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- TUJ2000 VOU7. Find the smallest 5 digit number which is divisible by 18, 24, and 32. |
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Answer» 2 I 18, 24, 32 L________ 3 I 9, 12, 16 L________ 3 I 3, 4, 16 L________ 2 I 1, 4, 16 L________ 2 I 1 , 2, 8 L________ 2 I 1 , 1, 4 L________ 2 I 1, 1, 2 L________ I 1, 1, 1 LCM= 2^{5} x 3^{2}=32x9=288 By using EUCLID DIVISION LEMMA,We get, a=bq+r 100,000=288(347)+64 100,000 - 64 = 99,936∴The largest 5-digit number that is divisible by 18,24 and 32 is 99,936 HOPE IT HELPS..! Read more on Brainly.in - https://brainly.in/question/2793943#readmore |
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