posted by Loyd on .
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
for cuberoot(2x5x5xP) to be an integer
so why is 20 not the answer?