Difference between revisions of "2024 SSMO Relay Round 5 Problems/Problem 1"
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} | ||
Revision as of 14:58, 10 September 2025
Problem
Let the super factorial be defined on positive integers as
Find the largest positive integer
such that that there are exactly
positive integers
such that
has fewer than
trailing zeroes.
Solution
Let be the number of trailing zeroes
has and
denote the number of trailing zeroes
has. Clearly,
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}
\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}
It is easy to see that for exactly
integers such that
has fewer than
trailing zeroes. Thus, the answer is
.
~SMO_Team