You may also like

problem icon


Make a cube out of straws and have a go at this practical challenge.

problem icon


Reasoning about the number of matches needed to build squares that share their sides.

problem icon

Tangram Paradox

How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!


Stage: 2 and 3 Challenge Level: Challenge Level:1
We received lots of solutions for this, with lots of good reasoning. Well done!

Many people began by experimenting with different numbers of frogs and trying to spot patterns. This is a great way to begin a problem like this!

Ellie from Ayr Grammar Primary School, Louis from Saint John the Baptist Catholic Primary School, Darcey, Maddy and Jessica from St John the Baptist School, Lisa from The Wheatlands and Kendyl from Kelly Elementary, Teton County School District, Wyoming came up with the following strategy for moving the frogs.

Here is Ellie’s explanation when there are $2$ red and $2$ blue frogs:

Move the blue frog closest to the middle along one place to the left. Then make the red frog which the blue frog is now next to, jump over the blue frog.  Then move red frog number $2$ along to the right once. Then make the blue frog closest to the left jump over the red frog and reach the left side. Then move the other blue frog to jump over the red frog. Next, move the red frog nearest the right to reach the very end. Make the other blue frog jump so that it is next to the other red frog. Lastly move the blue frog in the middle to the left once.

Micheal from Cloverdale Catholic School Canada used the same strategy and also said:

You just move each colour as many times as possible without ending up with two consecutive frogs of the same colour.

He then counted the number of moves when there are $9$ red frogs and $9$ blue frogs:

Blue moves one, Red moves $2$, blue moves $3$, red moves $4$, and so on. You will get the pattern:

blue $1$ – $3$ – $5$ – $7$ – $9$ – $9$ – $7$ – $5$ – $3$ – $1$
red   – $2$ – $4$ – $6$ – $8$ – $9$ – $8$ – $6$ – $4$ – $2$ –

A few people found this formula:

Minimum number of moves is $a+b+ab$

where $a$ is the number of red frogs and
            $b$ is the number of blue frogs.

Well done to Amy from Melbourn Village Collage, Kit from King Edward VI School Stratford, Rebbeca from St Stephens  School Australia and Bryant from the American International School of Lusaka in Zambia who found this.

Neveyan from Wallington County Grammar School wrote up their solution excellently;  click here to read their answer.

Well done everyone!