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 225>135

      We start with the larger number 225

      By Euclid Division Algorithm, we have

      225=1×135+90

      Since remainder 900 , we apply the division lemma to 135 and 90 to obtain

      135=90×1+45

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

      90=2×45+0

      Since the remainder is zero, the process stops

      Since the divisor at this stage is 45

      Here HCF(225,135)=HCF(135,90)=HCF(90,45)=45

      Therefore the HCF of 135 and 225 is 45.

  2. 196 and 38220

    • here 38220>196

      We start with the larger number 38220

      We apply the division lemma to 38220 and 196 to obtain

      38220=196×195+0

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

      Therefore 196=196×1+0

      Therefore HCF(38220,196)=196

  3. 867 and 255

  • Here 867>255

    We start with the larger number 867

    We apply the division lemma to 867 and 255 to obtain

    867=255×3+102

    Since remainder 1020 , we apply the division lemma to 255 and 102 to obtain

    255=102×2+51

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

    102=51×2+0

    Since the remainder is zero, the process stops.

    Therefore

    HCF(867,255)=HCF(255,102)=HCF(102,51)=51

Q-2 Show that any positive odd integer is of the form 6q+1 , or 6q+3 , or 6q+5 , where q is some integer

Solution:

Let a be any positive integer and b=6

Using Euclid division Algorithm, we have

a=bq+r{r0<b} (Equation 1)

Substituting b=6 in equation 1

a=6q+r Where r0<6

r=0,1,2,3,4,5

If r=0 , a=6q , 6q is divisible by 6

So, 6q is even

If r=1 , a=6q+1 , 6q+1 is not divisible by 2

If r=2 , a=6q+2 , 6q+2 is divisible by 2

So, 6q+2 is even

If r=3 , a=6q+3 , 6q+3 is not divisible by 2

If r=4 , a=6q+4 , 6q+4 is divisible by 2

So, 6q+4 is even

If r=5 , a=6q+5 , 6q+5 is not divisible by 2

Therefor the numbers 6q,6q+1,6q+2,6q+3,6q+4,6q+5 are either even or odd.

Here 6q,6q+2,6q+4 are even numbers

And 6q+1,6q+3,6q+5 are not exactly divisible by 2

Hence 6q+1,6q+3,6q+5 are odd numbers

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