Question

1. Use Euclid's division algorithm to find the HCF of :
135 and 225
) 196 and 38220
(11) 867 and 255​

Answer 1

Answer:

45,196,51

Step-by-step explanation:

225 = 135 × 1 + 90

135 = 90 × 1 + 45

90 = 45 × 2 + 0

When remainder = 0 , Divisor = HCF

therefore, HCF = 45

38220 = 196 × 195 + 0

When remainder = 0 , Divisor = HCF

therefore, HCF = 196

867 = 255 × 3 + 102

255 = 102 × 2 + 51

102 = 51 × 2 + 0

When remainder = 0 , Divisor = HCF

therefore, HCF = 51

Done!

Answer 2

a. 135 and 225

Euclid's division lemma :

Let a and b be any two positive Integers .

Then there exist two unique whole numbers q and r such that

a = bq + r ,

0 ≤ r <b

Now ,

Clearly, 225 > 135

Start with a larger integer , that is 225.

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

Euclid's division lemma :

Let a and b be any two positive Integers .

Then there exist two unique whole numbers q and r such that

a = bq + r ,

0 ≤ r <b

Now ,

Clearly, 38220 > 196

Start with a larger integer , that is 38220.

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.

f. 867 and 255

Euclid's division lemma :

Let a and b be any two positive Integers .

Then there exist two unique whole numbers q and r such that

a = bq + r ,

0 ≤ r <b

Now ,

Clearly, 867 > 255

Start with a larger integer , that is 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.