Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.1

Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.1

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.1

Exercise 2.1 Class 10 Maths Samacheer 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 a = bq + r, 0 ≤ r < b.
Here a = 3q + 2, where 0 ≤ q < 3, a leaves remainder 2 when divided by 3.
∴ 2, 5, 8, 11 ……………..

10th Maths Exercise 2.1 Samacheer Kalvi 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 = bq + r
532 = 21 × 25 + 7
Number of completed rows = 21
Number of flower pots left over = 7

Ex 2.1 Class 10 Samacheer 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) = n² – n.
We know that any positive integer is of the form 2q or 2q + 1 for some integer q. So, following cases arise.
Case I. When n = 2q.
In this case, we have
n² – n = (2q)² – 2q = 4q² – 2q = 2q(2q – 1)
⇒ n² – n = 2r, where r = q(2q – 1)
⇒ n² – n is divisible by 2.

Case II. When n = 2q + 1
In this case, we have
n² – n = (2q + 1)² – (2q + 1)
= (2q + 1)(2q + 1 – 1) = 2q(2q + 1)
⇒ n² – n = 2r, where r = q (2q + 1).
⇒ n² – n is divisible by 2.
Hence, n² – n is divisible by 2 for every positive integer n.
Hence it is Proved

10th Maths Exercise 2.1 Question 4.
When the positive integers a, b and c are divided by 13, the respective remainders are 9,7 and 10. Show that a + b + c is divisible by 13.
Answer:
Let the positive integer be a, b, and c
We know that by Euclid’s division lemma
a = bq + r
a = 13q + 9 ….(1)
b = 13q + 7 ….(2)
c = 13q + 10 ….(3)
Add (1) (2) and (3)
a + b + c = 13q + 9 + 13q + 7 + 13q + 10
= 39q + 26
a + b + c = 13 (3q + 2)
This expansion will be divisible by 13
∴ a + b + c is divisible by 13

10th Maths 2.1 Exercise 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 x².
Let x be an even integer.
x = 2q + 0
then x² = 4q² + 0
When x be an odd integer
When x = 2k + 1 for some interger k.
x² = (2k + 1 )²
= 4k² + 4k + 1
= 4k (k + 1) + 1
= 4q + 1
where q = k(k + 1) is some integer.
Hence it is proved.

10th Maths Exercise 2.1 In Tamil 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 × 1 + 72
The remainder 72 ≠ 0
Again applying Euclid’s division algorithm to the division of 340
340 = 72 × 4 + 52
The remainder 52 ≠ 0
Again applying Euclid’s division algorithm to the division 72 and remainder 52 we get
72 = 52 × 1 + 20
The remainder 20 ≠ 0
Again applying Euclid’s division algorithm
52 = 20 × 2 + 12
The remainder 12 ≠ 0
Again applying Euclid’s division algorithm
20 = 12 × 1 + 8
The remainder 8 ≠ 0
Again applying Euclid’s division algorithm
12 = 8 × 1 + 4
The remainder 4 ≠ 0
Again applying Euclid’s division algorithm
8 = 4 × 2 + 0
The remainder is zero
∴ 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 × 3 + 102
The remainder 102 ≠ 0
Using Euclid’s division algorithm
255 = 102 × 2 + 51
The remainder 51 ≠ 0
Again using Euclid’s division algorithm
102 = 51 × 2 + 0
The remainder is zero
∴ HCF = 51
∴ 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 × 1 + 576
The remainder 576 ≠ 0
Again using Euclid’s division algorithm
9648 = 576 × 16 + 432
The remainder 432 ≠ 0
Using Euclid’s division algorithm
576 = 432 × 1 + 144
The remainder 144 ≠ 0
Again using Euclid’s division algorithm
432 = 144 × 3 + 0
The remainder is 0
∴ 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 × 1 + 6
The remainder 6 ≠ 0
Using Euclid’s division algorithm
4 = 14 × 6 + 0
The remainder is 0
∴ HCF = 6
The HCF of 84 and 90 is 6
Find the HCF of 6 and 120
120 = 6 × 20 + 0
The remainder is 0
∴ HCF of 120 and 6 is 6
∴ HCF of 84, 90 and 120 is 6

Numbers And Sequences 10th Class 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.
1230 – 12 = 1218,
1926 – 12 = 1914
First we find the H.C.F. of 1218 & 1914 by Euclid’s division algorithm.
1914 = 1218 × 1 + 696
The remainder 696 ≠ 0.
Again using Euclid’s algorithm
1218 = 696 × 1 + 522
The remainder 522 ≠ 0.
Again using Euclid’s algorithm.
696 = 522 × 1 + 174
The remainder 174 ≠ 0.
Again by Euclid’s algorithm
522 = 174 × 3 + 0
The remainder is zero.
∴ The H.C.F. of 1218 and 1914 is 174.
∴ The required number is 174.

10th Maths 2.1 Question 8.
If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
Answer:
Find the HCF of 32 and 60
60 = 32 × 1 + 28 ….(1)
The remainder 28 ≠ 0
By applying Euclid’s division lemma
32 = 28 × 1 + 4 ….(2)
The remainder 4 ≠ 0
Again by applying Euclid’s division lemma
28 = 4 × 7 + 0….(3)
The remainder is 0
HCF of 32 and 60 is 4
From (2) we get
32 = 28 × 1 + 4
4 = 32 – 28
= 32 – (60 – 32)
4 = 32 – 60 + 32
4 = 32 × 2 -60
4 = 32 x 2 + (-1) 60
When compare with d = 32x + 60 y
x = 2 and y = -1
The value of x = 2 and y = -1

10th Maths Samacheer Kalvi 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 × y + 61
61 = 11 × 5 + 6 (∵ 88 is multiple of 11)
∴ 6 is the remainder. (When the number is divided by 88 giving the remainder 61 and when divided by 11 giving the remainder 6).

10th Samacheer Maths Exercise 2.1 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.