Q7 of 56 Page 17

Find the H.C.F of 52 and 117 and express it in form 52x + 117y?

By Euclid's Division Lemma 117 > 52

If we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = b q + r where 0 ≤ r ≤ b. and q is the quotient and r is remainder


117 = (52 × 2) + 13


52 = 13 × 4 + 0


As remainder is 0


So, H.C.F is 13


⇒ 13 = (117× 1) – (52 × 2)


⇒ 13 = – (52 × 2) + (117 × 1)


⇒ 13 = 52 x+ 117y


and x = (– 2), y = 1


∴ 52x + 117y can be expressed as 52 (–2) + 117 (1)


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