Question

Use Euclid's division algorithm to find the HCF of
(iv) 184, 230 and 276
(v) 136, 170 and 255

Answer 1

SOLUTION :

iv )

Given : Three positive integers are 184,230 and 276.  

First we take two positive integers 184 & 230

Here,  230 > 184.

Let a = 230 and b= 184

230 = 184 × 1+ 46

[By applying division lemma, a = bq + r]

Here, remainder = 46  ≠ 0, so take new dividend as 184 and new divisor as 46

Let a = 184 and b= 46

184 = 46 × 4 + 0

Here, remainder is zero and divisor is 46.

Now , we again apply division lemma to find the H.C.F of 46 & 276

Let a = 276 and b= 46

276 = 46 × 6 + 0

Here, remainder is zero and divisor is 46.

Hence ,H.C.F. of 184,230 and 276 is 46.

(v) Given : Three positive integers are 136, 170 and 255.

First we take two positive integers 136 & 170

Here,  170 > 136.

Let a = 170 and b= 136

170 = 136 × 1 + 34

[By applying division lemma, a = bq + r]

Here, remainder = 34  ≠ 0, so take new dividend as 136 and new divisor as 34

Let a = 136 and b= 34

136 = 34 × 4 + 0

Here, remainder is zero and divisor is 34..

Now , we again apply division lemma to find the H.C.F of 34 & 255.

Let a = 255 and b= 34

255 = 34 × 7  + 17

Here, remainder = 17  ≠ 0, so take new dividend as 34 and new divisor as 17.

Let a = 34 and b= 17

34 = 17 × 2 + 0

Here, remainder is zero and divisor is 17.

Hence ,H.C.F. of 136, 170 and 255  is 17.

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