Find the smallest natural number n such that 4410 x n is a perfect cube.
if 4410*n = n³
then
4410 = n² = perfect square
but
4410=21*21*10 is not a perfect square, so
n does not exist.
To find the smallest natural number n such that 4410 x n is a perfect cube, we need to find a number whose cube is divisible by 4410.
First, let's factorize 4410: 4410 = 2 x 3 x 5 x 7^2.
Since we need the smallest natural number, n, we can start by considering the smallest prime number factors (2, 3, 5).
Let's start with n = 1:
4410 x 1 = 4410 = 2 x 3 x 5 x 7^2
The cube of the smallest prime number factors is:
(2 x 3 x 5)^3 = 2^3 x 3^3 x 5^3 = 8 x 27 x 125 = 21600
We see that 4410 is not divisible by 21600, so n = 1 is not the smallest natural number we are looking for.
Next, we can try with n = 2:
4410 x 2 = 8820 = 2^2 x 3 x 5 x 7 x 7
The cube of the smallest prime number factors is:
(2^2 x 3 x 5 x 7)^3 = 2^6 x 3^3 x 5^3 x 7^3 = 64 x 27 x 125 x 343 = 10912000
We see that 4410 is not divisible by 10912000, so n = 2 is not the smallest natural number we are looking for.
Continuing this process, we find that n = 10 is the smallest natural number such that 4410 x n is a perfect cube.
4410 x 10 = 44100 = 2^2 x 3^2 x 5^2 x 7^2
The cube of the smallest prime number factors is:
(2^2 x 3^2 x 5^2 x 7^2)^3 = 2^6 x 3^6 x 5^6 x 7^6 = 64 x 729 x 15625 x 117649 = 2214330268000
So, the smallest natural number n such that 4410 x n is a perfect cube is n = 10.