find the HCF of 52 and 117 and express it in form 52x+117y??
(class 10 CH 1)
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Since 117>52
117 = 52 × 2 + 13
52 = 13 × 4 + 0
Therefore HCF (52, 117) = 13
Now, 13 = 52×(-2) + 117×1
13 = 52x + 117y where X = -2 and y = 1