### Picture Story

Can you see how this picture illustrates the formula for the sum of the first six cube numbers?

### Series Sums

Let S1 = 1 , S2 = 2 + 3, S3 = 4 + 5 + 6 ,........ Calculate S17.

### Attractive Tablecloths

Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?

# Big Fibonacci

##### Age 14 to 16 Short Challenge Level:

Let the first two terms of the sequence be $a$ and $b$ respectively.

Then the next three terms are $a+b$, $a+2b$, $2a+3b$. So $2a+3b = 2004$.

For $a$ to be as large as possible, we need $b$ to be as small as possible, consistent with both being positive integers.

If $b=1$ then $2a=2001$, but $a$ is an integer, so $b\not=1$.

However, if $b=2$ then $2a=1998$, so the maximum possible value of $a$ is $999$, giving us the first five terms:

$999$, $2$, $1001$, $1003$, $2004$

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.