Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$, $\left\lfloor\frac n5\right\rfloor$, and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$.
We need to find all numbers between $1$ and $600$ inclusive that are multiples of $4$, $5$, and/or $6$ which are also multiples of $4$, $5$, and/or $6$ when $1$ is added to them.
We begin by noting that the LCM of $4$, $5$, and $6$ is $60$.  We can therefore simplify the problem by finding all such numbers described above between $1$ and $60$ and multiplying the quantity of such numbers by $10$ ($600$/$60$ = $10$).
After making a simple list of the numbers between $1$ and $60$ and going through it, we see that the numbers meeting this condition are $4$, $5$, $15$, $24$, $35$, $44$, $54$, and $55$.  This gives us $8$ numbers.  $8$ * $10$ = $\boxed{080}$.

Solution 1.5
This is Solution 1 with a slick element included.  Solution 1 uses the concept that $60k+l$ is a solution for $n$ if $60k+l$ is a multiple of $3$, $4$, and/or $5$ and $60k+l+1$ is a multiple of $3$, $4$, and/or $5$ for positive integer values of $l$ and essentially any integer value of $k$.  But keeping the same conditions in mind for $k$ and $l$, we can also say that if $60k+l$ is a solution, then $60k-l-1$ is a solution!  Therefore, one doesn't have to go as far as determining the number of values between $1$ and $60$ and then multiplying by $10$.  One only has to determine the number of values between $1$ and $30$ and then multiply by $20$.  The values of $n$ that work between $1$ and $30$ are $4$, $5$, $15$, and $24$.  This gives us $4$ numbers.  $4$ * $20$ = $\boxed{080}$.

Note
Soon after the test was administered, a formal request was made to also accept $\boxed{081}$ as an answer and MAA decided to honor this request.  The gist of this request stated that the phrasing of the first part of the question could reasonably be interpreted to mean that one is given the condition to begin with that the integer is less than or equal to $600$.  In this case, if one was told that the values of $\left\lfloor \frac n4\right\rfloor$, $\left\lfloor\frac n5\right\rfloor$, and $\left\lfloor\frac n6\right\rfloor$ were $150$, $120$, and $100$ respectively, then the only possible choice for $n$ would be $600$ as $601$, $602$, and $603$ do not meet the condition as stated in the first part of the problem. If instead the problem asked for the numbers less than $600$ that met the second condition in the problem, there would have been one unique answer, $\boxed{080}$. ~burkinafaso ~sethl