(i) 210, 55
Step 1. Given integers are 210 and 55
such as 210 > 55
by Euclid division lemma
210 = 55 × 3 + 45 …(i)
Step 2. Since remainder 45 ≠ 0.
so for divisior 55 and remainder 45
by Euclid division lemma
55 = 45 × 1 + 10 …(ii)
Remainder ≠ 0
Step 3. For divisior 45 and remainder 10
by Euclid division lemma
45 = 10 × 4 + 5 …(iii)
remainder ≠ 0
Step 4. For new divisior 10 and remainder 5 by
Euclid division lemma
10 = 5 × 2 + 0 …(iv)
Hence, remainder = 0
So, HCF of 210, 55 is 5.
(ii) 420, 130
Step 1. For given integer 420 and 130.
By Euclid division lemma
420 = 130 × 3 + 30 …(i)
Step 2. Here remainder is not zero.
So for divisior 130 and remainder 30
By Euclid division lemma
130 = 30 × 4 + 10 …(ii)
Step 3. Here remainder is not zero.
So for divisior 30 and remiander 10
By Euclid division lemma
30 = 10 × 3 + 0 …(iii)
Here, remainder is zero.
Hence, HCF of 420 and 130 is 10.
(iii) 75, 243
Step 1. For given integer 75 and 243
243 > 75
By Euclid division lemma
243 = 75 × 3 + 18 …(i)
Step 2. Here remainder ≠ 0
So, using Euclid division lemma
for divisior 75 and remainder 18.
75 = 18 × 4 + 3 …(ii)
Step 3. Here remainder ≠ 0
for divisior 18 and remainder 3
18 = 3 × 6 + 0 …(iii)
Since, remainder = 0
Hence, H.C.F. of 75 and 243 = 3
(iv) 135, 225
Step 1. For given integer 135 and 225
225 > 135
By Euclid division lemma
225 = 135 × 1 + 90
Step 2. Here remainder is not zero.
So, for divisior 135 and remainder 90.
135 = 90 × 1 + 45
Step 3. Here remainder is not zero.
So for divisor 18 and remainder 3.
90 = 45 × 2 + 0
remainder = 0
Hence HCF of 135 and 225 = 45
(v) 196, 38220
Step 1. For a given integer 196 and 38220
38220 > 196
Step 2. Using Euclid division lemma
38220 = 196 × 195 + 0
Step 3. Since remainder = 0 and divisor = 196
Hence, HCF of 196 and 38220 = 196
(vi) 867, 255.
Step 1. For a given integer 867 and 255
867 > 255
Step 2. Using Euclid division lemma
867 = 255 × 3 + 102
Here, remainder ≠ 0
Divisor = 255, and remainder = 102
Step 3. Again, for divisor 255
and remainder 102 using Euclid Lemma.
255 = 102 × 2 + 51
Step 4. Here, remainder = 51 ≠ 0
again by using Euclid Divisor Lamma,
for divisor 102 and remainder 51.
Step 5. Here, the remainder is zero.
Hence, HCF (687, 255) = 51
