Question

Using Euclid's division algorithm find the HCF of the following numbers-
a.135 and 225
b.196 and 38220
c.280 and 12
d.867 and 254
e.288 and 120
f.867 and 255

Answer 1

a. 135 and 225

Clearly, 225 > 135

Applying the Euclid's division lemma to 225 and 135, we get

225 = 135 x 1 + 90

Since the remainder 90 ≠ 0, we apply the Euclid's division lemma to divisor 135 and remainder 90 to get

135 = 90 x 1 + 45

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

90 = 45 x 2 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 45 is the HCF of 225 and 135.

b. 196 and 38220

Clearly, 38220 > 196

Applying the Euclid's division lemma to 38220 and 196, we get

38220 = 196 x 195 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 196 is the HCF of 38220 and 196.

c. 280 and 12

Clearly, 280 > 12

Applying the Euclid's division lemma to 280 and 12, we get

280 = 12 x 23 + 4

Since the remainder 4 ≠ 0, we apply the Euclid's division lemma to divisor 12 and remainder 4 to get

12 = 4 x 3 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 4 is the HCF of 280 and 12.

d. 867 and 254

Clearly, 867 > 254

Applying the Euclid's division lemma to 867 and 254, we get

867 = 254 x 3 + 105

Since the remainder 105 ≠ 0, we apply the Euclid's division lemma to divisor 254 and remainder 105 to get

254 = 105 x 2 + 44

We consider the new divisor 105 and remainder 44 and apply the division lemma to get

105 = 44 x 2 + 17

We consider the new divisor 44 and remainder 17 and apply the division lemma to get

44 = 17 x 34 + 10

We consider the new divisor 17 and remainder 10 and apply the division lemma to get

17 = 10 x 1 + 7

We consider the new divisor 10 and remainder 7 and apply the division lemma to get

10 = 7 x 1 + 3

We consider the new divisor 7 and remainder 3 and apply the division lemma to get

7 = 3 x 2 + 1

We consider the new divisor 3 and remainder 1 and apply the division lemma to get

3 = 1 x 3 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 1 is the HCF of 867 and 254 .

e. 288 and 120

Clearly, 288 > 120

Applying the Euclid's division lemma to 288 and 120, we get

288 = 120 x 2 + 48

Since the remainder 48 ≠ 0, we apply the Euclid's division lemma to divisor 120 and remainder 48 to get

120 = 48 x 2 + 24

We consider the new divisor 48 and remainder 24 and apply the division lemma to get

48 = 24 x 2 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 24 is the HCF of 288 and 120.

f. 867 and 255

Clearly, 867 > 255

Applying the Euclid's division lemma to 867 and 255, we get

867 = 255 x 3 + 102

Since the remainder 102 ≠ 0, we apply the Euclid's division lemma to divisor 255 and remainder 102 to get

255 = 102 x 2 + 51

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

102 = 51 x 2 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 51 is the HCF of 867 and 255.

Answer 2

✰Question ✰

▪ Using Euclid's divisions algorithms find the HCF of the following numbers.

❪a❫ 135 and 225

❪b❫ 196 and 38220

❪c ❫ 280 and 12

❪d❫ 867 and 254

❪e❫ 288 and 120

❪f❫ 867 and 225

✰ Solution :-

❪a❫ 135 and 225

Since, 225 > 135 we apply the Euclid's divisions lemma to 225 and 135.

225 = 135 × 1 + 90.

since ,the remainder 90 ≠ 0 ,we apply the Euclid's division lemma to 135 and 90 to get

135 = 90× 1 + 45

we consider the new divisor 90 and remainder 45 and apply the division lemma we get,

90= 45× 2 = 0

here the remainder become zero .Since the divisor at stage is 45

∴ The HCF of 135 and 225 is 45.

❪b❫ 196 and 38220

Since, 38220 > 196 we apply the Euclid's divisions lemma to 38220 and 196.

38220=196 × 195 +0

Now,here the remainder is 0 .

So,the divisor at this stage is 195 .

∴ The HCF of 38220 and 196 is 195.

❪c❫ 280 and 12

Here, 280 > 12

So, by applying Euclid's divisions lemma .

280 = 12 × 23 +4

since ,the remainder 4 ≠ 0 ,we apply the Euclid's division lemma to 12 and 4 to get

12 = 4 × 3 = 0

∴The HCF of 280 and 12 is 4.

❪d❫ 867 and 254

Since, 867 > 254 we apply the Euclid's divisions lemma to 867 and 254

867=254 × 3 +105

since ,the remainder 105 ≠ 0 ,we apply the Euclid's division lemma to 254 and 105 to get

254 = 105 × 2 + 44

we consider the new divisor 105and remainder 44and apply the division lemma we get,

105 =44×2+17

we consider the new divisor 44 and remainder 17and apply the division lemma we get,

44=17× 2 +10

we consider the new divisor 17and remainder 10 and apply the division lemma we get,

17 =10×1+7

we consider the new divisor 10 and remainder 7 and apply the division lemma we get,

10=7 × 1 +3

we consider the new divisor 7 and remainder 3 and apply the division lemma we get,

7=3×2+1

we consider the new divisor 3 and remainder 1 and apply the division lemma we get,

3=3×1 +0

Here, the remainder at this stage is

∴The HCF of 867 and 254 is 1.

❪e❫ 288 and 120

Here, 228> 120

By applying Euclid's division lemma we get ,

228=120× 2 +48

since ,the remainder 48 ≠ 0 ,we apply the Euclid's division lemma to 120 and 48 to get

120= 48× 2 +24

we consider the new divisor 48 and remainder 24 and apply the division lemma we get,

48=24×2 +0

Here the remainder is 0

∴The HCF of 288 and 120 is 24.

❪f❫ 867 and 225

here, 867 > 225

By applying Euclid's division lemma we get

867 =225× 3 +102

since ,the remainder 102 ≠ 0 ,we apply the Euclid's division lemma to 225 and 102 to get

225 =102 × 2 +51

we consider the new divisor 102 and remainder 51 and apply the division lemma we get,

102=51×2 +0

Here, the remainder is 0

∴The HCF of 867 and 225 is 51.

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