# NCERT Class 10 Solutions: Real Numbers (Chapter 1) Exercise 1.1 – Part 1

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Q-1 Use Euclid’s division algorithm to find the HCF of:

1. 135 and 225

2. 196 and 38220

3. 867 and 255

Solution:

1. 135 and 225

• Here

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.

2. 196 and 38220

• here

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

3. 867 and 255

• Here

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

Q-2 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

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