Brilli's decagon hopping workout

Relevant wiki: Application of Divisibility Rules

If Brilli travels a number of spaces that is divisible by 10, then he will end up back where he started from, at vertex 1.

For his first 10 hops, he travels $10\times 1=10$ spaces. This brings him back to vertex 1.

For his next 10 hops, he travels $10\times 2=20$ spaces. 20 is divisible by 10, so this brings him back to vertex 1.

For his next 10 hops, he travels $10\times 3=30$ spaces. 30 is divisible by 10, so this brings him back to vertex 1.

For his next 10 hops, he travels $10\times 4=40$ spaces. 40 is divisible by 10, so this brings him back to vertex 1.

Because Brilli is making 10 hops each time, he will always be going a number of spaces that is divisible by 10.

At the end of Brilli's 100 hops, he will be back on vertex 1.

$10(1+2+...+10) (mod 10) = 0$ $\rightarrow$ vertex 1

roots of unity

After every complete rotation it will come on one only example if it is taking 1 step then after taking 10 steps it will come to one Similarly if it is taking 2 steps at a time then after ten steps again it will come to one Similarly after taking 10 steps 10 times it will come to 1 and stop

basically every set of hops he ends up back on space 1

However many numbers, n. It is always sum(1 to n) * 10, and therefore divisible by 10. Which is home.

hint: roots of unity

First, we know that we can multiply 10 times 1=10, 10 times 2=20, 10 times 3=30, and so on up to 10 times 10=100. If we divide each result by 10, we are left with 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. We can add these up quickly using the formula n(n+1)/2, which is used in determine the result of the arithmetic series 1+2+3+.....+(n-1)+n. Our result is 55.

Since it requires 11 jumps to get back to his starting spot (vertex 1), and 55/11=5 with no remainder, then he will be standing on vertex 1 when he completes his workout.

Although this problem can probably be solved with Roots of Unity, this solution is faster.