Melody

What is the average of all positive integers that have four digits when written in base 3, but two digits when written in base 5? Write your answer in base 10.

hint:

4 digits base 3 go from      3^3 =?   to       (3^4)-1=?

2 digit base 5 go from        5^1 =?   to       (s^2)-1=?

where is the overlap? 

How do you find the average?

Thanks very much Alan,

I finally managed it with a different method.

\(\begin{array}{r@{~}c@{~}l l} 3x+5y-6z &=&2, & \textrm{(i)} \\ 5xy-10yz-6xz &=& -41, & \textrm{(ii)} \\ xyz&=&6. & \textrm{(iii)} \end{array} \)

Let 

\(a=3x \qquad\rightarrow \qquad x=\frac{a}{3}\\ b=5y \qquad\rightarrow \qquad y=\frac{b}{5}\\ c=-6z \qquad\rightarrow \qquad z=\frac{c}{-6}\\\)

The above equations become 

\(a+b+c=2\\ ab+bc+ac=-123\\ abc=-540\)

This gave me a hint:

If a,b and c are the roos of a monic cubic equations then we have.

Note: I used g because x has a different meaning in this question.

\(g^3-(a+b+c)g^2+(ab+bc+ac)g-(abcd)=0\\ g^3-(2)g^2+(-123)g-(-540)=0\\ g^3-2g^2-123g+540=0\\ \)

I used the wolfram|alpha calculator to factoize this and got

\((g-9)(g-5)(g+12)=0\)

so        a,b,c = -12,5,9     (any order)

These can be arranged in 3! = 6 ways so there will be 6 answers.

To show all the possible answers I made up a Excel table:

Alan thinks that there are 9 solutions but i don't know what the others could be ....

Notice that my last solution is the one that Alan found.  

I have forgotten how to do this.  

I would like to be reminded of the method though.  Any takers?

I think so too.   

I have not solved this but this is how I have been thinking.

I started with a circle of radius 1

The area is pi and the circumference is 2pi   so

\(Ratio=\frac{A}{P}=\frac{1}{2}\)

I inscribed a regular polygon in the circle as shown.  In this case, since it is a trangle, n will be 3, but the formulas are general

I worked out x using the cosine rule.  and I worked out that if the triangle is to have an area of pi (like the circle) the dilation factor is as given.

So now the equivalaent ratio is   \(R_2=\frac{\pi}{P}\)   

My intentions was to show that   P decreased as n increased hense   R₂  increases as n increases.

However it got to complicated and i couldn't make it work.    

Maybe I piced the wrong values as my starting point   idk  

I did not use your apothem idea.  Maybe that would have worked out better ....

On the way there

let the initial uphill bit be a metres

let the flat bit be b metres

and let the downhile bit be c metres

\(speed=\frac{distance}{time}\\ so\\ time=\frac{distance}{speed}\\\)

\(time = \frac{a}{100}+\frac{b}{120}+\frac{c}{150}\quad+ \quad \frac{c}{100}+\frac{b}{120}+\frac{a}{150}\\ time =\frac{1}{10}[ \frac{a+c}{10}+\frac{2b}{12}+\frac{c+a}{15}]\quad\\ time =\frac{1}{10}[ \frac{30(a+c)}{300}+\frac{50b}{300}+\frac{20(c+a)}{300}]\quad\\ time =\frac{1}{3000}[ 30a+30c+50b+20c+20a]\quad\\ time =\frac{1}{3000}[ 50a+50b+50c]\quad\\ time =\frac{1}{60}[ a+b+c]\quad\\\)

\(averaage \;Speed=\frac{distance}{time}\\ averaage \;Speed=\frac{2(a+b+c)}{\frac{1}{60}(a+b+c)}\\ averaage \;Speed=120 \;m/hour \)

LaTex:

time = \frac{a}{100}+\frac{b}{120}+\frac{c}{150}\quad+ \quad \frac{c}{100}+\frac{b}{120}+\frac{a}{150}\\
time =\frac{1}{10}[ \frac{a+c}{10}+\frac{2b}{12}+\frac{c+a}{15}]\quad\\
time =\frac{1}{10}[ \frac{30(a+c)}{300}+\frac{50b}{300}+\frac{20(c+a)}{300}]\quad\\
time =\frac{1}{3000}[ 30a+30c+50b+20c+20a]\quad\\
time =\frac{1}{3000}[ 50a+50b+50c]\quad\\
time =\frac{1}{60}[ a+b+c]\quad\\

why should you get extra credit for asking other people to do the work?

Seems to me that an F would be more appropriate.

It depends on many things.

Your pre knowledge

Your ability

Your memory

etc.

For me, I think it was my memory that let me down the most.

I could understand fairly easily i guess but I had to do hours of repetitious exerceises almost every night in order to cement the techniques/ methods/ formulas etc into my brain.

If you do not understand the pre-learning then a good private tutor could help enormously.

A,B and C represent the areas

\(\frac{C}{A+B+C}=\frac{2}{5}\qquad (1)\\~\\ B=\frac{3}{7}(A+B)\\ \frac{7B-3C}{3}=A \qquad (2)\\ sub \;\;2\;\; into \;\;1 \\ \frac{C}{\frac{7B-3C}{3}+B+C}=\frac{2}{5}\\ \frac{3C}{7B-3C+3B+3C}=\frac{2}{5}\\ \frac{3C}{10B}=\frac{2}{5}\\ \frac{C}{B}=\frac{20}{15}\\ \frac{C}{B}=\frac{4}{3}\\ \frac{B}{C}=\frac{3}{4}\\\)

LaTex:

\frac{C}{A+B+C}=\frac{2}{5}\qquad (1)\\~\\
B=\frac{3}{7}(A+B)\\
\frac{7B-3C}{3}=A  \qquad (2)\\
sub \;\;2\;\; into \;\;1 \\
\frac{C}{\frac{7B-3C}{3}+B+C}=\frac{2}{5}\\
\frac{3C}{7B-3C+3B+3C}=\frac{2}{5}\\
\frac{3C}{10B}=\frac{2}{5}\\
\frac{C}{B}=\frac{20}{15}\\
\frac{C}{B}=\frac{4}{3}\\
\frac{B}{C}=\frac{3}{4}\\

I m not sure what I am meant to do to show this.

To me it is just obvious.   

Anyway ......if you let

\(z_1=a(cos\theta +isin\theta )\\ z_2=a(cos\alpha +isin\alpha )\\ \text{midpoint will be average of the real parts + average of the imaginary part}\\ midpoint=\frac{a(cos\theta +\cos\alpha)}{2}+\frac{a(sin\theta +\sin\alpha)}{2}\\ midpoint=\frac{a(cos\theta +\cos\alpha)+a(sin\theta +\sin\alpha)}{2}\\ midpoint=\frac{z_1+z_2}{2}\\\)