27823: An Egg Drop Number

Back on the 7th of May 2021, I made a post titled The Egg Drop Numbers in which I remarked:

On June 6th, 2025 (the anniversary of D-Day by the way), I will enjoy the next egg drop number, $\mathbf{\text{27823}}$, which is the maximum number of floors for which someone, armed with 9 eggs, will be able to determine the critical floor with at most 15 trials.

Today is that day and so here is a post regarding the number created using Gemini 2.5 Pro and the following prompt:

Can you explain the mathematics behind the following statement: "the egg drop number, $\mathbf{\text{27823}}$, is the maximum number of floors for which someone, armed with 9 eggs, will be able to determine the critical floor with at most 15 trials." This has to do the famous egg breaking problem but can your explanation focus specifically on the $\mathbf{\text{27823}}$.

Here is Gemini's response: 

$\mathbf{\text{The Mathematics of the Egg Dropping Puzzle: The Case of 27,823}}$

The solution to the famous egg dropping puzzle, when determining the maximum number of floors that can be tested, can be found using combinatorial mathematics, specifically with binomial coefficients.

$\mathbf{\text{The General Formula}}$

The maximum number of floors, $N$, that can be determined with a given number of eggs, $e$, and a maximum number of trials, $t$, is given by the following summation:$$N(e,t)=\sum _{i=1}^e(\genfrac{}{}{0ex}{}{t}{i})$$In this formula, the term $(\genfrac{}{}{0ex}{}{t}{i})$ represents the binomial coefficient "t choose i", which calculates the number of unique ways to choose $i$ items from a larger set of $t$ items. The formula for the binomial coefficient is:$$(\genfrac{}{}{0ex}{}{t}{i})=\frac{t!}{i!(t-i)!}$$Each term in the summation represents the number of possible outcomes for a given number of egg breaks within the allowed trials.

$\mathbf{\text{Calculating for 9 Eggs and 15 Trials}}$

We are given the specific parameters:

• Number of eggs, $e=9$
• Maximum number of trials, $t=15$

Substituting these values into the general formula, we get:

$$N(9,15)=\sum _{i=1}^9(\genfrac{}{}{0ex}{}{15}{i})$$This expands to the following sum:$$\begin{array}{rl}N(9,15)=& (\genfrac{}{}{0ex}{}{15}{1})+(\genfrac{}{}{0ex}{}{15}{2})+(\genfrac{}{}{0ex}{}{15}{3})+(\genfrac{}{}{0ex}{}{15}{4})+(\genfrac{}{}{0ex}{}{15}{5})+\\ & (\genfrac{}{}{0ex}{}{15}{6})+(\genfrac{}{}{0ex}{}{15}{7})+(\genfrac{}{}{0ex}{}{15}{8})+(\genfrac{}{}{0ex}{}{15}{9})\end{array}$$Next, we calculate the value of each individual binomial coefficient:

• ${\displaystyle (\genfrac{}{}{0ex}{}{15}{1})=15}$
• ${\displaystyle (\genfrac{}{}{0ex}{}{15}{2})=\frac{15!}{2!13!}=105}$
• ${\displaystyle (\genfrac{}{}{0ex}{}{15}{3})=\frac{15!}{3!12!}=455}$
•  ${\displaystyle (\genfrac{}{}{0ex}{}{15}{4})=\frac{15!}{4!11!}=1,365}$
• ${\displaystyle (\genfrac{}{}{0ex}{}{15}{5})=\frac{15!}{5!10!}=3,003}$
• ${\displaystyle (\genfrac{}{}{0ex}{}{15}{6})=\frac{15!}{6!9!}=5,005}$
• ${\displaystyle (\genfrac{}{}{0ex}{}{15}{7})=\frac{15!}{7!8!}=6,435}$
• ${\displaystyle (\genfrac{}{}{0ex}{}{15}{8})=\frac{15!}{8!7!}=6,435}$
•  ${\displaystyle (\genfrac{}{}{0ex}{}{15}{9})=\frac{15!}{9!6!}=5,005}$

Finally, we sum these values to find the total number of floors:$$15+105+455+1,365+3,003+5,005+6,435+6,435+5,005=27,823$$Thus, the maximum number of floors for which one can determine the critical floor with 9 eggs and at most 15 trials is precisely $\mathbf{\text{27,823}}$.