Question 1190223
<font face="Times New Roman" size="+2">


If 5700 men have rations for 66 days, then there must be a total of 5700 times 66 or 376200 rations available.


On the first day, 5700 rations are used, but on the second day, 20 less are used, and on the third day, 20 less than that.


The total rations used is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=0}^n\,5700-20i]


This is an arithmetic sequence where the first term is *[tex \Large 5700], the common difference is *[tex \Large -20], there are *[tex \Large n\ +\ 1] terms (zero start), and the last term is *[tex \Large 5700-20n].


Applying the formula for the sum of an arithmetic series:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=0}^n\,5700-20i\ =\ (5700\,-\,10n)(n\,+\,1)]


Set this equal to the known total rations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (5700\,-\,10n)(n\,+\,1)\ =\ 376200]


And solve the quadratic for the smaller root

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>