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

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

26

step-by-step explanation:

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl