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Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

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Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

This is a tricky problem to solve...

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