Express the hcf of 65 and 117 in terms Of 65m+117n and find m and n
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firstly.. let's find the hcf of 65 & 117....
117 = 65×1+52
65 = 52×1+13
52= 13×4+0
the remainder at this stage is 0......
so, 13 is the hcf of 65 & 117.
117 = 65×1+52
65 = 52×1+13
52= 13×4+0
the remainder at this stage is 0......
so, 13 is the hcf of 65 & 117.
tamanna4946:
sorry I mistakenly submitted the unfinished answer..
remaining portion is....
13=65-52×1
13=65-(117-65×1)×1
13=65-117×1+65×1
13=2×65-117×1
65m+117n=65×2+117×(-1)
so
m=2
n=-1
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