# 10th Maths Chapter 2. Numbers And Sequences Exercise 2.2

10th Standard Maths Chapter 2 Exercise 2.2 Numbers And Sequences Guide Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

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

**Solution:
**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.

**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 = 2a × 3b then find a and b.

**Solution:**

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.

**Solution:
**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.

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?

**Solution:**

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. 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