A stick is divided by red marks into 7 equal segments and by green marks into 13 equal segments. Then it is cut into 20 equal pieces. Prove that any piece (except the 2 end pieces) contain exactly one mark (which may be green or red)
My textbook previously introduces the concept of the mediant of 2 fractions and uses it to prove as follows:$$\text{every red mark is of the form }\frac{x}{7}$$ $$\text{every green mark is of the form }\frac{y}{13}$$ $$\operatorname{mediant}\left(\frac{x}{7},\frac{y}{13}\right) = \frac{x+y}{20}$$ so between any 2 green and red segments, there is a cut point and therefore, 2 segments of different colour cannot belong to the same piece.
Moreover, 2 segments of the same colour cannot belong to the same piece as both $\frac{1}{7}\text{ }$&$\text{ }\frac{1}{13}$ are greater than $\frac{1}{20}$
However, I want to know if there is another, simpler proof that does not require any knowledge of the mediant, especially if the denominator of the cut points is not equal to the sum of the denominators of the marks
Suppose the stick is $20(13)(7)$ long We have 20 pieces of length $13(7)$ with places {0,1,2,3,...,19} And 13 red pieces of length $20(7)$ And 7 green pieces of length $20(13)$
The first red will be on place 1 $(20=1(13)+7)$the second red will be on place 3 $(40=3(13)+1$ the third red will be on place 4 $floor(60/13)=4$ In general the $n$ red mark will be on the $[20(n)/13]$ place
For the green the $m$ green mark will be on the $[20(m)/7]$
Now you need to show the for $n=1\cdots 13$ and $m=1\cdots 7$
$$\left[\frac{20(n)}{13}\right]\ne \left[\frac{20(m)}{7}\right]$$