Find the HCF of 52 and 117 and expressit in form 52x+117y
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Solution:-
By Euclid's Division Lemma 117 > 52
117 = (52 × 2) + 13 (52 is the divisor)
52 = 13 × 4 + 0 ; The division process ends here, as remainder is 0. So, HCF is 13 (Here, 13 is divisor)
13 can also be expressed as 52x + 117y i.e. as 52 (-2) + 117 (1)
By Euclid's Division Lemma 117 > 52
117 = (52 × 2) + 13 (52 is the divisor)
52 = 13 × 4 + 0 ; The division process ends here, as remainder is 0. So, HCF is 13 (Here, 13 is divisor)
13 can also be expressed as 52x + 117y i.e. as 52 (-2) + 117 (1)
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