Euclid's Division Lemma - Math with JM - home

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Euclid’s Division Lemma• Let a and b be any two positive integers.Thenthere exist unique integers q and r such thata=bq+r, 0≤r

Application of the Euclids DivisionLemma for H.C.F• Let 2 numbers be a and b.• Apply Euclids Division Lemma toa and b and obtain two wholenumbers q 1 and r 1 such thata = bq 1 +r 1 . 0≤r 1

Application of the Euclids DivisionLemma for H.C.F• If r 1 = 0, b is the H.C.F of a and b.• If r 1 ≠ 0 apply Euclids divisionlemma to b and r 1 and obtain 2whole numbers q 1 and r 2 suchthat b = q 2 r 1 +r 2

Application of the Euclids DivisionLemma for H.C.F• If r 2 =0 then r 1 is the H.C.F of a andb.• If r 2 ≠0 then apply euclids divisionlemma to r 1 and r 2 and continuethe steps till the remainder is zero.• The divisor at this stage is theH.C.F of a and b

Lets Practice-1Find HCF OF 455 and 42 by Euclid’s division algorithm455= 42 x 10 +3542 = 35 x 1 + 735 = 7 x 5 + 0Since the Remainder has become 0, and wecannot proceed further, therefore the HCF ofgiven numbers is the divisor in last step or theremainder of second last step i.e 7 (Answer)

Lets Practice-2•Using EuclidsDivision Lemma findH.C.F of 180670 and2937.

By Euclids division lemma we have180670= 2937 X 61 + 15132397 = 1513 X 1 + 14241513 = 1424 X 1 + 891424 = 89 X 16 + 0∴ HCF(180670, 2937) = HCF (2937, 1513)= HCF (1513, 1424)= HCF ( 1424, 89)= 89Hence HCF(180670, 2937) = 89

Class AssignmentUsing Euclid’s Division Algorithmfind HCF of1. 420 and 1302. 12576 and 4052Home AssignmentEx: 1.1 Question 1

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