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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