Bunuel wrote:
What is the smallest positive integer n for which n!/9^9 is an integer?
A. 36
B. 39
C. 54
D. 78
E. 81
Official solution from Veritas Prep.
This is a number theory question – note the term “positive integer” and the fact that the question is essentially asking about the divisibility of \(n!\). As such, we will begin by prime factoring \(9^9\) as \((3^2)^{9}=3^{18}\). The real question is how far we have to go to find eighteen copies of 3.
The cop-out answer would be to simply take \(n=18∗3=54\); \(54!\) definitely contains at least eighteen factors of 3, since it contains \(3∗1, 3∗2, 3∗3, …, 3∗18\). However, \(54!\) actually contains far more than just the eighteen 3s, since several factors contain “bonus” 3s. For instance, \(54!\) contains 9, which is \(3^2\), and 27, which is \(3^3\), and even 54 itself, which is \(2∗3^3\). The correct answer must be less than 54.
At this point we could calculate the exact number of 3s in \(54!\) (it has twenty-six of them, as it turns out) and adjust from there, but it might be simpler to just try answers A and B, since one or the other must be correct.
\(36!\) contains twelve multiples of 3 (\(3∗1\) through \(3∗12\)), but it also contains four multiples of 9 (\(9∗1\) through \(9∗4\)), each of which provides an additional factor of 3, since \(9=3^2\). And \(36!\) Even contains one multiple of \(3^3=27\), which provides an “additional” additional 3. Taken together, \(36!\) contains \(12+4+1=17\) factors of 3, just short of our goal.
We can see at this point that \(39!\) provides exactly one more 3 than does \(36!\), since \(39!=39∗38∗37∗36!\).
Therefore, \(39!\) contains exactly 18 factors of 3, and answer B is correct.
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