kushal.adhia wrote:If x is prime and nx is both the square of an integer and the cube of an integer, where n is a positive integer, what is the greatest possible value of (1/nx)?

A. 1
B. 1/32
C. 1/64
D. 1/81
E. 1/729

How do u even go about solving this???

Kushal

The question asks for the greatest possible value of (1/nx).
Which can be rephrased as: What is the least possible value of (nx)?

Now x is a prime and nx is both the square of an integer and the cube of an integer, where n is a positive integer.
Thus the question becomes: What is the smallest positive integer which is both the square of an integer and the cube of an integer?

As x is prime and we have to minimize nx, x should be minimum.
Thus, x must be equal to 2. As nx is simultaneously square of an integer and the cube of an integer, nx must be of the form nx = (a^6)*(b^6)... , where a, b... are primes.

So minimum value of nx = 2^6 = 64.

Largest possible value of (1/nx) = 1/64.

The correct answer is C.

kushal.adhia wrote:

As nx is simultaneously square of an integer and the cube of an integer, nx must be of the form nx = (a^6)*(b^6)... , where a, b... are primes.

How do u get the 6 in the equation?

Thanks

Suppose n is an integer which is simultaneously square of an integer (Say, a) and the cube of an integer (Say, b), then n must be such that, n = a^2 and n = b^3. Thus, a^2 = b^3 => a = b^(3/2) and b = a^(2/3).

As a and b must be integer, b must be square (or any even power) of an integer. Otherwise b^(3/2) cannot be an integer. Also a must be cube (or any multiple power of 3) of an integer, otherwise a^(2/3) cannot be an integer. Therefore, we can write a and b as,

• (1) a = c^3m
  (2) b = d^2m

Where, m is an integer.

Now, n = a^2 = (c^3m)^2 = c^6m
Or in the same way, n = d^6m

Thus for n to be simultaneously square of an integer and the cube of an integer, n must be a power of 6 (or any multiple of 6) of an integer. As the question asks for minimum such value, the power is 6.

Hope this helps.