Difference between revisions of "2024 SSMO Relay Round 5 Problems/Problem 1"
(Created page with "==Problem== Let the super factorial <math>!(n)</math> be defined on positive integers as <math>\prod_{i=1}^n i!.</math> Find the largest positive integer <math>k</math> such...") |
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==Solution== | ==Solution== | ||
| + | Let <math>f(n)</math> be the number of trailing zeroes <math>!(n)</math> has and <math>g(n)</math> denote the number of trailing zeroes <math>n!</math> has. Clearly, <math>f(n) = f(n-1)+g(n).</math> 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 <math>k=12,</math> exactly <math>12</math> integers such that <math>!(n)</math> has fewer than <math>12</math> trailing zeroes. Thus, the answer is <math>\boxed{12}</math>. | ||
| + | |||
| + | ~SMO_Team | ||
Latest revision as of 14:59, 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}
It is easy to see that for 
exactly 
integers such that has fewer than trailing zeroes. Thus, the answer is 
.
~SMO_Team
