### Plants

Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?

### Forgot the Numbers

On my calculator I divided one whole number by another whole number and got the answer 3.125 If the numbers are both under 50, what are they?

### Shapes in a Grid

Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?

# Junior Frogs

##### Stage: 2 Challenge Level:

Towards the end of August we had some last minute solutions sent in from different countries, namely - England, New Zealand, Australia, China and Denmark. We had a very precise, accurate and well-explained solution sent in earlier, by Hannah from Leicester High School for Girls in the UK. This represents the most exhaustive solution that I have come across for my particular activities. Well done Hannah! You have good reason to be proud of your work. I hope you will continue to work on some of the activities from this site.

To keep this simple, let's assign a letter to each of our adventurous amphibians: from left to right, their names are A, B, C and D. At the start, you are able to move any of the $4$ amphibians. It doesn't matter whether you move a toad or a frog first, because the layout is symmetrical. Let's say we move a frog first. Imagine a situation where A jumps over B , we'd be in a right mess because no amphibian would be able to move anymore. So this means we have to slide B onto the lily pad directly beside it. Now, we can either move A, which would bring us back to the situation mentioned at the beginning, or we could, more sensibly, make C jump over B. Just to recap, our current position is A, C, B, (space), D.

At this point we can either move B or D. We'll have to think ahead. Say we move B onto the vacant lily pad. Now we could move either A or D. However, both these moves would bring us to a dead end and a serious congestion. This means we will have to slide D onto the lily pad beside it. From here, the only thing we can do is make B jump over D. Now we can move either A or D. Moving D would bring us to a congestion that just won't do, so we are obliged to jump A over C. There is only one move we can make at this point: sliding C onto the left-most lily pad. After this, there still is only one possible move: jumping D over A. From here on it's quite obvious: just slide A nicely in beside B, and we've exchanged the positions of the frogs and toads.

This is not the only solution: the exact reflection of what I've just explained, where the toad moves first, is the other answer.    Altogether, this has taken $8$ moves in total, according to the NRICH interactivity. Each amphibiotic species made exactly two slides and two jumps. I can be sure that this is the lowest possible number of moves because I have worked this out systematically. However, I have found that solving this puzzle in more than $8$ moves is impossible; from several trials I have discovered that doing it in $8$ moves is the only solution. This means that $8$ moves is also the highest as well as lowest possible number of moves for two frogs and two toads.

This time the outcome is quite different: the interactivity states that frogs made four slides and four jumps, and that toads made two slides and five jumps. This means that toads made fewer moves than frogs. The total number of moves is $15$. I tried this again starting with the toads, and this time the frogs made two slides and five jumps, and the toads made four of each. I can be assured that this is the shortest possible number of moves because I worked it out systematically and always chose the one move that would work. I think the key is to make an alternating pattern of frogs and toads, and from then on it's easy.

I worked out the smallest number of moves for one frog and one toad, as well as four frogs and four toads, to see if there was a pattern and whether I could come up with a formula.

Here are my results:

$1$ of each type of amphibian, $3$ moves

$2$ of each type of amphibian, $8$ moves

$3$ of each type of amphibian, $15$ moves

$4$ of each type of amphibian, $24$ moves.

I noticed that the number of moves is the number of each type of amphibian multiplied by itself plus 2. Now, that's a bit of a mouthful, so here's the formula that I came up with:  $n(n + 2)$ or $n² + 2n$.

This might encourage others to write explanations in their own words, describing what they have done and where they got to.