10th Maths Answers Chapter 2 Numbers and Sequences Ex 2.2

10th Maths Answers Chapter 2 Numbers and Sequences Ex 2.2

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

Answer:

4ⁿ = (2 × 2)ⁿ = 2ⁿ × 2ⁿ

2 is a factor of 4ⁿ.

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

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

Example:

4² = 16

4³ = 64

4⁴ = 256

4⁵ = 1,024

4⁶ = 4,096

4⁷ = 16,384

4⁸ = 65, 536

4⁹ = 262,144

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

Answer:

2ⁿ is always even for any values of n.

[Example. 2² = 4, 2³ = 8, 2⁴ = 16 etc]

5m is always odd and it ends with 5.

[Example. 5² = 25, 5³ = 125, 5⁴ = 625 etc]

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

[Example. 2³ × 5³ = 8 × 125 = 1000

2² × 5² = 4 × 25 = 100]

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

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

Answer:

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.

4.If 13824 = 2ᵃ × 3ᵇ then find a and b.

Answer:

If 13824 = 2ᵃ × 3ᵇ

Using the prime factorisation tree

 

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

= 2⁹ × 3³ = 2ᵃ × 3ᵇ

∴ a = 9, b = 3.

5.If p₁x₁ × p₂x₂ × p₃x₃ × p₄x₄ = 113400 where p₁, p₂, p₃, p₄ are primes in ascending order and x₁, x₂, x₃, x₄ are integers, find the value of P₁ ,P₂, P₃, P₄ and x₁, x₂, x₃, x₄.

Answer:

If p₁x₁ × p₂x₂ × p₃x₃ × p₄x₄ = 113400

p₁, p₂, p₃, P₄ are primes in ascending order, x₁, x₂, x₃, x₄ are integers.

using Prime factorisation tree.

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

= 23 × 34 × 52 × 7

= p₁x₁ × p₂x₂ × p₃x₃ × p₄x₄

∴ p₁= 2, p₂ = 3, p₃ = 5, p₄ = 7, x₁ = 3, x₂ = 4, x₃ = 2, x₄ = 1.

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

Answer:

408 and 170. 

408 = 2³ × 3¹ × 17¹

170 = 2¹ × 5¹ × 17¹

∴ H.C.F. = 2¹ × 17¹ = 34.

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

∴ L.C.M. = 2³ × 3¹ × 5¹ × 17¹

= 2040.

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

Answer:

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

24 = 2³ × 3

15 = 3 × 5

36 = 2² × 3²

∴ L.C.M = 2³ × 3² × 5¹

= 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.

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

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

Answer:

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