Question

using euclid division algorithm find weather. the pair of no 847 , and 2160 are co prime or not​

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, 2160 > 847

Start with a larger integer , that is 2160.

Applying the Euclid's division lemma to 2160 and 847, we get

2160 = 847 × 2 + 466

Since the remainder 466 ≠ 0, we apply the Euclid's division lemma to divisor 847 and remainder 466 to get

847 = 466 × 1 + 381

We consider the new divisor 466 and remainder 381 and apply the division lemma to get

466 = 381 × 1 + 85

We consider the new divisor 381 and remainder 85 and apply the division lemma to get

381 = 85 × 4 + 41

We consider the new divisor 85 and remainder 41 and apply the division lemma to get

85 = 41 × 2 + 3

We consider the new divisor 41 and remainder 3 and apply the division lemma to get

41 = 3 × 13 + 2

We consider the new divisor 3 and remainder 2 and apply the division lemma to get

3 = 2 × 1 + 1

We consider the new divisor 2 and remainder 1 and apply the division lemma to get

2 = 1 × 2 + 0

Now, the remainder at this stage is 0.

Co-primes are 2 numbers which have only one as a common factor.

So, the divisor at this stage, ie, 1 is the HCF of 2160 and 847.

Answer 2

Question:

Using Euclid division algorithm find whether the pair of number 847 and 2160 are co-prime or not.

Answer:

The pair of numbers 847 and 2160 are co prime.

Given:

The given pair of number : 847 and 2160 .

To Find:

Whether the pair of number 847 and 2160 are co-prime or not.

Solution:

We know,

Euclid's division lemma :

Let m and n be any two positive Integers .

Then there exist two unique whole numbers x and y such that

m = nx + y ,

0 ≤ y < n

Now ,

➛ Clearly, 2160 > 847

Start with a larger integer , that is 2160.

➛ Applying the Euclid's division lemma to 2160 and 847, we get

➡ 2160 = 847 × 2 + 466

➛Since the remainder 466 ≠ 0, we apply the Euclid's division lemma to divisor 847 and remainder 466 to get

➡847 = 466 × 1 + 381

➛ We consider the new divisor 466 and remainder 381 and apply the division lemma to get

➡ 466 = 381 × 1 + 85

➛We consider the new divisor 381 and remainder 85 and apply the division lemma to get

➡381 = 85 × 4 + 41

➛ We consider the new divisor 85 and remainder 41 and apply the division lemma to get

➡ 85 = 41 × 2 + 3

➛ We consider the new divisor 41 and remainder 3 and apply the division lemma to get

➡41 = 3 × 13 + 2

➛ We consider the new divisor 3 and remainder 2 and apply the division lemma to get

➡ 3 = 2 × 1 + 1

➛ We consider the new divisor 2 and remainder 1 and apply the division lemma to get

➡2 = 1 × 2 + 0

Now, the remainder at this stage is 0.

➛Co-primes are 2 numbers which have only one as a common factor.

➛ So, the divisor at this stage, ie, 1 is the HCF of 2160 and 847.

Hence, the HCF of numbers 847 and 2160 is 1 are co prime.

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OR

➪ 2160 = ( 847 × 2 ) + 466

➪ 847 = ( 466 × 1 ) + 381

➪ 466 = ( 381 × 1 ) + 85

➪ 381 = ( 85 × 4 ) + 41

➪ 85 = ( 41 × 2 ) + 3

➪41 = ( 3 × 13 ) + 2

➪ 3 = ( 2 × 1 ) + 1

➪ 2 = ( 1 × 2 ) + 0

As 1 is the H.C.F of 847 and 2160,

∴ 847 and 2160 are co-primes as they have only 1 as their H.C.F.