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:

Image describing euclid division algorithm

Algorithm for Euclid Division

Image describing euclid division algorithm

  1. 135 and 225

    • Here Equation

      We start with the larger number 225

      By Euclid Division Algorithm, we have

      Equation

      Since remainder Equation , we apply the division lemma to 135 and 90 to obtain

      Equation

      We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain

      Equation

      Since the remainder is zero, the process stops

      Since the divisor at this stage is 45

      Here Equation

      Therefore the HCF of 135 and 225 is 45.

  2. 196 and 38220

    • here Equation

      We start with the larger number 38220

      We apply the division lemma to 38220 and 196 to obtain

      Equation

      We apply Euclid Division Algorithm of Divisor 196 and the remainder 0.

      Therefore Equation

      Therefore Equation

  3. 867 and 255

  • Here Equation

    We start with the larger number 867

    We apply the division lemma to 867 and 255 to obtain

    Equation

    Since remainder Equation , we apply the division lemma to 255 and 102 to obtain

    Equation

    We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain

    Equation

    Since the remainder is zero, the process stops.

    Therefore

    Equation

Q-2 Show that any positive odd integer is of the form Equation , or Equation , or Equation , where q is some integer

Solution:

Let a be any positive integer and Equation

Using Euclid division Algorithm, we have

Equation (Equation 1)

Substituting Equation in equation 1

Equation Where Equation

Equation

If Equation , Equation , 6q is divisible by 6

So, 6q is even

If Equation , Equation , Equation is not divisible by 2

If Equation , Equation , Equation is divisible by 2

So, Equation is even

If Equation , Equation , Equation is not divisible by 2

If Equation , Equation , Equation is divisible by 2

So, Equation is even

If Equation , Equation , Equation is not divisible by 2

Therefor the numbers Equation are either even or odd.

Here Equation are even numbers

And Equation are not exactly divisible by 2

Hence Equation are odd numbers

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