### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### Have You Got It?

Can you explain the strategy for winning this game with any target?

### Counting Factors

Is there an efficient way to work out how many factors a large number has?

# Tricky Fractions

##### Age 11 to 14 ShortChallenge Level

Anwser: 341

The first pair of terms add up to $\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}$

The next pair of terms add up to $\dfrac18-\dfrac1{16}=\dfrac{1}{16}$

The next pair of terms ($\frac1{32}-\frac1{64}$) add up to $\dfrac1{64}$

... the last pair of terms add up to $\frac1{1024}$

The whole sum is $\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dfrac{1}{1024}$

Common denominator: $1024$

Each fraction is $4$ times smaller than the last so whole sum is equal to

$\dfrac{256+64+16+4+1}{1024}=\dfrac{341}{1024}$

Multiplying the sequence by 2
If $$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\dots-\frac{1}{1024}$$then \begin{align}\frac{2x}{1024}=&2\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\dots-\frac{1}{1024}\right)\\=&1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\dots-\frac{1}{512}\end{align}Notice that most of the terms in $\dfrac{2x}{1024}$ are the same as the terms in $\dfrac{x}{1024}$  \begin{align}\frac{2x}{1024}+\frac{x}{1024}=1-&\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\dots-\frac{1}{512}\\+&\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\dots-\frac{1}{1024}\\ \Rightarrow \frac{3x}{1024}=1+&\text{ }0\quad+\ \quad\dots\ \quad+\quad 0\text{ }- \frac{1}{1024}=\frac{1023}{1024}&\\ \therefore x=1023&\div3=341 \end{align}

You can find more short problems, arranged by curriculum topic, in our short problems collection.