Class 10 Mathematics - Chapter Real Numbers NCERT Solutions | Use Euclid’s division algorithm to

Welcome to the NCERT Solutions for Class 10th Mathematics - Chapter Real Numbers. This page offers a step-by-step solution to the specific question from Exercise 1, Question 1: use euclid rsquo s division algorithm to find the....
Question 1

Use Euclid’s division algorithm to find the HCF of :
   (i) 135 and 225      (ii) 196 and 38220      (iii) 867 and 255

Answer

(i) Here, we have to find H.C.F of 135 and 225

First divide divide the larger integer smaller integer

Since, 225 > 135

Therefore, by Euclid’s Division algorithm

225 = 135 × 1 + 90                                       (i)

Here 90 ≠ 0, so proceed the same procedure further

Again by E.D.L,              (E.D.L = Euclid’s division algorithm)

135 = 90 × 1 + 45                                         (ii)

As we know, 45 ≠ 0 therefore, again by E.D.L

90 = 45 × 2 + 0                                             (iii)

Here, r = 0 so we cannot proceed further. The divisor at this Stage is 45.

From (i), (ii) and (iii)

H.C.F (225, 135) = H.C.F (135, 90) = H.C.F (90, 45) = 45.

 

(ii) Here, we have to find H.C.F of 38220 and 196

First divide the larger integer smaller integer

Since, 3822 > 196

Therefore by Euclid’s Division Algorithm

38220 = 196 × 195 + 0

Here, r = 0 so we cannot proceed further. The divisor at this Stage is 196.

Hence, H.C.F (38220, 196) = 196.

 

(iii) Here, we have to find H.C.F of 867 and 255

First divide the larger integer smaller integer

Since, 867 > 255

Therefore, by Euclid’s Division algorithm

867 = 255 × 3 + 102                           (i)

Remainder 102 ≠ 0, so proceed the same procedure further using E.D.L

255 = 102 × 2 + 51                            (ii)

Here, 51 ≠ 0 again using E.D.L = 51 × 2

102 = 51 × 2 + 0                                (iii)

Here, r = 0 so we cannot proceed further. The divisor at this Stage is 51.

From (i), (ii) and (iii)

H.C.F (867, 255) = H.C.F (255, 102) = H.C.F (102, 51) = 51.

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