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Rama09 [41]
3 years ago
8

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

Mathematics
1 answer:
Olegator [25]3 years ago
7 0

Answer:

18245

Step-by-step 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

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