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45 votes
45 votes
Find the least number which when divisible by 20, 24, 32 and 38 leaves a remainder 5 in cach case



User Abolfazl Miadian
by
2.7k points

1 Answer

18 votes
18 votes

Answer:

18245

Explanation:

We have to use L.C.M,

L.C.M(20,24,32,38)

2|20,24,32,38

2| 10 ,12 ,16 ,19

2| 5 , 6 , 8 , 19

2| 5 , 3 , 4 , 19

2| 5 , 3 , 2 , 19

L.C.M = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 x 19

= 18240

Now for each case remainder is 5,

So the number is 18240+5

=> 18245

User Kary
by
2.9k points