Sally The Frog is Back (A Question About Her Life)!

This is what you need to know:

2014 = 19x53x2.

57 = 19x3.

gcd(10,57)=1.

The problem can be related to modular arithmetic:

$10x+57y \equiv z \mod(2014)$ .

We want to minimize x while making sure that z takes all the values that it can, i.e, z goes from 0 to 2013.

Let the vertex from which she starts jumping be vertex 0 and the last one (the first vertex counter clockwise to vertex 0) be vertex 2013. It can be seen pretty easily that if Sally starts jumping using steps of 57 alone, she can cover 106 distinct vertices before landing back at vertex 0. Here, it is equivalent to keeping x=0 and letting y go from 0 to 105. Note that all the vertices (z's values) that are multiples of 19 would have been covered in this journey.

Now that she is back at vertex 0, she needs to jump to another vertex of the polygon that is not a multiple of 19. And from there, she can once again cover another 106 different vertices (again, by covering 57 steps each jump). She has no option but to jump to vertex 10. And once she gets back to vertex 10, she can go to vertex 20 and do the same and move on to vertex 30 and so on. She only has to do this till she jumps to the 180th vertex. This is because she would have covered $106\times(18+1) = 2014$ distinct vertices in this manner. Note that jumping to vertex 190 would be jumping to a vertex whose number is a multiple of 19 and recall that all the vertices that are multiples of 19 were covered from when she started doing this process from vertex 0. Hence she only has to use 18 leaps of size 10 in order to cover all the vertices.