Difference between revisions of "2024 SSMO Relay Round 5 Problems/Problem 1"

 
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Line 10: Line 10:
 
g(n)&0&0&0&0&1&1&1&1&1&2&2&2&2&2\\
 
g(n)&0&0&0&0&1&1&1&1&1&2&2&2&2&2\\
 
f(n)&0&0&0&0&1&2&3&4&5&7&9&11&13&15\\
 
f(n)&0&0&0&0&1&2&3&4&5&7&9&11&13&15\\
\end{array}
 
 
 
\begin{array}{ccccccccccccc}
 
  & 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & . & 0 \\
 
  & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & . & 0 \\
 
  &  & 1 & 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & . & 0 \\
 
  &  &  & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & . & 0 \\
 
  &  &  &  & 2 & 5 & 0 & 0 & 0 & 0 & 0 & . & 0 \\
 
  &  &  &  &  & 3 & 0 & 0 & 0 & 0 & 0 & . & 0 \\
 
  &  &  &  &  &  & 3 & 5 & 0 & 0 & 0 & . & 0 \\
 
  &  &  &  &  &  &  & 4 & 0 & 0 & 0 & . & 0 \\
 
  &  &  &  &  &  &  &  & 4 & 5 & 0 & . & 0 \\
 
+ &  &  &  &  &  &  &  &  & 5 & 0 & . & 0 \\
 
  & 6 & 1 & 7 & 2 & 8 & 3 & 9 & 5 & 0 & 5 & . & 5 \\
 
 
\end{array}
 
\end{array}
  

Latest revision as of 14:59, 10 September 2025

Problem

Let the super factorial $!(n)$ be defined on positive integers as $\prod_{i=1}^n i!.$ Find the largest positive integer $k$ such that that there are exactly $k$ positive integers $n$ such that $!(n)$ has fewer than $k$ trailing zeroes.

Solution

Let $f(n)$ be the number of trailing zeroes $!(n)$ has and $g(n)$ denote the number of trailing zeroes $n!$ has. Clearly, $f(n) = f(n-1)+g(n).$ Consider the following table:

\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} n&1&2&3&4&5&6&7&8&9&10&11&12&13&14\\\hline g(n)&0&0&0&0&1&1&1&1&1&2&2&2&2&2\\ f(n)&0&0&0&0&1&2&3&4&5&7&9&11&13&15\\ \end{array}

It is easy to see that for $k=12,$ exactly $12$ integers such that $!(n)$ has fewer than $12$ trailing zeroes. Thus, the answer is $\boxed{12}$.

~SMO_Team