Question

find the smallest 4 digit number divisible by both 72 and 120
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Answer 1

by euclids division lemma

120=72×1+48

r. is not =0. so again apply lemma

72 =48×1+24

r. is again not= 0

so again apply lemma

48=24×2+0

r= 0

so the smallest number is 24

hope it helps