This took me way longer than 5 minutes to figure this one out. Very tricky!

So \(n\) is a perfect cube, a perfect square, and is an integer.

\(n\) can therefore be expressed in the following ways:

\(n^{\frac{x}{2}}\) must be an integer.

\(n^{\frac{x}{3}}\) must be an integer.

What this boils down to is:

\(\frac{x}{2}\) is an integer.

\(\frac{x}{3}\) is an integer.

If \(\frac{x}{2}\) and \(\frac{x}{3}\) aren't integers, then \(n\) can't be an integer (unless \(n = 1\), but we know that's not the case here).

Example:
n = 20, x = 2

\(20^1\)
\(20^{\frac{2}{3}}\)

Since both have to be satisfied, \(x = 2\) doesn't work.

So what is the least number \(x\) can equal in order to be divisible by 2 and 3? Their \(LCM\) of \(6\).

\(n^{\frac{6}{2}}\)
\(n^{\frac{6}{3}}\)

Simplifying:

\(n^{3}\)
\(n^{2}\)

This works! So if \(n\) is a number raised to the power of 6, \(n\) remains an integer after square rooting and cube rooting.

So, in order for n to be a perfect cube and a perfect square, n, at a minimum, must have a power of 6.

In other words, \(n = y^6\), where y is some arbitrary integer.


Now \(n\) is also divisible by 20. That means that:

\(n = 2\) * \(2\) * \(5\) * \(k\)

where \(k\) is an integer.

We know that \(n\) has a power of 6, so \(k\) must account for that power of 6. We have a 2 and 5 as the prime factors, and those together equal 10. If k = \(10^5\), then that would account for that power of 6 in \(n\):

\(n\) = \(2\) * \(5\) * \(2\) * \(k\)
\(n\) = \(10\) * \(2\) * \(10^5\)
\(n\) = \(10^6\) * \(2\)

We know that \(10^6\) is divisible by 20, so that satisfies the divisibility by 20 restriction.

Since we want the smallest possible integer \(n\), we can ignore the 2, meaning that \(10^6\) is our answer.

How many digits does \(10^6\) have?

1,000,000 = 7 digits.

The answer is E


To double check, take the square root and the cube root of 1,000,000, and you'll find they're both integers! 1,000,000 is also divisible by 20, so that wraps up the question restrictions.