If n is a positive integer, and (45*14*7^n - 15*7^(n - 1)*54)^(1/2) is

Official Solution:


If \(n\) is a positive integer, and \(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}\) is NOT an integer, what is the value of \(n\)?

Simplify given expression:

\(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}=\)

\(=\sqrt{2*3^2*5*7^2*7^{(n-1)} - 2*3^4*5*7^{(n - 1)}}=\)

\(=\sqrt{2*3^2*5*7^{(n-1)}(7^2 - 3^2)}=\)

\(=\sqrt{2*3^2*5*7^{(n-1)}(2^3*5)}=\)

\(=\sqrt{2^4*3^2*5^2*7^{(n-1)}}=\)

\(=2^2*3*5*\sqrt{7^{(n-1)}}\)

So, we are given that \(2^2*3*5\sqrt{7^{(n-1)}}\) is NOT an integer.

Notice that since \(n\) is a positive integer, then \(7^{(n-1)}\) is also a positive integer.

Now, the square root of any positive integer is either an integer or an irrational number. Meaning that, \(\sqrt{positive \ integer}\) cannot be a fraction, for example it cannot equal to \(\frac{1}{2}, \ \frac{3}{7}, \ \frac{19}{2}, \ \frac{1}{60}\) ... It MUST be an integer (1, 2, 3, ...) or an irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{7}\), ...). Thus, since \(7^{(n-1)}\) is a positive integer then \(\sqrt{7^{(n-1)}}\) is either an integer itself or an irrational number.

Therefore, for \(2^2*3*5*\sqrt{7^{(n-1)}}\) NOT to be an integer, \(\sqrt{7^{(n-1)}}\) must NOT be an integer. \(\sqrt{7^{(n-1)}}\) is NOT an integer for (positive) EVEN values of \(n\).

So, from above, \(n\) is a positive even integer: 2, 4, 6, 8, ...

(1) \(n\) is a prime number

There is only one even prime, namely 2. So, \(n = 2\). Sufficient.

(2) \(n < 4\)

There is only one positive even number less than 4, namely 2. So, \(n = 2\). Sufficient.


Answer: D