**Samacheer Kalvi 10th Maths Book Back Solution:**

Tamil Nadu **10th Maths Book Back Answers Unit 2 – Numbers and Sequences Ex 2.2** are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete** Samacheer Kalvi Books Back Answers/Solutions** are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 2 Numbers and Sequences consists of 5 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Numbers and Sequences Ex 2.2 Answers in English.

**10th Maths Book Back Answers/Solutions:**

TN Samacheer Kalvi 10th Maths Unit 2 Chapter 2 Book Back Exercise given below. The 10th Maths Book Back Solutions Guide is uploaded below,

**Chapter 2**

** Numbers and Sequences Ex 2.2**

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

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.

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.

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.

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.

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.

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.

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.

**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

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