Find the smallest number which leaves the remainders 13 41 and 29 at the end when divided by 20 48 a...........

Quantitative Aptitude ( CCS) LCM, HCF

Find the smallest number which leaves the remainders 13, 41 and 29 at the end when divided by 20, 48 and 36 respectively

755

713

725

777

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Explanation:

Here least number is needed that means we need the Least Common Multiple i.e. LCM
We must now first find LCM of 20, 36 and 48
∴ LCM = 4 x 3 x 5 x 3 x 4 = 720
But this is not the answer because there are remainders as per the given condition.
If we observe closely, the difference between the given numbers and remainders is same
20 - 13 = 7; 48 - 1 = 7; 36 - 29 = 7
Difference is same = 7
So simply subtract this difference from LCM.
Number = 720 - 7 = 713

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