This problem is quite a tricky one but can be solved in several different ways.

Congratulations to Jeremy, formerly of Epsom College, who sent in his solution which involves some concise mathematical reasoning combined with a little trial and error:

Preamble: The "sum of the digits test" says that, if the sum of the digits of a whole number is divisble by 3, so is the number.

The sum of the 4 amounts (in pence) is 711. The sum of these digits is divisble by 3, so 3 divides 711 (leaving 237).

By the same reasoning, 237 is also divisible by 3 (leaving 79, a prime number).

So a trial solution is (in pounds) 1 x 3 x 3 x 0.79 (we need 4 factors).

However, this does not add up correctly (it adds to 779 pence - but it is close).

Now we need to multiply each of the amounts by a multiplier r1, r2,r3,r4 so that r1.r2.r3.r4 = 1.

The multipliers are simple fractions like 1/2 so that the new number is still a whole number of pence in each case.

For the first 3 digits (1,3,3) we will only get totals of 150 p, 100p, 50p, 25p, 20 p, 10 p etc when we apply the multiplier - yet the sum must end in a 1p.

So we will try r4 = 4 (ie. take 4 lots of.79), as this ends in a 6

(4 is the only multiple of 9 that does end in a 6 among the digits 1 - 9)

and when combined with a 25 p adjustment this gives a total ending in 1p.

The only multiple of 9 that ends in a 1 is 81 and 9 lots of 0.79 is too many to work, so this rules out the other adjustments to £1, £3 and £3 which are multiples of 10.

(£1 x 5/4) x (£3 x 1/2) x (£3 x 2/5) x (£0.79 x 4) gives the correct answer:

£1.25 x £1.50 x £1.20 x £3.16 = £7.11

£1.25 + £1.50 + £1.20 + £3.16 = £7.11

Note that the product of the multipliers equals one: 5/4 x 1/2 x 2/5 x 4 = 1

Another approach which relies on systematic working uses the following:

It's probably easier to work with whole numbers, so let's work in pence, rather than pounds. Let's call the prices of the 4 items a, b, c and d.

The sum of the items is £7.11 or 711p,so:

$a + b + c + d = 711$

$\frac{a}{100} \times \frac{b}{100} \times \frac{c}{100} \times \frac{d}{100} = 7.11$

It follows that

$711 \times 10^{6} = 79 \times 3^{2} \times 2^{6} \times 5^{6}$

We also know that:

a,b,c and d are all less than 708 (each item must be equal to or more than 1p)
Either 3 values are odd and1 even, or 1 is odd and 3 are even (a + b+ c + d = 711 and 711 is odd)