If m = 3^n, what is the greatest value of n for which m is a

Let's talk strategy here. Many explanations on this forum focus blindly on the math. While that isn't necessarily bad, remember: the GMAT is a critical-thinking test. While there might be equations you could memorize that give you solutions for problems like this, those equations break down as soon as the GMAT words the question a little bit differently. For those of you studying for the GMAT, you will want to internalize strategies that actually minimize the amount of math that needs to be done, making it easier to manage your time. The tactics I will show you here will be useful for numerous questions, not just this one. My solution is going to walk through not just what the answer is, but how to strategically think about it. Ready? Here is the full "GMAT Jujitsu" for this question:

The Approach

Questions like this are actually designed to test your ability to recognize patterns behind otherwise messy math. \(25!\) is a big number. Obviously, there is no way you are going to multiply it out. The key is to focus on exactly what the question is asking, looking for logical leverage to strategically attack the question. In this case, the problem asks, "if \(m = 3^n\), what is the greatest value of \(n\) for which \(m\) is a factor of \(25!\)?" In other words, "how many times could we factor a \(3\) out of \(25\) factorial?"

Of course, a possible trap would be to just think of the multiples of \(3\)s in the factorial:

\(25*\) \((24)\) \(*23*22*\) \((21)\) \(*20*19*\) \((18)\) \(*17*16*\) \((15)\) \(*14*13*\) \((12)\) \(*11*10*\) \((9)\) \(*8*7*\) \((6)\) \(*5*4*\) \((3)\) \(*2*1\)

There are eight multiples of \(3\) in \(25!\), but this is a classic trap (and, incidentally, the most-commonly picked wrong answer.) The way around this bad logic is to realize that the number of "multiples of \(3\)" embedded in a factorial isn't necessarily the same thing as the number of \(3\)s. Some numbers contain more than one factor of \(3\). For example, \(18 = 2*3*3\). \(18\) can contribute two \(3\)s to the total amount. Since the question asks us for the maximum number of \(3\)s we could factor out of \(25!\), we need to look for all the possible \(3\)s. \(9\), of course, contains two \(3\)s (\(9=3*3\)). \(18\) also contains two \(3\)s (\(18=2*3*3\)).
\[
\begin{matrix}
\text{Factor:} & \text{3} & \text{6} & \text{9} & \text{12} & \text{15} & \text{18} & \text{21} & \text{24} \\ \hline
\text{# of 3s} & \text{1} & \text{1} & \text{2} & \text{1}& \text{1}& \text{2}& \text{1}& \text{1} & \text{= 10}
\end{matrix}
\]
Thus, \(25!\) contains \(10\) total \(3\)s, and the answer is B.

A Hypothetical

Some other answers in this forum have suggested a formula to figure this out: \(\frac{25}{3}+\frac{25}{3^2}=8+2=10\) (taking only quotients into account.) While this equation technically works for this particular problem, such a formula completely implodes if the problem were to ask, "if \(m = 4^n\), what is the greatest value of \(n\)...?" The formula would give the solution: \(\frac{25}{4}+\frac{25}{4^2}=6+1=7\). But this is not even close to the right answer. Since \(4=2*2\), a factor containing a single \(2\) can still make a difference, even if it doesn't contain a \(4\). The number of \(2\)s in this hypothetical question would be:
\[
\begin{matrix}
\text{Factor:} & \text{2} & \text{4} & \text{6} & \text{8} & \text{10} & \text{12} & \text{14} & \text{16} & \text{18} & \text{20} & \text{22} & \text{24} \\ \hline
\text{# of 2s} & \text{1} & \text{2} & \text{1} & \text{3}& \text{1}& \text{2}& \text{1}& \text{4}& \text{1}& \text{2}& \text{1}& \text{3} & \text{= 22}
\end{matrix}
\]
With \(22\) possible \(2\)s embedded in \(25!\), that would mean we could extract \(11\) possible \(4\)s... much more than the "equation" calculated.

Takeaways

Now, let’s look back at this problem through the lens of strategy. Your job as you study for the GMAT isn't to memorize the solutions to specific questions; it is to internalize strategic patterns that allow you to solve large numbers of questions. This problem can teach us patterns seen throughout the GMAT. The primary pattern this problem demonstrates is "Mathugliness" -- whereby the GMAT tries to "flex on you," using obnoxious or repetitive math to make the problem seem harder than it actually is. When you encounter Mathugliness, get excited. Look for leverage in the problem that will allow you to solve the problem conceptually, instead of working out all the math. You just need to look at how the problem is structured and determine what would limit or define the value you are looking for. Determine what those limits are, and you have your answer. It is much safer to look for patterns in how factors are distributed, instead of blindly trusting in an equation. Make a table, organize your thoughts, and focus on what the question is asking. The GMAT is notorious for crafting questions in a way that trips up students who blindly rely on "tricks." (Incidentally, can you see the pattern in the table above for the "# of 2s"? Patterns turn "inefficient" math into great critical-thinking opportunities.) And that is thinking like the GMAT.