=2 \times 2 \times 3=2^{2} \times 3 $

$\Rightarrow$ $12^{n}=(2^{2} \times 3)^{n}=2^{2 n} \times 3^{n}$ [since, there is no term contains 5]

Hence, there is no value of $n \in N$ for which $12^{n}$ ends with digit zero or five.

Solution

We have to find the LCM of $40 cm, 42 cm$ and $45 cm$ to get the required minimum distance. For this,

$ \begin{aligned} 40 & =2 \times 2 \times 2 \times 5, \\ 42 & =2 \times 3 \times 7 \\ 45 & =3 \times 3 \times 5 \\ 45) & =2 \times 3 \times 5 \times 2 \times 2 \times 3 \times \\ & =30 \times 12 \times 7=210 \times 12 \\ & =2520 \end{aligned} $

$ \begin{aligned} & \text{ and } \quad 45=3 \times 3 \times 5 \\ & \therefore \quad LCM(40,42,45)=2 \times 3 \times 5 \times 2 \times 2 \times 3 \times 7 \end{aligned} $

Minimum distance each should walk $2520 cm$. So that, each can cover the same distance in complete steps.

Solution

Denominator of the rational number $\frac{257}{5000}$ is 5000 .

Now, 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 $2^{m} \times 5^{n}$, where $m=3$ and $n=4$ are non-negative integers.

$\therefore \quad$ Rational number $=\frac{257}{5000}=\frac{257}{2^{3} \times 5^{4}} \times \frac{2}{2}$

[since, multiplying numerator and denominater by 2 ]

$=\frac{514}{2^{4} \times 5^{4}}=\frac{514}{(10)^{4}}$

$=\frac{514}{10000}=0.0514$

Hence, which is the required decimal expansion of the rational $\frac{257}{5000}$ and it is also a terminating decimal number.

Solution

Let us suppose that $\sqrt{p}+\sqrt{q}$ is rational.

Again, let $\sqrt{p}+\sqrt{q}=a$, where $a$ is rational.

Therefore,

$ \sqrt{q}=a-\sqrt{p} $

On squaring both sides, we get

$ q=a^{2}+p-2 a \sqrt{p} \quad[\because(a-b)^{2}=a^{2}+b^{2}-2 a b] $

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

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

Solution

Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers ’ $a$ ’ and 6 , there exist non-negative integers $q$ and $r$ such that

$\Rightarrow \quad a^{3}=(6 q+r)^{3}=216 q^{3}+r^{3}+3 \cdot 6 q \cdot r(6 q+r)$

$\Rightarrow \quad a^{3}=(216 q^{3}+108 q^{2} r+18 q r^{2})+r^{3}$

$[\because(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)]$

where, $0 \leq r<6$

Case I When $r=0$, then putting $r=0$ in Eq. (i), we get

$ a^{3}=216 q^{3}=6(36 q^{3})=6 m $

where, $m=36 q^{3}$ is an integer.

Case II When $r=1$, then putting $r=1$ in Eq. (i), we get

$ a^{3}=(216 q^{3}+108 q^{3}+18 q)+1=6(36 q^{3}+18 q^{3}+3 q)+1 $

$\Rightarrow \quad a^{3}=6 m+1$, where $m=(36 q^{3}+18 q^{3}+3 q)$ is an integer.

Case III When $r=2$, then putting $r=2$ in Eq. (i), we get

$ \begin{aligned} a^{3} & =(216 q^{3}+216 q^{2}+72 q)+8 \\ a^{3} & =(216 q^{3}+216 q^{2}+72 q+6)+2 \\ \Rightarrow \quad a^{3} & =6(36 q^{3}+36 q^{2}+12 q+1)+2=6 m+2 \end{aligned} $

where, $\quad m=(36 q^{2}+36 q^{2}+12 q+1)$ is an integer.

Case IV When $r=3$, then putting $r=3$ in Eq. (i), we get

$ \begin{aligned} & \qquad \begin{aligned} a^{3} & =(216 q^{3}+324 q^{2}+162 q)+27=(216 q^{3}+324 q^{2}+162 q+24)+3 \\ & =6(36 q^{3}+54 q^{2}+27 q+4)+3=6 m+3 \end{aligned} \\ & \text{ where, } m=(36 q^{2}+54 q^{2}+27 q+4) \text{ is an integer. } \end{aligned} $

Case V When $r=4$, then putting $r=4$ in Eq. (i), we get

$ \begin{aligned} a^{3} & =(216 q^{2}+432 q^{2}+288 q)+64 \\ & =6(36 q^{3}+72 q^{2}+48 q)+60+4 \\ & =a^{3} 6(36 q^{3}+72 q^{2}+48 q+10)+4=6 m+4 \end{aligned} $

where, $m=(36 q^{3}+72 q^{2}+48 q+10)$ is an integer.

Case VI When $r=5$, then putting $r=5$ in Eq. (i), we get

$ \begin{aligned} & a^{3}=(216 q^{3}+540 q^{2}+450 q)+125 \\ \Rightarrow \quad & a^{3}=(216 q^{3}+540 q^{2}+450 q)+120+5 \\ \Rightarrow \quad & a^{3}=6(36 q^{3}+90 q^{2}+75 q+20)+5 \\ \Rightarrow \quad & a^{3}=6 m+5 \end{aligned} $

where, $m=(36 q^{3}+90 q^{2}+75 q+20)$ is an integer.

Hence, 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 forms $6 m, 6 m+1,6 m+2,6 m+3,6 m+4$ and $6 m+5$ i.e., $6 m+r$.

Thinking Process

Since, $n$ is positive integer put $n=1,2,3, \ldots$ in the given numbers and makes the order triplets, then we see that any one digit in a triplet is divisible by 3.

Solution

Let $a=n, b=n+2$ and $c=n+4$

$\therefore \quad$ Order triplet is $(a, b, c)=(n, n+2, n+4)$

Where, $n$ is any positive integer i.e., $n=1,2,3, \ldots$

At $n=1 ;(a, b, c)=(1,1+2,1+4)=(1,3,5)$

At $n=2 ;(a, b, c)=(2,2+2,2+4)=(2,4,6)$

At $n=3 ;(a, b, c)=(3,3+2,3+4)=(3,5,7)$

At $n=4 ;(a, b, c)=(4,4+2,4+4)=(4,6,8)$

At $n=5 ;(a, b, c)=(5,5+2,5+4)=(5,7,9)$

At $n=6 ;(a, b, c)=(6,6+2,6+4)=(6,8,10)$

At $n=7 ;(a, b, c)=(7,7+2,7+4)=(7,9,11)$

At $n=8 ;(a, b, c)=(8,8+2,8+4)=(8,10,12)$

We observe that each triplet consist of one and only one number which is multiple of 3 i.e., divisible by 3 .

Hence, one and only one out of $n,(n+2)$ and $(n+4)$ is divisible by 3 , where, $n$ is any positive integer.

Alternate Method

On dividing ’ $n$ ’ by 3 , let $q$ be the quotient and $r$ be the remainder.

Then, $n=3 q+r$, where, $0 \leq r<3$

$\Rightarrow \quad n=3 q+r$, where, $r=0,1,2$

$\Rightarrow \quad n=3 q$ or $n=3 q+1$ or $n=3 q+2$

Case I If $n=3 q$, then $n$ is only divisible by 3 .

but $n+2$ 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 only divisible by 3 , but $n$ and $n+4$ are not divisible by 3 .

So, in this case, $(n+2)$ is divisible by 3 .

Case III When $n=3 q+2$, then $(n+4)=3 q+6=3(q+2)$, which is only divisible by 3 , but $n$ and $(n+2)$ are not divisible by 3 .

So, in this case, $(n+4)$ is divisible by 3 .

Hence, one and only one out of $n,(n+2)$ and $(n+4)$ is divisible by 3 .

Solution

Any three consecutive positive integers must be of the form

$n,(n+1)$ and $(n+2)$, where $n$ is any natural number. i.e., $n=1,2,3, \ldots$

Let, $\quad a=n, b=n+1$ and $c=n+2$

$\therefore \quad$ Order triplet is $(a, b, c)=(n, n+1, n+2)$, where $n=1,2,3, \ldots$

At $n=1 ;(a, b, c)=(1,1+1,1+2)=(1,2,3)$

At $n=2 ;(a, b, c)=(1,2+1,2+2)=(2,3,4)$

At $n=3 ;(a, b, c)=(3,3+1,3+2)=(3,4,5)$

At $n=4 ;(a, b, c)=(4,4+1,4+2)=(4,5,6)$

At $n=5 ;(a, b, c)=(5,5+1,5+2)=(5,6,7)$

At $n=6 ;(a, b, c)=(6,6+1,6+2)=(6,7,8)$

At $n=7 ;(a, b, c)=(7,7+1,7+2)=(7,8,9)$

At $n=8 ;(a, b, c)=(8,8+1,8+2)=(8,9,10)$

We observe that each triplet consist of one and only one number which is multiple of 3 i.e., divisible by 3 .

Hence, one of any three consecutive positive integers must be divisible by 3 .

Solution

Let

$$ \begin{align*} \Rightarrow & a=n^{3}-n \Rightarrow \quad a=n \cdot(n^{2}-1) \\ \Rightarrow & a=n \cdot(n-1) \cdot(n+1) \quad[\because(a^{2}-b^{2})=(a-b)(a+b)] \\ \Rightarrow & a=(n-1) \cdot n \cdot(n+1) \tag{i} \end{align*} $$

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 also divisible by 3 .

III. If one of the factor of any number is an even number, then it is also divisible by 2 .

$ \begin{matrix} \therefore & a=(n-1) \cdot n \cdot(n+1) \\ & \text{ Now, sum of the digits } = n-1+n+n+1=3.n\\ &=\text{ multiple of } 3, \text{ where } n \text{ is any positive integer. } \end{matrix} $

and $(n-1) \cdot n \cdot(n+1)$ will always be even, as one out of $(n-1)$ or $n$ or $(n+1)$ must of even.

Since, conditions II and III is completely satisfy the Eq. (i).

Hence, by condition I the number $n^{3}-n$ is always divisible by 6 , where $n$ is any positive integer.

Hence proved.

Solution

Given numbers are $n,(n+4),(n+8),(n+12)$ and $(n+16)$, where $n$ is any positive integer.

Then, let $n=5 q, 5 q+1,5 q+2,5 q+3,5 q+4$ for $q \in N \quad$ [by Euclid’s algorithm]

Then, in each case if we put the different values of $n$ in the given numbers. We definitely get one and only one of given numbers is divisible by 5 .

Hence, one and only one out of $n, n+4, n+8, n+12$ and $n+16$ is divisible by 5 .

Alternate Method

On dividing on $n$ by 5 , let $q$ be the quotient and $r$ be the remainder.

Then $n=5 q+r$, where $0 \leq r<5$.

$ \begin{matrix} \Rightarrow & n=5 q+r, \text{ where } r=0,1,2,3,4 \\ \Rightarrow & n=5 q \text{ or } 5 q+1 \text{ or } 5 q+2 \text{ or } 5 q+3 \text{ or } 5 q+4 . \end{matrix} $

Case I If $n=5 q$, then $n$ is only 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 only divisible by 5 .

So, in this case, $(n+4)$ is divisible by 5 .

Case III If $n=5 q+3$, then $n+2=5 q+3+12=5 q+15=5(q+3)$, which is divisible by 5 .

So, in this case $(n+12)$ is only divisible by 5 .

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

So, in this case, $(n+16)$ is only divisible by 5 .

Hence, 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.