You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.2

**10th Maths Exercise 2.2 Samacheer Kalvi Question 1.**

For what values of natural number n, 4n can end with the digit 6?

Solution:

4^{n} = (2 × 2)^{n} = 2^{n} × 2^{n}

2 is a factor of 4^{n}.

So, 4^{n} is always even and end with 4 and 6.

When n is an even number say 2, 4, 6, 8 then 4^{n} can end with the digit 6.

Example:

4^{2} = 16

4^{3} = 64

4^{4 }= 256

4^{5} = 1,024

4^{6} = 4,096

4^{7} = 16,384

4^{8} = 65, 536

4^{9} = 262,144

**Ex 2.2 Class 10 Samacheer Question 2.**

If m, n are natural numbers, for what values of m, does 2^{n} × 5^{m} ends in 5?

Answer:

2^{n} is always even for any values of n.

[Example. 2^{2} = 4, 2^{3} = 8, 2^{4} = 16 etc]

5^{m} is always odd and it ends with 5.

[Example. 5^{2} = 25, 5^{3} = 125, 5^{4} = 625 etc]

But 2^{n} × 5^{m} is always even and end in 0.

[Example. 2^{3} × 5^{3} = 8 × 125 = 1000

2^{2} × 5^{2} = 4 × 25 = 100]

∴ 2^{n} × 5^{m} cannot end with the digit 5 for any values of m.

**Exercise 2.2 Class 10 Maths Samacheer Question 3.**

Find the H.C.F. of 252525 and 363636.

Solution:

To find the H.C.F. of 252525 and 363636

Using Euclid’s Division algorithm

363636 = 252525 × 1 + 111111

The remainder 111111 ≠ 0.

∴ Again by division algorithm

252525 = 111111 × 2 + 30303

The remainder 30303 ≠ 0.

∴ Again by division algorithm.

111111 = 30303 × 3 + 20202

The remainder 20202 ≠ 0.

∴ Again by division algorithm

30303 = 20202 × 1 + 10101

The remainder 10101 ≠ 0.

∴ Again using division algorithm

20202 = 10101 × 2 + 0

The remainder is 0.

∴ 10101 is the H.C.F. of 363636 and 252525.

**10th Maths Exercise 2.2 Question 4.**

If 13824 = 2^{a} × 3^{b} then find a and b.

Solution:

If 13824 = 2^{a} × 3^{b}

Using the prime factorisation tree

13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

= 2^{9} × 3^{3} = 2^{a} × 3^{b}

∴ a = 9, b = 3.

**10th Maths Exercise 2.2 In Tamil Question 5.**

If p_{1}^{x}_{1 }× p_{2}^{x}_{2} × p_{3}^{x}_{3} × p_{4}^{x}_{4 }= 113400 where p_{1}, p_{2}, p_{3}, p_{4} are primes in ascending order and x_{1}, x_{2}, x_{3}, x_{4} are integers, find the value of P_{1}, P_{2}, P_{3}, P_{4} and x_{1}, x_{2}, x_{3}, x_{4}.

Solution:

If p_{1}^{x}_{1 }× p_{2}^{x}_{2} × p_{3}^{x}_{3} × p_{4}^{x}_{4 }= 113400

p_{1}, p_{2}, p_{3}, P_{4} are primes in ascending order, x_{1}, x_{2}, x_{3}, x_{4} are integers.

using Prime factorisation tree.

113400 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 7

= 23 × 34 × 52 × 7

= p_{1}^{x}_{1 }× p_{2}^{x}_{2} × p_{3}^{x}_{3} × p_{4}^{x}_{4}

∴ p_{1}= 2, p_{2} = 3, p_{3} = 5, p_{4} = 7, x_{1} = 3, x_{2} = 4, x_{3} = 2, x_{4} = 1.

Question 6.

Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of arithmetic.

Solution:

408 and 170.

408 = 2^{3} × 3^{1} × 17^{1}

170 = 2^{1} × 5^{1} × 17^{1}

∴ H.C.F. = 2^{1} × 17^{1} = 34.

To find L.C.M, we list all prime factors of 408 and 170, and their greatest exponents as follows.

∴ L.C.M. = 2^{3} × 3^{1} × 5^{1} × 17^{1}

= 2040.

**10th Maths 2.2 Exercise Question 7.**

Find the greatest number consisting of 6 digits which is exactly divisible by 24, 15, 36?

Solution:

To find L.C.M of 24, 15, 36

24 = 2^{3} × 3

15 = 3 × 5

36 = 2^{2} × 3^{2}

∴ L.C.M = 2^{3} × 3^{2} × 5^{1}

= 8 × 9 × 5

= 360

If a number has to be exactly divisible by 24, 15, and 36, then it has to be divisible by 360. Greatest 6 digit number is 999999.

Common multiplies of 24, 15, 36 with 6 digits are 103680, 116640, 115520, …933120, 999720 with six digits.

∴ The greatest number consisting 6 digits which is exactly divisible by 24, 15, 36 is 999720.

**10th Maths Ex 2.2 Question 8.**

What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?

Answer:

Find the L.C.M of 35, 56, and 91

35 – 5 × 7 56

56 = 2 × 2 × 2 × 7

91 = 7 × 13

L.C.M = 23 × 5 × 7 × 13

= 3640

Since it leaves remainder 7

The required number = 3640 + 7

= 3647

The smallest number is = 3647

**10th Maths 2.2 Question 9.**

Find the least number that is divisible by the first ten natural numbers.

Solution:

The least number that is divisible by the first ten natural numbers is 2520.

Hint:

1,2, 3,4, 5, 6, 7, 8,9,10

The least multiple of 2 & 4 is 8

The least multiple of 3 is 9

The least multiple of 7 is 7

The least multiple of 5 is 5

∴ 5 × 7 × 9 × 8 = 2520.

L.C.M is 8 × 9 × 7 × 5

= 40 × 63

= 2520