NCERT Class 10 Solutions: Real Numbers (Chapter 1) Exercise 1.1 – Part 1
Q1 Use Euclid’s division algorithm to find the HCF of:

135 and 225

196 and 38220

867 and 255
Solution:

135 and 225

Here
We start with the larger number 225
By Euclid Division Algorithm, we have
Since remainder , we apply the division lemma to 135 and 90 to obtain
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
Since the remainder is zero, the process stops
Since the divisor at this stage is 45
Here
Therefore the HCF of 135 and 225 is 45.


196 and 38220

here
We start with the larger number 38220
We apply the division lemma to 38220 and 196 to obtain
We apply Euclid Division Algorithm of Divisor 196 and the remainder 0.
Therefore
Therefore


867 and 255

Here
We start with the larger number 867
We apply the division lemma to 867 and 255 to obtain
Since remainder , we apply the division lemma to 255 and 102 to obtain
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
Since the remainder is zero, the process stops.
Therefore
Q2 Show that any positive odd integer is of the form , or , or , where q is some integer
Solution:
Let a be any positive integer and
Using Euclid division Algorithm, we have
(Equation 1)
Substituting in equation 1
Where
If , , 6q is divisible by 6
So, 6q is even
If , , is not divisible by 2
If , , is divisible by 2
So, is even
If , , is not divisible by 2
If , , is divisible by 2
So, is even
If , , is not divisible by 2
Therefor the numbers are either even or odd.
Here are even numbers
And are not exactly divisible by 2
Hence are odd numbers