arithmetic
posted by Loyd .
Find the smallest positive integer P such that the cube root of 400 times P is an integer.

This sounds awkward, but is actually easy. Consider the number 400P. It must end in zero  actually, it must end in 00, since it has 100 as a factor.
Therefore its cube root must end in zero. Check this for yourself: the last digit of a cube can be determined by the last digit of the cube root. 1x1x1 = 2, 2x2x2=8 and so on, regardless of the higher digits.
So for the cube root we're already down to factors of 10  we only have to look at 10, 20, 30... and their cubes all end in 000 because of the three 10s in the factors.
Does 10 work? No, 1000 is a cube of 10, but 400 doesn't divide evenly.
Now try 20. 
20 * 20 = 400, so 20 is only the square root of 400, not the cube root.
Unfortunately, I can only tell you that 20 is not the answer, but I do not know the answer. Sorry. 
400 = 2x2x2 x 2 x 5x5
every triplet of "same factors" will produce a cube
so to make the right side of the above statement into a perfect cube, we need another 2x2x5 or 20
So let's multiply both sides by 20
400x20 = 2x2x2 x 2x2x2 x5x5x5
400(20) = 8000
400P = 8000
so P = 20 
I think the easiest is to say the
cuberoot(400P)
=cuberoot(2x2x2x2x5x5xP)
=2cuberoot(2x5x5xP)
for cuberoot(2x5x5xP) to be an integer
then P=2x2x5=20
so why is 20 not the answer?