Euclid's division of 65and 117
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65=5×13
117=3×3×13
HCF=13
117=3×3×13
HCF=13
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Given integers are 65 and 117 such that 117>65 .
Applying division lemma , we get
117 = 65 × 1 + 52 .......(1)
65 = 52 × 1 + 13 .......(2)
52 = 13 × 4 + 0
At this stage , the remainder is 0 . So, the last divisor or non zero remainder at the earlier stage that is 13 is the HCF of 117 and 65 .
From (2) we have
13 = 65 -52 × 1
13 = 65-(117-65 ×1 )
13 = 65-117 + 65 × 1
13 = 65 × 2 + 117 (-1 )
13 = 65 - 117 - 65 × 1
13 = 65m + 117 n where m = 2 and n = -1
Applying division lemma , we get
117 = 65 × 1 + 52 .......(1)
65 = 52 × 1 + 13 .......(2)
52 = 13 × 4 + 0
At this stage , the remainder is 0 . So, the last divisor or non zero remainder at the earlier stage that is 13 is the HCF of 117 and 65 .
From (2) we have
13 = 65 -52 × 1
13 = 65-(117-65 ×1 )
13 = 65-117 + 65 × 1
13 = 65 × 2 + 117 (-1 )
13 = 65 - 117 - 65 × 1
13 = 65m + 117 n where m = 2 and n = -1
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