# 10th Maths – Chapter 2. Numbers And Sequences – Exercise 2.1

10th Standard Maths Chapter 2 Exercise 2.1 Numbers And Sequences Guide Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

### 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 ……………..

### 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

### 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, the 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

### 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

### 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.

### 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

### 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.

### 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

### 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).

### 10. Prove that two consecutive positive integers are always coprime.

**Solution:**

- Let the consecutive positive integers be x and x + 1.
- The two number are co – prime both the numbers are divided by 1.
- If the two terms are x and x + 1 one is odd and the other one is even.
- HCF of two consecutive number is always 1.
- 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.