Challenge Level

$\begin{align} & \text {old price}\times\tfrac{104}{100}=\text{multiple of 100}\\

\therefore & \text{old price}\times\tfrac{26}{25} = \text{multiple of 100}\\

\therefore & \tfrac{\text{old price}}{25}\times2\times13=\text{multiple of 100}\\

\therefore &\tfrac{\text{old price}}{25}=\text{multiple of 50}

\end{align}$

So the smallest possible $\text{old price}$ is $50\times 25$

Which gives a new price of $50\times 26=1300$ ie. $ £13$

For $n$ to be 1, the original price would need to be about 96p.

4% of 96p is 0.04$\times$96p = 3.84p, so 4% inflation from 96p is 96p + 3.84p = 99.84p

4% of 97p is 0.04$\times$97p = 3.88p, so 4% inflation from 97p is 97p + 3.88p = 100.88p

Those are either side of £1, so $n$ cannot be 1.

For $n$ to be 2, the original price would need to be about £1.92.

4% of £1.92 is 0.04$\times$ £1.92 = £0.0768, so 4% inflation from £1.92 is £1.92 + £0.0768 = £1.9968

4% of £1.93 is 0.04$\times$ £1.93 = £0.0772, so 4% inflation from £1.93 is £1.93 + £0.0772 = £2.0072

Those are either side of £2, so $n$ cannot be 2.

Continuing in this way is slow, but eventually we get that $n$ = 13.

The price increase is 4% of the original price. If the new price is a whole number of pounds, then it is also a whole number of pennies. So since the old price was also a whole number of pennies, the increase must be a whole number of pennies too.

This means that 4% of the old price is a whole number of pennies.

4% is equivalent to $\frac1{25}$, so $\frac1{25}$ of the old price is a whole number of pennies, so the old price must be a multiple of 25p.

This means that the old price must be in the form £X.00, £X.25, £X.50 or £X.75.

For $n$ to be as small as possible, the increase should also be as small as possible, so first we should try an increase of 25p, from £($n-1$).75 to £$n$.

That means that 4%, or $\frac{1}{25}$, of the original amount should be 25p, so the original amount was 25p$\times$25 = £6.25. However £6.25 + 25p = £6.50, which is not a whole number of pounds.

Next try an increase of 50p, from £($n-1$).50 to £$n$.

That means that 4%, or $\frac{1}{25}$, of the original amount should be 50p, so the original amount was 50p$\times$25 = £12.50. £12.50 + 50p = £13.00. So $n$ = 13.

Let the old price be £$a$, where £$a$ is an exact number of pence (so it has at most 2 decimal places). Then if the new price is £$n$, $n$ is a 4% increase of $a$.

This means that $a+\frac{4}{100}a=n\Rightarrow\frac{100}{100}a+\frac{4}{100}a=n\Rightarrow \frac{104}{100}a=n$.

So $104a=100n\Rightarrow a=\frac{100}{104}n=0.9\dot{6}1538\dot{4}n$, and $a$ should have at most 2 d.p.

$0.9\dot{6}1538\dot{4}\times2=1.\dot92307\dot6$, which has more than 2 d.p., so $n$ cannot be $2$.

$0.9\dot{6}1538\dot{4}\times3=2.8\dot84615\dot3$, which has more than 2 d.p., so $n$ cannot be $3$.

$0.9\dot{6}1538\dot{4}\times4=3.\dot84615\dot3$, which has more than 2 d.p., so $n$ cannot be $4$.

Continuing in this way until we get a possible value for $a$, $n=13$ ( since $0.9\dot{6}1538\dot{4}\times4=12.5$ exactly).

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