## Samacheer Kalvi 10th Maths Book – Chapter 2 Exercise 2.2: Book Back Question and Answers

TN Samacheer Kalvi 10th Maths Book uploaded as PDF and its book back question & answers solutions Guide PDF given below. Class 10th Standard New Syllabus Maths Books consist of 8 units. Each chapter is well designed and organized by the TN textbook corporation. Complete solutions for all 8 Units’ book back questions uploaded and available on our site. For Samacheer kalvi 10th Maths – Numbers and Sequences Ex 2.2 book back question and answers solutions are given below:

Complete Samacheer Kalvi 10th Maths Guide available in PDF for free download. Check Samacheer Kalvi’s 10th Maths Unit 2 Chapter 1 Book Back Questions and Answers below:

TN Samacheer Kalvi 10th Maths Unit 2 Book Back Exercise given below.

**Chapter 2**

** Numbers and Sequences Ex 2.2**

- 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. The 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 the 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 the 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

**Other Important Links for 10th Maths Book Back Answers Solutions:**

For 10th Maths Book Unit 1 book back question and answers, check the link – **Samacheer kalvi 10th Maths Unit 2 Numbers and Sequences**

Click Here for Complete Samacheer Kalvi 10th Book Back Solution Guide PDF – **Samacheer Kalvi 10th Maths Book Back Answers**