Melody

The box starts with x  notes.   (the money value is never important)

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He takes     \(\frac{x}{3}+2 = \frac{x+6}{3}\)

and leaves  \(\frac{2x}{3}-2=\frac{2x-6}{3}\\\)

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from the ones left 

he takes  \(\frac{1}{2}*\frac{2x-6}{3}+2=\frac{x-3}{3}+\frac{6}{3}=\frac{x+3}{3}\\ \)

and leaves   \(\frac{x-3}{3}-\frac{6}{3}=\frac{x-9}{3}\)

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from the ones left 

he takes  \(\frac{1}{2}*\frac{x-9}{3}+2\)

and leaves ...

AND you should be able to take it from there

Huh?  I think we need more explanation.

There is no diagram  

Find how much the coins are worth altogether.   

So value of the nickels + value of the quarters

then

divide the answer by 40 because you have 40 coins.

Good question.   

Thanks Geno and  Ginger.

It is good to be reminded of Morgan Tud    

The queen misses those days and that scoundrel      

There might be an easier way to do this but this is what I have learnt. (I think from Heureka)  Thanks Heureka

How many integers n satisfy 0 < n < 60 and 4n=6 (mod 7)

4n=6  (mod 7)

\(4n=6\mod 7\\ 4n=7k+6\\ 4n=8k+4-k+2\\ n=(2k+1)+\frac{2-k}{4}\\ let \;\;a=\frac{2-k}{4}\\ \text{after rearranging }k=2-4a\\ \text{substituting back in}\\ n=2(2-4a)+1+a\\ n=5-7a\\ \text{n is between 0 and 60}\\ n=5,12,19,26,33,40,47,54 \)

So I found 8.

Since f(1)=32

and

f(0)=47

I think that the sum of the coefficients is  32-47 = -15

This is my logic

Since f(0)=47,    the constant must be 47    this is not a coefficient

Consider the polynomial to be

\(f(x)=ax^n+bx^{n-1}+cx^{n-2}+\dots+47\\ f(1)=a+b+c+\dots +47=32\\ so\\ a+b+c+\dots +47=32\\ a+b+c+\dots +(\text{coefficient of x)}=32-47\\\)

umm  maybe

A ray is the locus of collinear points radiating from a given start point.