Students can download Maths Chapter 2 Numbers and Sequences Ex 2.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Question 1.

Find all positive integers which when divided by 3 leaves remainder 2.

Answer:

All the positive integers when divided by 3 leaves remainder 2

By Euclid’s division lemma

a = bq + r, 0 < r < b

a = 3q + r where 0 < q < 3

a leaves remainder 2 when divided by 3

∴ The positive integers are 2, 5, 8, 11,…

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.

Solution:

Using Euclid’s division algorithm,

a = 21q + r, we get 532 = 21 × 25 + 7.

The remainder is 7.

No. of completed rows = 25, left over flower pots = 7 pots.

Question 3.

Prove that the product of two consecutive positive integers is divisible by 2.

Answer:

Let n – 1 and n be two consecutive positive integers, then the product is n (n – 1)

n(n – 1) = n^{2} – n

We know that any positive integer is of the form 2q or 2q + 1 for same integer q

Case 1:

when n = 2 q

n^{2} – n = (2q)^{2} – 2q

= 4q^{2} – 2q

= 2q (2q – 1)

= 2 [q (2q – 1)]

n^{2} – n = 2 r

r = q(2q – 1)

Hence n^{2} – n. divisible by 2 for every positive integer.

Case 2:

when n = 2q + 1

n^{2} – n = (2q + 1 )^{2} – (2q + 1 )

= (2q + 1) [2q + 1 – 1]

= 2q (2q + 1)

n^{2} – n = 2r

r = q (2q + 1)

n^{2} – n divisible by 2 for every positive integer.

Question 4.

When the positive integers be 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.

Solution:

Let the positive integers be a, b, and c.

a = 13 q + 9

b = 13q + 1

c = 13 q + 10

a + b + c = 13q + 9 + 13q + 7 + 13q + 10

= 39q + 26

= 13 (3q + 2)

which is divisible by 13.

Question 5.

Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.

Answer:

Let the integer be ” x ”

The square of its integer is “x^{2} ”

Let x be an even integer

x = 2q + 0

x2 = 4q^{2}

When x is an odd integer

x = 2k + 1

x^{2} = (2k + 1)^{2}

= 4k^{2} + 4k + 1

= 4k (k + 1) + 1

= 4q + 1 [q = k(k + 1)]

It is divisible by 4

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

(ii) 867 and 255

(iii) 10224 and 9648

(iv) 84, 90 and 120

Solution:

To find the H.C.F. 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

340 = 72 × 4 + 52

The remainder 52 ≠ 0.

Again applying Euclid’s division algorithm

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.

Therefore H.C.F. of 340 and 412 is 4.

(ii) To find the H.C.F. of 867 and 255, using Euclid’s division algorithm.

867 = 255 × 3 + 102

The remainder 102 ≠ 0.

Again 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.

Therefore the H.C.F. of 867 and 255 is 51.

(iii) To find H.C.F. 10224 and 9648. Using Euclid’s division algorithm.

10224 = 9648 × 1 + 576

The remainder 576 ≠ 0.

Again using Euclid’s division algorithm

9648 = 576 × 16 + 432

Remainder 432 ≠ 0.

Again applying Euclid’s division algorithm

576 = 432 × 1 + 144

Remainder 144 ≠ 0.

Again using Euclid’s division algorithm

432 = 144 × 3 + 0

The remainder is zero.

There H.C.F. of 10224 and 9648 is 144.

(iv) To find H.C.F. of 84, 90 and 120.

Using Euclid’s division algorithm

90 = 84 × 1 + 6

The remainder 6 ≠ 0.

Again using Euclid’s division algorithm

84 = 6 × 14 + 0

The remainder is zero.

∴ The H.C.F. of 84 and 90 is 6. To find the H.C.F. of 6 and 120 using Euclid’s division algorithm.

120 = 6 × 20 + 0

The remainder is zero.

Therefore H.C.F. of 120 and 6 is 6

∴ H.C.F. of 84, 90 and 120 is 6.

Question 7.

Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.

Answer:

Find the HCF of ( 1230 – 12) and (1926- 12)

i.e HCF of 1218 and 1914

By Euclid’s division algorithm

1914 = 1218 × 1 + 696

The remainder 696 ≠ 0

By Euclid’s division algorithm

1218 = 696 × 1 + 522

The remainder 522 ≠ 0

Again by Euclid’s division algorithm

696 = 522 × 1 + 174

The remainder 174 ≠ 0 Again by Euclid’s division algorithm

522 = 174 × 3 + 0

The remainder is zero

∴ HCF of 1218 and 1914 is 174

The largest value is 174

Question 8.

If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.

Solution:

Applying Euclid’s divison lemma to 32 and 60, we get

60 = 32 × 1 + 28 ……………. (i)

The remainder is 28 ≠ 0.

Again applying division lemma

32 = 28 × 1 + 4 ……………. (ii)

The remainder 4 ≠ 0.

Again applying division lemma

28 = 4 × 7 + 0 ………….. (iii)

The remainder zero.

∴ H.C.F. of 32 and 60 is 4.

From (ii), we get

32 = 28 × 1 + 4

⇒ 4 = 32 – 28 × 1

⇒ 4 = 32 – (60 – 32 × 1) × 1

⇒ 4 = 32 – 60 + 32

⇒ 4 = 32 × 2+(-1) × 60

∴ 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?

Answer:

Let the positive integer be “x”

x = 88 × y + 61 (a = pq + r)

since 88 is a multiple of 11

61 = 11 × 5 + 6

∴ The remainder is 6

Question 10.

Prove that two consecutive positive integers are always coprime.

Solution:

Let the numbers be I, I + 1:

They are co-prime if only +ve integer that divides both is 1.

I is given to be +ve integer.

So I = 1, 2, 3, ….

∴ One is odd and the other one is even. Hence H.C.F. of the two consecutive numbers is 1. Hence the result.