Combinatorics problem : circle hopscotch

[bio]

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

 

[Opalg]

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)

Your answer is correct. You can use algebra to show that 11 4-hops and 2 3-hops is the only possibility. In fact, suppose $x$ is the number of 4-hops and $y$ is the number of 3-hops. You are told that $4x+3y = 50$, and $x+y = 13$. Solve those equations to get $x=11$ and $y=2$.

There are many possible answers for the route taken. You have given a route where all the 4-hops occur first, followed by the two 3-hops. The 3-hops could have occurred at any stage during the route. As you say, this gives many possible combinations. But that is not what the question asks for. It just asks for the number of 4-hops and the number of 3-hops.

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)

The total number of jumps must add up to 50. You have given one way to achieve that, namely 11 4-hops and 2 3-hops. That is the solution with the largest possible number of 4-hops, and the smallest possible number of 3-hops. To find the other solutions, use the fact that $4\times3 = 3\times4$. So if you reduce the number of 4-hops by 3, you must increase the number of 3-hops by 4. That way, you can list all the possible combinations.

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?

This is a tricky one. The obvious way to minimise the number of numbers landed on is to find a route where you land on the same numbers on each of the two circuits. For example, you could use 4 4-hops and 3 3-hops, to land on the numbers 4,8,12,16,19,22,0. That way, you have completed one circuit, landing on 7 small circles and finishing on the starting circle. You can then repeat the exact same route for the second circuit, landing on the same 7 small circles again.

Is 7 the smallest possible number of circles a student can land on in one game? Yes, because it would not be possible to get round using only 6 (or fewer) landing places. I'll leave you to figure out why.

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!

I don't understand this part of the problem, because it doesn't tell you what Jo's strategy is. After she completes the first circuit by passing 0, does she continue to use only 3-hops, or does she then take another 5 4-hops? It's not clear to me.

 

[bio]

I don't understand this part of the problem, because it doesn't tell you what Jo's strategy is. After she completes the first circuit by passing 0, does she continue to use only 3-hops, or does she then take another 5 4-hops? It's not clear to me.

I'm assuming that after he gets to or passes 0, he does 54-hops again although I'm not certain. However, if you can, can you do both please?

 

[Opalg]

I'm assuming that after he gets to or passes 0, he does 54-hops again although I'm not certain. However, if you can, can you do both please?

Assuming she repeats the five 4-hops on each lap, when will she first land on circle $0$? On her first lap, she lands on 4,8,12,16,20 with the five 4-hops. Then two 3-hops take her to $23$ and $1$. On her second lap, repeating the same sequence of hops she will always be one circle ahead of where she was on the first lap. So the five 4-hops will take her to $21$, and two 3-hops take her on to $2$. The third lap will be a bit different, because the five 4-hops will take her to $22$, and then she only needs one 3-hop to reach $0$.

Therefore Jo reaches $0$ after every three laps, during which she has jumped a total of $15$ 4-hops and $5$ 3-hops. (Notice that $15\times4 + 5\times3 = 60 + 15 = 75 = 3\times25.$ This confirms that she has completed exactly three laps.)

In the same time, Mike has jumped $15$ 3-hops and $5$ 4-hops (because he always does the opposite number of hops to Jo). Since $15\times3 + 5\times4 = 45 + 20 = 65$, he will be $10$ circles short of where Jo is when she completed her third lap. Each time Jo completes three laps and gets back to $0$, Mike will have fallen another ten circles behind her.

So when Jo completes her fifteenth lap, Mike will be $50$ circles behind her. But $50$ is an exact multiple of $25$. This tells you that Mike will then be on the same circle as Jo, in other words they will both be at $0$. The answer to d) is therefore that they will both meet at $0$ when Jo has completed $15$ laps and Mike has completed $13$ laps (because he is then $50$ circles, or two whole laps, behind Jo).

Mathematically, this is an exercise about remainders when you divide by 25. Each hop means that you add 3 or 4 to the circle number, but when you reach 25 you replace it by 0. So you can use ordinary addition, but each time a number gets gets bigger than 25, you subtract 25 and just look at the remainder. This is called modular arithmetic, and it is an important topic in mathematics.