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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
