Question

HCF of 134791 and 6341 by euclid division Lemma please please please please solve this problem ​

Answer 1

Answer:

1

Step-by-step explanation:

Euclid's division lemma:

Two positive integers a and b, there exist whole numbers q and r satisfying a = bq + r, 0 ≤ r < b.

Now,

a = 134791 and b = 6341 {a > b}

(i) 134791 = 6341 * 21 + 1630 {Remainder ! = 0}

(ii) 6341 = 1630 * 3 + 1451 {Remainder ! = 0}

(iii) 1630 = 1451 * 1 + 179 {Remainder ! = 0}

(iv) 1451 = 179 * 8 + 19 {Remainder ! = 0}

(v) 179 = 19 * 9 + 8 {Remainder ! = 0}

(vi) 19 = 8 * 2 + 3 {Remainder ! = 0}

(vii) 8 = 3 * 2 + 2 {Remainder ! = 0}

(viii) 3 = 2 * 1 + 1 {Remainder ! = 0}

(ix) 2 = 1 * 2 + 0 {Remainder = 0}

∴ HCF of 134791 and 6341 is 1

Hope it helps!

Answer 2

Appply the division lemma

6341 = 6339 × 1 + 2

Since the remainder not equal to zero,

Apply lemma again

6339 = 2 × 3169 + 1

2 = 1 × 2 + 0

The remainder has now become zero.

HCF ( 6341 , 6339 ) = 1

Now we have to find HCF of 1 and 134791

134791 = 1 × 134791 + 0

HCF ( 1, 134791 ) = 1

Therefore ,

HCF ( 134791 , 6341 , 6339 ) = 1