# Combinatorics problem:circle hopscotch; the faster it's answered the better

Can anyone help me with the following scenario:

A hopping circuit is painted on a school playground. It consists of 25 small circles, with the numbers 0 ( at the 12 o' clock place) to 24, arranged as a big circle. Each student jumps either 3 or 4 spaces clockwise(so a student can end up either in circle 3 or 4 on their first hop). Students must go twice around the circuit and start and end at 0 to complete a game. All students must list in order the numbers they land on and record the total number of hops they take.

a) In one game a student took 13 hops. Write down a possible list of numbers he landed on.

I worked out that to clear 50 numbers with 13 numbers you van have 11 4-hops & 2 3-hops. This makes the numbers 4,8,12,16,20,24,3,7,11,15,19,22,0. But is 11 4-hops and 2 3-hops the only combination? And is there some sort of relativity formula to shorten the answer a bit( I can assume there will be tens and hundreds of combinations)

b) Find all possible combinations of the number of 3-hops and number of 4-hops in a game.

Again, I can assume there will be quite a few. Is there a relativity formula for this too?( You aren't limited to 13 hops for this question)

c)What is the smallest number of different numbers a student can land on in one game? Explain your answer

I'm totally stuck on this one . I don't know how are you supposed to prove that you have the smallest amount of different numbers?

d) Jo and Mike decide to play a longer version of the game. They both play simultaneously and start at 0. When Jo jumps 4 circles, Mike jumps 3 and vice-versa. Jo first 5 hops are 4-hops. After that, she takes 3-hops until she reaches or passes 0. How many laps will each of them take until they next meet at 0?

Again, I'm totally stumped. How are you meant to find that out? I've tried tracking the pattern but to no avail Thanks!

P.S. I'm only a Year 7, so please don't post answers with calculus in them. I won't understand a single bit. Algebra 1 is fine, though Note by Www Www
3 years, 11 months ago

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Are you familiar with Diophantine equations? Solving these problems is equivalent to solving a few linear Diophantine equations. Let $x$ be the number of 3-hops and $y$ be the number of 4-hops.

For (a), you need to simultaneously solve $x+y=13$ and $3x+4y=50$ over the non-negative integers. Rewriting the second equation in terms of $y$ and noting that $y$ is a non-negative integer, we get $x\equiv 2\pmod 4$. The first equation provides the bounds $0\leq x\leq 13$, so possible solutions for $x$ are $2,6,10$. Correspondingly, the possible solutions for $y$ are $11,8,5$ respectively. It's now evident that only $(x,y)=(2,11)$ satisfies the first equation and hence is the only solution. As for the number of configurations using this number of hops (you wrote down one), there are $\dbinom {13}{11}=\dbinom {13}2$ ways since we can, out of all the 13 hops, select 11 of the (not necessarily consecutive) hops as 4-hops and the rest will be 3-hops. Although, I'm not sure how many of those configurations are distinct.

For (b), we need the solutions of $3x+4y=50$ over the non-negative integers. From the work done in (a), we note that the solutions $x,y$ are $(2,11),(6,8),(10,5)$ with a total of $13,14,15$ hops respectively.

For (c), the answer seems to be $7$. Note that a player lands on at least 6 different numbers in the first lap alone and through a tedious analysis, it can be shown that if the player lands on only 6 different numbers in the first lap, he/she ends up stepping on a number during the second lap different from the 6 numbers. Hence, the player steps on at least 7 different numbers in one game. We shall now show that a game exists in which the player steps on exactly 7 different numbers. To do that, we find solutions $(x,y)$ to $3x+4y=25$ which are $(3,4),(7,1)$. The first solution steps on 7 different numbers in a game and the second on 8. The numbers stepped on the first lap are repeated for the second lap. So, with 3 3-hops, 4 4-hops in the first lap and repeating the same for the second lap (with 14 hops in total), the numbers stepped on are $3,6,9,13,17,21,0$ (7 different numbers).

For (d), I have a problem with the wording of the question. What does Jo do after passing 0 for the first time (i.e., completing the first lap but not the game). Does she continue with the 3-hops, or does she change to 4-hops? And if so, when does she change back to 3-hops? The question says "she takes 3-hops until she reaches or passes 0." but doesn't​ say what she does after passing 0 but not reaching 0. Can you clarify this?

- 3 years, 11 months ago