find the HCF by euclids division algorithm of the numbers 92690 7378 and 7161.
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For every point of integers a and b there exist unique integer q and r such that a = bq + r
where 0 ≤ r < b
So here and a > b
a = 92690 And b = 7378 ,So that
92690 = 7378 × 13 + 4154
7378 = 4154 × 1 + 3224
4154 = 3224 × 1 + 930
3224 = 930 × 3 + 434
934 = 434 × 2 + 62
434 = 62 × 7 + 0
Here r = 0 So H.C.F. of 92690 and 7378 is 62
Now apply Euclid division lemma on 62 and 7161
Here a = 7161 and b = 62 ,So that a> b
7161 = 62 × 115 + 31
62 = 31 × 2 + 0
Here r = 0 , So h.C.F. of 62 and 7161 is 31 .
∴ H.C.F. of 92690 , 7378 and 7161 is = 31
(Very tiring question)
( Not from exam point of view)
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