### Sixational

The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.

### Squaresearch

Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?

### Loopy

Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?

# Three Frogs

##### Stage: 4 Challenge Level:

Many thanks for your solutions. Some of you were able to get a feel for the idea that the number of jumps would need to be even because you need to reverse the process to get back to where you started. This did not entirely cover all possibilities as it is possible not to retrace your steps but to continue with a set of "complementary" moves that get you back to the start. Because there are 3! possible arrangements of the frogs - this route would take six moves, which still gives an even number. The diagram below might help explain this:

Josh Smith of Wymondham High School noticed the relevance of the multiple of two and Ali of Bourne Grammar School thought that the 3! was important.

Those of you who gave the answer that it was possible because the number of hops needs to be a multiple of three gave no clear explanation of your reasoning so it is difficult for me to tell where your argument went wrong. Could it be that you thought that, as there were three frogs, there needs to be a multiple of three moves? But why?

Andrei of school 205 Bucharest offered the following solution:

Let the three frogs be A, B and C. In the initial configuration they are:

A - B - C

After moving them, they will be in the following positions:

0: A - B - C

1: B - A - C

2: B - C - A

3: C - B - A

4: C - A - B

5: A - C - B

6: A - B - C

I tried several combinations but in all of them the number of moves needed to return to the initial position was even. This happens because the frogs move to a new position, and then they return to it (making a minimum of 2 moves).

In this case it is impossible, after 999 moves, the frogs to be in the same positions as in the beginning.