GingerAle

Here’s a neat and orderly way to transcribe this as a power of (2):

\(\hspace {15em} \huge2 \large^ {\log_2\left(\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{3}{m}}\right)^3\right)}\\\)

GA 

--. .-

Fun fact: sum of powers of 2 starting from 2^0 up to 2^n = 2^n-1

It’s probably not fun if it’s not a fact.... But it’s definitely NOT a fact.

The formula for the sum of powers of (2) is \(2^{(n+1)}-1 \text { where (n) is an integer}\)

... now I get the trick (think about it).

OK, I’ve thought about it... 

Your formula (2^n – 1) gives the correct number of games played when (n=8), but how does the problem solver find the (8)? ...And after finding it, how does it help in solving the problem?

This is a binary question. There are 256 Players. Each game eliminates one player. How many players must be eliminated until only one remains? Answer: 255. Because 256 – 255 = 1

For this kind of question, the answer is always one less than the number of players. 

No; aces only becomes Faces by adding an “F”

Solution:

\(\text {52-card standard deck. }\\ \text {12 face cards}\\\)

\(\text {There are } \binom{52}{5}\text { ways to draw five cards.}\\ \text {There are }\binom{12}{5} \text { ways to draw five face cards.}\\ \, \Large P_{\small 5 \text { face cards}} = \dfrac {\binom{12}{5}}{\binom{52}{5}}= \dfrac {33}{108290} \approx 0.03047\%\)

GA

--. .-

See post #11, below.

Note that Geno’s data set plots y ≈ 0.0535715 for P(8). This value is very close to 0.05 for P(8) on Melody’s graph, so it’s probably correct. So, the answer to your question is, The polynomial at P(8) has y ≈ 0.0535715

This math problem demonstrates how extrapolating (infer by extending known information) a point from a polynomial equation created from a limited and narrow data set will have an error –sometimes a significant error –for the location of a nearby data point.  The reason is the created polynomial does not include data points for the change in the curve.

Using Melody’s Geobraga graph, overlay g(n)=((n)/(n^(2)-1)). This will show the points on the original plot compared to the polynomial based on the sample points. Note that the graphs align perfectly until after the seventh point, where they begin to diverge. By the eighth point, the polynomial graph shows a sudden divergence down toward the x-axis, instead of the asymptotic glide toward the x-axis. So, for the polynomial, at P(8) y ≈ 0.0535715 compared to y =8/63 (≈ 0.12698) for the original equation.

GA

--. .-

The determinate of a square matrix [A] is equal to zero if and only if the determinate of its reduced row echelon form [B] is equal to zero. It’s important to note that matrix [A] and reduced-form [B] do not have the same determinates, but they are zero at the same time. Because they are zero at the same time its undefined.

A graph of the \(y =\frac{n}{n^2-1}\) equation shows two asymptotes and a root of zero. The inverse of this equation \(y= \frac{n^2-1}{n}\)  will give two roots (1) and (-1) (the singularities of its inversion) and an asymptote at zero.  This suggests the fifth-degree polynomial equation that represents a finite portion of that equation has one real root and four complex roots.  Creating this polynomial equation by using (7) points of \(y =\frac{n}{n^2-1}\), and then extending (interpolating) it linearly yields a point that is notably different from the base equation.  That may be (one of) the point of the exercise –an example of catastrophe ...  

My attempt at using Gauss-Jordan elimination was a catastrophe...

I’m interested in the official AoPS solution. I’m sure it will be over my head, but it will give me something to look up to.... 

GA

--. .-

Here’s a faster solution method:

\(\large \text{half-life} = \large \dfrac{29} {Log_2(6272) – Log_2(196)} = 5.8 \text { minutes} \)

GA

--. .-