Exercise 2.1 - Chapter 2 Numbers and Sequences 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

$\operatorname{Ex} 2.1$  : Chapter 2 - Numbers and Sequences - 10th Maths Guide Samacheer Kalvi Solutions
Question 1.
Find all positive integers which when divided by 3 leaves remainder $2 .$
Answer:
The positive integers when divided by 3 leaves remainder 2 .
By Euclid's division lemma $\mathrm{a}=\mathrm{bq}+\mathrm{r}, 0 \leq \mathrm{r}<\mathrm{b}$.
Here $\mathrm{a}=3 \mathrm{q}+2$, where $0 \leq \mathrm{q}<3$, a leaves remainder 2 when divided by 3 .
$\therefore 2,5,8,11 \ldots \ldots \ldots . \ldots . .$

Question 2.
A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21
flower pots. Find the number of completed rows and how many flower pots are left over?
Answer:
Here $a=532, b=21$
Using Euclid's division algorithm
$a=b q+r$
$532=21 \times 25+7$
Number of completed rows $=21$
Number of flower pots left over $=7$
 

Question 3.
Prove that the product of two consecutive positive integers is divisible by $2 .$
Solution:
Let $n-1$ and $n$ be two consecutive positive integers. Then their product is $(n-1) n$.
$(n-1)(n)=n^{2}-n$.
We know that any positive integer is of the form $2 \mathrm{q}$ or $2 \mathrm{q}+1$ for some integer $\mathrm{q}$. So, following cases arise.
Case I. When $\mathrm{n}=2 \mathrm{q}$.

In this case, we have
$n^{2}-n=(2 q)^{2}-2 q=4 q^{2}-2 q=2 q(2 q-1)$
$\Rightarrow \mathrm{n}^{2}-\mathrm{n}=2 \mathrm{r}$, where $\mathrm{r}=\mathrm{q}(2 \mathrm{q}-1)$
$\Rightarrow \mathrm{n}^{2}-\mathrm{n}$ is divisible by 2 .
Case II. When $\mathrm{n}=2 \mathrm{q}+1$
In this case, we have
$\begin{aligned}
&\mathrm{n}^{2}-\mathrm{n}=(2 \mathrm{q}+1)^{2}-(2 \mathrm{q}+1) \\
&=(2 \mathrm{q}+1)(2 \mathrm{q}+1-1)=2 \mathrm{q}(2 \mathrm{q}+1) \\
&\Rightarrow \mathrm{n}^{2}-\mathrm{n}=2 \mathrm{r} \text {, where } \mathrm{r}=\mathrm{q}(2 \mathrm{q}+1) \\
&\Rightarrow \mathrm{n}^{2}-\mathrm{n} \text { is divisible by } 2 .
\end{aligned}$
Hence, $\mathrm{n}^{2}-\mathrm{n}$ is divisible by 2 for every positive integer $\mathrm{n}$.
Hence it is Proved
 

Question $4 .$
When the positive integers $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are divided by 13 , the respective remainders are 9,7 and 10 .
Show that $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is divisible by 13 .
Answer:
Let the positive integer be $\mathrm{a}, \mathrm{b}$, and $\mathrm{c}$
We know that by Euclid's division lemma
$a=b q+r$
$a=13 q+9 \ldots .(1)$
$b=13 q+7 \ldots .(2)$
$\mathrm{c}=13 \mathrm{q}+10 \ldots(3)$
Add (1) (2) and (3)
$a+b+c=13 q+9+13 q+7+13 q+10$
$=39 \mathrm{q}+26$
$a+b+c=13(3 q+2)$
This expansion will be divisible by 13
$\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}$ is divisible by 13

Question $5 .$
Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4 .
Solution:
Let $x$ be any integer.
The square of $x$ is $x^{2}$.
Let $x$ be an even integer.
$x=2 q+0$
then $x^{2}=4 q^{2}+0$
When $x$ be an odd integer
When $x=2 k+1$ for some interger $k$.
$\begin{aligned}
&x^{2}=(2 k+1)^{2} \\
&=4 k^{2}+4 k+1 \\
&=4 k(k+1)+1 \\
&=4 q+1
\end{aligned}$
where $q=k(k+1)$ is some integer.
Hence it is proved.
 

Question $6 .$
Use Euclid's Division Algorithm to find the Highest Common Factor (H.C.F) of
(i) 340 and 412
Answer:
To find the HCF of 340 and 412 using Euclid's division algorithm. We get
$412=340 \times 1+72$
The remainder $72 \neq 0$
Again applying Euclid's division algorithm to the division of 340
$340=72 \times 4+52$
The remainder $52 \neq 0$
Again applying Euclid's division algorithm to the division 72 and remainder 52 we get
$72=52 \times 1+20$
The remainder $20 \neq 0$
Again applying Euclid's division algorithmThe remainder $20 \neq 0$
Again applying Euclid's division algorithm
$52=20 \times 2+12$
The remainder $12 \neq 0$
Again applying Euclid's division algorithm
$20-12 \times 1$ 1 8
The remainder $8 \neq 0$
Again applying Euclid's division algorithm
$12=8 \times 1+4$
The remainder $4 \neq 0$
Again applying Euclid's division algorithm
$8=4 \times 2+0$
The remainder is zero
$\therefore$ HCF of 340 and 412 is 4
(ii) 867 and 255
Answer:
To find the HCF of 867 and 255 using
Euclid's division algorithm. We get
$867=255 \times 3+102$
The remainder $102 \neq 0$
Using Euclid's division algorithm
$255=102 \times 2+51$
The remainder $51 \neq 0$
Again using Euclid's division algorithm
$102=51 \times 2+0$
The remainder is zero
$\therefore \mathrm{HCF}=51$
$\therefore$ HCF of 867 and 255 is 51

(iii) 10224 and 9648
Answer:
Find the HCF of 10224 and 9648 using Euclid's division algorithm. We get
$10224=9648 \times 1+576$
The remainder $576 \neq 0$
Again using Euclid's division algorithm
$9648=576 \times 16+432$
The remainder $432 \neq 0$
Using Euclid's division algorithm
$576=432 \times 1+144$
The remainder $144 \neq 0$
Again using Euclid's division algorithm
$432=144 \times 3+0$
The remainder is 0
$\therefore \mathrm{HCF}=144$
The HCF of 10224 and 9648 is 144
(iv) 84,90 and 120
Answer:
Find the HCF of 84,90 and 120 using Euclid's division algorithm
$90=84 \times 1+6$
The remainder $6 \neq 0$
Using Euclid's division algorithm
$4=14 \times 6+0$
The remainder is 0
$\therefore \mathrm{HCF}=6$
The HCF of 84 and 90 is 6
Find the HCF of 6 and 120
$120=6 \times 20+0$
The remainder is 0
$\therefore$ HCF of 120 and 6 is 6
$\therefore$ HCF of 84,90 and 120 is 6
 

Question $7 .$
Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
Solution:
The required number is the H.C.F. of the numbers.

$\begin{aligned}
&1230-12=1218 \\
&1926-12=1914
\end{aligned}$
First we find the H.C.F. of $1218 \& 1914$ by Euclid's division algorithm.
$1914=1218 \times 1+696$
The remainder $696 \neq 0$.
Again using Euclid's algorithm
$1218=696 \times 1+522$
The remainder $522 \neq 0$.
Again using Euclid's algorithm.
$696=522 \times 1+174$
The remainder $174 \neq 0$.
Again by Euclid's algorithm
$522=174 \times 3+0$
The remainder is zero.
$\therefore$ The H.C.F. of 1218 and 1914 is 174 .
$\therefore$ The required number is 174 .
 

Question 8.
If $\mathrm{d}$ is the Highest Common Factor of 32 and 60 , find $x$ and $y$ satisfying $\mathrm{d}=32 \mathrm{x}+60 \mathrm{y}$.
Answer:
Find the HCF of 32 and 60
$60=32 \times 1+28 \ldots .(1)$
The remainder $28 \neq 0$
By applying Euclid's division lemma
$32=28 \times 1+4 \ldots(2)$
The remainder $4 \neq 0$
Again by applying Euclid's division lemma
$28=4 \times 7+0 \ldots(3)$
The remainder is 0
HCF of 32 and 60 is 4
From (2) we get
$32=28 \times 1+4$
$4=32-28$
$=32-(60-32)$
$4=32-60+32$
$4=32 \times 2=60$
$4=32 \times 2+(-1) 60$
When compare with $\mathrm{d}=32 \mathrm{x}+60 \mathrm{y}$
$x=2$ and $y=-1$
The value of $x=2$ and $y=-1$

Question $9 .$
A positive integer when divided by 88 gives the remainder 61 . What will be the remainder when the same number is divided by 11 ?
Solution:
Let a (+ve) integer be $x$.
$x=88 \times y+6 l$
$61=11 \times 5+6(\because 88$ is multiple of 11$)$
$\therefore 6$ is the remainder. (When the number is divided by 88 giving the remainder 61 and when divided by 11 giving the remainder 6 ).
 

Question $10 .$
Prove that two consecutive positive integers are always coprime.
Answer:
1. Let the consecutive positive integers be $x$ and $x+1$.
2. The two number are co - prime both the numbers are divided by 1 .
3. If the two terms are $x$ and $x+1$ one is odd and the other one is even.
4. HCF of two consecutive number is always 1 .
5. Two consecutive positive integer are always coprime.
Fundamental Theorem of Arithmetic
Every composite number can be written uniquely as the product of power of prime is called fundamental theorem of Arithmetic.