Q: 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?

HCF $(616,32)$ will give the maximum number of columns in which they can march. We can use Euclid's algorithm to find the HCF. $616=32 \times 19+8$ $32=8 \times 4+0$ The $\operatorname{HCF}(616,32)$ is 8. Therefore, they can march in 8 columns each.

Q: Use Euclid's division lemma to show that the square of any positive integer is either of form $3 m$ or $3 m+1$ for some integer $m$. [Hint: Let $x$ be any positive integer then it is of the form $3 q, 3 q+1$ or $3 q+2$. Now square each of these and show that they can be rewritten in the form $3 m$ or $3 m+1$.]

Let $a$ be any positive integer and $b=3$. Then $a=3 q+r$ for some integer $q \geq 0$ And $r=0,1,2$ because $0 \leq r Therefore, $a=3 q$ or $3 q+1$ or $3 q+2$ Or, $$ \begin{aligned} a^{2} & =(3 q)^{2} \text { or }(3 q+1)^{2} \text { or }(3 q+2)^{2} \\ & =(3 q)^{2} \text { or } 9 q^{2}+6 q+1 \text { or } 9 q^{2}+12 q+4 \\ & =3 \times\left(3 q^{2}\right) \text { or } 3 \times\left(3 q^{2}+2 q\right)+1 \text { or } 3 \times\left(3 q^{2}+4 q+1\right)+1 \\ & =3 k_{1} \text { or } 3 k_{2}+1 \text { or } 3 k_{3}+1 \end{aligned} $$ Where $k_{1}, k_{2}$, and $k_{3}$ are some positive integers Hence, it can be said that the square of any positive integer is either of the form $3 m$ or $3 m+1$.

Q: Use Euclid's division lemma to show that the cube of any positive integer is of the form $9 m, 9 m+1$ or $9 m+8$.

Let $a$ be any positive integer and $b=3$ $a=3 q+r$, where $q \geq 0$ and $0 \leq r $a=3 q$ or $3 q+1$ or $3 q+2$ Therefore, every number can be represented as these three forms. There are three cases. Case 1: When $a=3 q$, $a^{3}=(3 q)^{3}=27 q^{3}=9\left(3 q^{3}\right)=9 m$ Where $m$ is an integer such that $m=3 q^{3}$ Case 2: When $a=3 q+1$, $a^{3}=(3 q+1)^{3}$ $a^{3}=27 q^{3}+27 q^{2}+9 q+1$ $a^{3}=9\left(3 q^{3}+3 q^{2}+q\right)+1$ $a^{3}=9 m+1$ Where $m$ is an integer such that $m=\left(3 q^{3}+3 q^{2}+q\right)$ Case 3: When $a=3 q+2$, $a^{3}=(3 q+2)^{3}$ $a^{3}=27 q^{3}+54 q^{2}+36 q+8$ $a^{3}=9\left(3 q^{3}+6 q^{2}+4 q\right)+8$ $a^{3}=9 m+8$ Where $m$ is an integer such that $m=\left(3 q^{3}+6 q^{2}+4 q\right)$ Therefore, the cube of any positive integer is of the form $9 m, 9 m+1$, or $9 m+8$. \section*{Mathematics} (Chapter - 1) (Real Numbers) (Class X) \section*{Exercise 1.2}

Question 1

Use Euclid's division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255

Accepted Answer

(i) 135 and 225

Since $225>135$, we apply the division lemma to 225 and 135 to obtain $225=135 imes 1+90$

Since remainder $90 eq 0$, we apply the division lemma to 135 and 90 to obtain
$135=90 imes 1+45$
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
$90=2 imes 45+0$
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45 ,
Therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
$38220=196 imes 195+0$
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii) 867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain $867=255 imes 3+102$

Since remainder $102 eq 0$, we apply the division lemma to 255 and 102 to obtain
$255=102 imes 2+51$
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
$102=51 imes 2+0$
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51 , Therefore,
HCF of 867 and 255 is 51.

Question 2

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

Accepted Answer

Let $a$ be any positive integer and $b=6$.
Then, by Euclid's algorithm, $a=6 q+r$ for some integer $q \geq 0$, and $r=0,1,2,3,4,5$ because $0 \leq r<6$.
Therefore, $a=6 q$ or $6 q+1$ or $6 q+2$ or $6 q+3$ or $6 q+4$ or $6 q+5$
Also, $6 q+1=2 imes 3 q+1=2 k_{1}+1$, where $k_{1}$ is a positive integer $6 q+3=(6 q+2)+1=2(3 q+1)+1=2 k_{2}+1$, where $k_{2}$ is an integer $6 q+5=(6 q+4)+1=2(3 q+2)+1=2 k_{3}+1$, where $k_{3}$ is an integer Clearly,
$6 q+1,6 q+3,6 q+5$ are of the form $2 k+1$, where $k$ is an integer.
Therefore, $6 q+1,6 q+3,6 q+5$ are not exactly divisible by 2 .

Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form $6 q+1$, or $6 q+3$, or $6 q+5$

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?

Accepted Answer

HCF $(616,32)$ will give the maximum number of columns in which they can march.

We can use Euclid's algorithm to find the HCF.
$616=32 imes 19+8$
$32=8 imes 4+0$
The $\operatorname{HCF}(616,32)$ is 8.
Therefore, they can march in 8 columns each.

Question 4

Use Euclid's division lemma to show that the square of any positive integer is either of form $3 m$ or $3 m+1$ for some integer $m$.
[Hint: Let $x$ be any positive integer then it is of the form $3 q, 3 q+1$ or $3 q+2$. Now square each of these and show that they can be rewritten in the form $3 m$ or $3 m+1$.]

Accepted Answer

Let $a$ be any positive integer and $b=3$.
Then $a=3 q+r$ for some integer $q \geq 0$
And $r=0,1,2$ because $0 \leq r<3$
Therefore, $a=3 q$ or $3 q+1$ or $3 q+2$ Or,
$$
\begin{aligned}
a^{2} & =(3 q)^{2} ext { or }(3 q+1)^{2} ext { or }(3 q+2)^{2} \
& =(3 q)^{2} ext { or } 9 q^{2}+6 q+1 ext { or } 9 q^{2}+12 q+4 \
& =3 imes\left(3 q^{2} ight) ext { or } 3 imes\left(3 q^{2}+2 q ight)+1 ext { or } 3 imes\left(3 q^{2}+4 q+1 ight)+1 \
& =3 k_{1} ext { or } 3 k_{2}+1 ext { or } 3 k_{3}+1
\end{aligned}
$$

Where $k_{1}, k_{2}$, and $k_{3}$ are some positive integers
Hence, it can be said that the square of any positive integer is either of the form $3 m$ or $3 m+1$.

Question 5

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

Accepted Answer

Let $a$ be any positive integer and $b=3$
$a=3 q+r$, where $q \geq 0$ and $0 \leq r<3$
$a=3 q$ or $3 q+1$ or $3 q+2$
Therefore, every number can be represented as these three forms.
There are three cases.
Case 1: When $a=3 q$,
$a^{3}=(3 q)^{3}=27 q^{3}=9\left(3 q^{3}\right)=9 m$
Where $m$ is an integer such that $m=3 q^{3}$
Case 2: When $a=3 q+1$,
$a^{3}=(3 q+1)^{3}$
$a^{3}=27 q^{3}+27 q^{2}+9 q+1$
$a^{3}=9\left(3 q^{3}+3 q^{2}+q\right)+1$
$a^{3}=9 m+1$
Where $m$ is an integer such that $m=\left(3 q^{3}+3 q^{2}+q\right)$
Case 3: When $a=3 q+2$,
$a^{3}=(3 q+2)^{3}$
$a^{3}=27 q^{3}+54 q^{2}+36 q+8$
$a^{3}=9\left(3 q^{3}+6 q^{2}+4 q\right)+8$
$a^{3}=9 m+8$
Where $m$ is an integer such that $m=\left(3 q^{3}+6 q^{2}+4 q\right)$
Therefore, the cube of any positive integer is of the form $9 m, 9 m+1$, or $9 m+8$.

\section*{Mathematics}
(Chapter - 1) (Real Numbers)
(Class X)

\section*{Exercise 1.2}