Find the smallest positive integer n for which 126n cube root is a whole number
To find the smallest positive integer 'n' for which 126n's cube root is a whole number, we need to find the smallest value of 'n' that makes 126n a perfect cube.
Let's start by writing down the prime factorization of 126:
126 = 2^1 * 3^2 * 7^1
To make 126n a perfect cube, we need to make the exponents of each prime factor divisible by 3. Since the exponent of 2 is already 1, it needs to be divisible by 3, so 'n' must be divisible by 2^2 = 4.
Let's check the next prime factor, 3. The exponent of 3 is 2, and to make it divisible by 3, we need to increase it by 1, resulting in an exponent of 3.
Considering the final prime factor, 7, its exponent is already 1, which is not divisible by 3. Hence, we need to increase it to 3 to make it a perfect cube.
Therefore, the prime factorization of the smallest possible value of 126n is:
126n = 2^2 * 3^3 * 7^3
Now, we can find the value of 'n' by dividing the prime factorization by the original number. So, dividing 126n by 126, we get:
n = 2^2 * 3^3 * 7^3 / 126
Simplifying further, we get:
n = 2^2 * 3^2 * 7^2
Calculating the value, we have:
n = 4 * 9 * 49
n = 1764
Therefore, the smallest positive integer 'n' for which 126n's cube root is a whole number is 1764.
any perfect cube has 3 equal factors
e.g. 216 = 6 x 6 x 6
126n = 2 x 3 x 3 x 7 n
so to have triples, we need 2x2, 3, and 7x7
n = 4(3)(49) = 588
our number 126n would be 74088
check:
74088 = (2x2x2)(3x3x3)(7x7x7)