prove that hcf of 65 and 117 is of the form 65m+117n
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By Euclid's division algorithm
117 = 65x1 + 52.
65 = 52x1 + 13
52 = 13x4 + 0
Therefore 13 is the HCF (65, 117).
Now work backwards:
13 = 65 + 52x(-1)
13 = 65 + [117 + 65x(-1)]x(-1)
13 = 65x(2) + 117x(-1).
∴ m = 2 and n = -1.
117 = 65x1 + 52.
65 = 52x1 + 13
52 = 13x4 + 0
Therefore 13 is the HCF (65, 117).
Now work backwards:
13 = 65 + 52x(-1)
13 = 65 + [117 + 65x(-1)]x(-1)
13 = 65x(2) + 117x(-1).
∴ m = 2 and n = -1.
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