Question

Using eculids division find the HCF for 1218 & 1914

Answer 1

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, 1914 > 1218

Start with a larger integer , that is 1914.

Applying the Euclid's division lemma to 1914 and 1218, we get

1914 = 1218 × 1 + 696

Since the remainder 696 ≠ 0, we apply the Euclid's division lemma to divisor 1218 and remainder 696 to get

1218 = 696 × 1 + 522

We consider the new divisor 696 and remainder 522 and apply the division lemma to get

696 = 522 × 1 + 174

We consider the new divisor 522 and remainder 174 and apply the division lemma to get

522 = 174 × 3 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 174 is the HCF of 1914 and 1218.

Answer 2

To Find :-

• HCF of 1218 and 1914 .

Solution:-

Since 1914 >1218 ,we apply the division lemma to 1914 and 1218 to get

1914 = 1218 × 1 + 696

since the remainder 696 ≠ 0 ,we apply the division lemma to 1218 and 696 ,to get

1218 = 696 × 1 + 522

We consider the new divisor 696 and the remainder 522 and apply the division lemma to get

696= 522 × 1 + 174

We consider the new divisor 522 and the remainder 174 and apply the division lemma to get

522 = 174 × 3 +0

The remainder has now become 0 ,so our procedure stops. Since the divisor at this stage is 174 , the HCF of 1218 and 1914 is 174.

∴ 174 is the HCF of 1218 and 1914 .