NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 | CBSE class 10 maths chapter 1 Ex 1.1

To Prove: Any Positive odd integer is of the form \( 6q + 1, 6q + 3, 6q +5\)
Proof: To prove the statement by Euclid's lemma we have to consider divisor as 6 and then find out the possible remainders when divided by 6 By taking, '\( a \)' as any positive integer and \( b=6 \).
Applying Euclid's algorithm
\(
a=6 q+r
\)
As divisor is 6 the remainder can take only 6 values from 0 to 5
Here, \( r= \) remainder \( =0,1,2,3,4,5 \) and \( q \geq 0 \)
So, total possible forms are \( 6 q+0,6 q+1,6 q+2,6 q+3,6 q+4 \) and \( 6 q+5 \)
\( 6 q+0, \) (6 is divisible by 2, its an even number)
\( 6 \mathrm{q}+1 \), (6 is divisible by 2 but 1 is not divisible by 2, its an odd number)
\( 6 \mathrm{q}+2 \), (6 and 2 both are divisible by 2, its an even number)
\( 6 \mathrm{q}+3 \), (6 is divisible by 2 but 3 is not divisible by 2, its an odd number)
\( 6 \mathrm{q}+4 \), (6 and 4 both are divisible by 2, its an even number)
\( 6 \mathrm{q}+5 \), (6 is divisible by 2 but 5 is not divisible by 2, its an odd number)
Therefore, odd numbers will be in the form \( 6 q+1 \), or \( 6 q+3 \), or \( 6 q+5 \) Hence, Proved.

Suppose, both groups are arranged in 'n' columns, for completely filling each column,
The maximum no of columns in which they can march is the highest common factor of their number of members. i.e. \( \mathrm{n}=\operatorname{HCF}(616,32) \)
By using, Euclid's division algorithm
\( 616=32 \times 19+8 \)
Remainder \( \neq 0 \)
So, again Applying Euclid's division algorithm
\( 32=8 \times 4+0 \)
HCF of \( (616,32) \) is 8.
So, They can march in 8 columns each.

To Prove: Square of any number is of the form 3 m or \( 3 \mathrm{~m}+1 \)
Proof: to prove this statement from Euclid's division lemma, take any number as a divisor, in question we have 3 m and \( 3 \mathrm{~m}+1 \) as the form
So, By taking, '\( a \)' as any positive integer and \( b=3 \).
Applying Euclid's algorithm \( \mathrm{a}=\mathrm{bq}+\mathrm{r} \).
\(
a=3 q+r
\)
Here, \( \mathrm{r}= \) remainder \( =0,1,2 \) and \( \mathrm{q} \geq 0 \) as the divisor is 3 there can be only 3 remainders, 0, 1 and 2.
So, putting all the possible values of the remainder in, \( a=3 q+r \)
\( \mathrm{a}=3 \mathrm{q} \) or \( 3 \mathrm{q}+1 \) or \( 3 \mathrm{q}+2 \)
And now squaring all the values,
When \( a=3 q \) Squaring both sides we get,
\(a^{2}=(3 q)^{2}\)
\(a^{2}=9 q^{2}\)
\(a^{2}=3\left(3 q^{2}\right)\)
\(a^{2}=3 k_{1}\)
Where \( \mathrm{k}_{1}=3 \mathrm{q}^{2} \)
When \( \mathrm{a}=3 \mathrm{q}+1 \) Squaring both sides we get, \( \mathrm{a}^{2}=(3 \mathrm{q}+1)^{2} \)
\(a^{2}=9 q^{2}+6 q+1\)
\(a^{2}=3\left(3 q^{2}+2 q\right)+1\)
\(a^{2}=3 k^{2}+1\)
Where \( \mathrm{k}_{2}=3 \mathrm{q}^{2}+2 \mathrm{q} \)
When \( \mathrm{a}=3 \mathrm{q}+2 \)
Squaring both sides we get, \( \mathrm{a}^{2}=(3 \mathrm{q}+2)^{2} \)
\(
\begin{array}{l}
\mathrm{a}^{2}=9 \mathrm{q}^{2}+12 q+4 \\
\mathrm{a}^{2}=9 \mathrm{q}^{2}+12 q+3+1 \\
\mathrm{a}^{2}=3\left(3 \mathrm{q}^{2}+4 q+1\right)+1 \\
\mathrm{a}^{2}=3 \mathrm{k}_{3}+1
\end{array}
\)
Where \( k_{3}=3 q^{2}+4 q+1 \)
Where \( \mathrm{k}_{1}, \mathrm{k}_{2} \) and \( \mathrm{k}_{3} \) are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3 m or \( 3 \mathrm{~m}+1 \).

Let a be any positive integer. Then, it is of the form 3q or, \( 3 q+1 \) or, \( 3 q+2 \).
We know that according to Euclid's division lemma:
\( \mathrm{a}=\mathrm{bq}+\mathrm{r} \) So, we have the following cases:
Case I When \( \mathrm{a}=3 \mathrm{q} \)
In this case, we have
\( a^{3}=(3 q)^{3}=27 q^{3}=9\left(3 q^{3}\right)=9 m \), where \( m=3 q 3 \)
Case II When \( \mathrm{a}=3 \mathrm{q}+1 \)
In this case, we have
\(
\begin{array}{l}
a^{3}=(3 q+1)^{3} \\
\Rightarrow 27 q^{3}+27 q^{2}+9 q+1 \\
\Rightarrow 9 q\left(3 q^{2}+3 q+1\right)+1 \\
\Rightarrow a^{3}=9 m+1, \text { where } m=q\left(3 q^{2}+3 q+1\right)
\end{array}
\)
Case III When \( \mathrm{a}=3 \mathrm{q}+2 \)
In this case, we have
\(
\begin{array}{l}
a^{3}=(3 q+1)^{3} \\
\Rightarrow 27 q^{3}+54 q^{2}+36 q+8 \\
\Rightarrow 9 q\left(3 q^{2}+6 q+4\right)+8 \\
\Rightarrow a^{3}=9 m+8, \text { where } m=q\left(3 q^{2}+6 q+4\right)
\end{array}
\)
Hence, \( a^{3} \) is the form of \( 9 m \) or, \( 9 m+1 \) or, \( 9 m+8 \)