Deepak's Classes

Send your query on 8882984633

NCERT
Class Xth
CBSE Board
Chapter 01 - PART 04

Exercise 1.1

1. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Solution:
225 = 135 X 1 + 90
135 = 90 X 1 + 45
90 = 45 X 2 + 0 HCF is 45

(ii) 38220 = 196 X 195 + 0 HCF is 195
(iii) 867 = 255 X 3 + 102
255 = 102 X 2 + 51
102 = 51 X 2 + 0 HCF is 51

2. IF q is an integer. Show that any positive odd integer is of the form 6q+1or, 6q+3 or 6q+5

Solution:

Let’s check what we get if we multiply any integer (q) with 6
6X1 = 6, 6X2=12, 6X3=18, 6X4=24, 6X5=30, 6X999 = 5994 So it is seen that no matter what q’s value is, even/odd, if we multiply it with 6 we will get an even number.

Always remember

Any integer X Any even integer = An even Integer
Odd Integer X Odd Integer = Odd Integer
Even Integer X Even Integer = Even Integer

Even Integer + Odd Integer = Even Integer
Odd Integer + Odd Integer = Even Integer
Even Integer + Even Integer = Even Integer

Every number we get is an even number ( Divisible by 2 completely)
Let’s check what we get when we add any odd number in an even number

2 + 1 = 3
2 + 5 = 7
2 + 3 = 5

Hence it is seen that any positive odd integer is of the form 6q+1or, 6q+3 or 6q+5.
3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? 

Solution:
616 = 32 X 19 + 8
32 = 8 X 4 + 0
8 Columns

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. 

Solution:
Let “x” be an integer 3q, 3q+1, 3q+2
for q a positive integer
x = 3X1=3
x = 3X1+1 = 4
x = 3X1+2 = 5
Square of x = 9,16,25
Square of x can be written as = (3X3), (3X5+1), (3X8+1)
We can conclude that square of any positive integer can be written as 3m or 3m+1.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m,9m+1or9m+8.

Solution:

If “n” is a positive integer it’s cube will be n^3
Euclid’s division lemma says
n = qb+r
for r = 0 & b = 9
n^3 = (q + 0)^3
= 729q^3
= 9 (81q^3)
Here m = 81q^3
for r = 1 & b = 9
n^3 = (9q + 1)^3
= 729q^3 + 1 + 27q(9q+1)
= 729q^3 + 1 + 729 q^2 + 27 q
= (729q^3 + 729 q^2 + 27 q ) + 1
= 9(81q^3+81q^2+3q) + 1
Here m = 81q^3+81q^2+3q
for r = 2 & b = 9
n^3 = ( 9q+2)^3
= (9q)^3+ 8+ 54q(9q+2)
=(9q)^3+8+486q^2+108q
=9(81q^3+54q^2+12q)+8
Here m = (81q^3+54q^2+12q)

NCERT Book
CBSE Board
Class Xth
Chapter 01- Part 02

Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.

Method 01

q can be any positive integer we will check the given conditions for a sequence of possible q values
suppose q = 1 or 2 or 3 or 4 or 5 ………..
As said
For even numbers

2 = 2 X 1 + 0
4 = 2 X 2 + 0
6 = 2 X 3 + 0
998 = 2 X 499 + 0
It is clearly visible that every even number can be represented as multiple of 2 only i.e.2q

For odd numbers

1 = 2 X 0 + 1
3 = 2 X 1 + 1
5 = 2 X 2 + 1
11111 = 2 X 5555 + 1
So we can also say that every odd number can be presented in 2q + 1 form

Method 02

Euclid’s Division Lemma

a = bq+r

it is given that value of b is 2
r must be greater or equal to 0 and
smaller than b so
r is either 1 or 0

so for any even number/integer a must be 2q since even number is completely divisible by 2 like 28 = 2 X 14, 8 = 2X 4 so on
and for any odd integer r = 1
since a can not be completely divisible by 2
like 3 = 2X1+1
5 = 2X2+1
So every odd number can be represented as 2q+1
and every even number can represented as 2q