Question

1. 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.

2. Given that HCF (306, 657) = 9, find LCM (306, 657).​

Answer 1

$\large\bf{\underline{\underline{Answer\:1}}}$

Let$\red{\bf {'a'}}$be any positive integer.

On dividing it by 3 , let 'q' be the quotient and 'r' be the remainder.

$\red{\bf {Such\: that}}$

$\underline{\boxed{\sf\purple{a = 3q + r}}}$

$\red{\bf {where}}$

$\underline{\boxed{\sf\purple{r = 0 ,1 , 2}}}$

$\red{\bf {When}}$

$\underline{\boxed{\sf\purple{r = 0}}}$

$\mapsto\bf{r = 25\:m}$∴ a = 3q

$\red{\bf {When}}$

$\underline{\boxed{\sf\purple{r = 1}}}$

$\mapsto\bf{∴ a = 3q + 1}$

$\red{\bf {When}}$

$\underline{\boxed{\sf\purple{r = 2}}}$

$\mapsto\bf{∴ a = 3q + 2}$

$\red{\bf {When}}$

$\underline{\boxed{\sf\purple{a = 3q}}}$

$\textbf{On squaring both the sides,}$

$\begin{gathered} {a}^{2} = 9 {q}^{2} \\ {a}^{2} = 3 \times (3 {q}^{2} ) \\ {a}^{2} = 3 \\ where \: m = 3 {q}^{2} \end{gathered}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\pink{\bigstar}★ \large\underline{\boxed{\bf\pink{When, \:a = 3q + 1}}}$

$\textbf{On squaring both the sides ,}$

$\begin{gathered} {a}^{2} = (3q + 1)^{2} \\ {a}^{2} = 9 {q}^{2} + 2 \times 3q \times 1 + {1}^{2} \\ {a}^{2} = 9 {q}^{2} + 6q + 1 \\ {a}^{2} = 3(3 {q}^{2} + 2q) + 1 \\ {a}^{2} = 3m + 1 \\ where \: m \: = 3 {q}^{2} + 2q\end{gathered}$

$\red{\bf {When\:, a \:= \:3q\: + \:2}}$

$\textbf{On squaring both the sides,}$

$\begin{gathered} {a}^{2} = (3q + 2)^{2} \\ {a}^{2} = 3 {q}^{2} + 2 \times 3q \times 2 + {2}^{2} \\ {a}^{2} = 9 {q}^{2} + 12q + 4 \\ {a}^{2} = (9 {q}^{2} + 12q + 3) + 1 \\ {a}^{2} = 3(3 {q}^{2} + 4q + 1) + 1 \\ {a}^{2} = 3m + 1 \\ where \: m \: = 3 {q}^{2} + 4q + 1\end{gathered}$

$\bold{Therefore ,}$

the square of any positive integer is either of the form$\red{\bf {3m\: or\: 3m\:+\:1.}}$

$\large\bf{\underline{\underline{Answer\:2}}}$

LCM of $\red{\bf {306}}$ and $\red{\bf {657}}$ is $\red{\bf {22338}}$

$\large\bf{\underline{\underline{Let\: us\: see}}}$

how to find the HCF of$\red{\bf {306}}$and $\red{\bf {657}}$

$\large\bf{\underline{\underline{Explanation}}}$

We will use the formula to find the LCM of 306 and 657.

LCM (a, b) = (a × b) / HCF (a, b)

Given two numbers 306 and 657 and their HCF is 9.

$\bold{To \:find :}$

LCM (306,657 )

Using the formula,

$\bold{We have}$

$\sf{LCM\: (306, 657)}$

$\boxed{\red{\sf \frac{= (306 × 657)}{HCF (306, 657)}}}$

$\boxed{\red{\sf \frac{= 201042}{9}}}$

$\mapsto\bf{= 22338}$

$\underline{\bf{Thus,}}$

$\textbf{LCM of 306 and 657 is 22338}$

Thankyou :)