First we can consider the sum of the prices, working in pence.

Let the prices be a, b, c and 150 pence. Then a + b + c = 711 - 150 = 561

Now consider the product of the prices, in pounds.

Then a/100 x b/100 x c/100 = 7.11 ÷ 1.5 = 4.74

So abc = 4740000, which can be written using prime factors

abc = 2⁵ x 3 x 5⁴ x 79

Therefore, one of these prices (let’s say a) must be a multiple of 79 (pence); either 79, 158, 237, 316, 395 or 474. It can’t be the next multiple (7 x 79) as there is no factor of 7 in the product abc.

Now consider the “5”s in the prime factorization. They must be shared between the three values a, b and c.

Option 1

If a is a multiple of 5 then a = 395, and bc = 2⁵ x 3 x 5³

We also get b + c = 561 - 395 = 166, so it cannot be the case that both b and c are multiples of 5.

Therefore one of b and c (let’s say b) must be a multiple of 125. b cannot be anything other than 125, as 2 x 125 = 250, and we know b < 166.

This gives us a = 395, b = 126 and c = 166 - 125 = 41.

But 41 is not in the prime factorisation of abc, so a cannot be 395.

Option 2

a is not a multiple of 5, so either one of b and c is a multiple of 5⁴ (= 625) or both b and c are multiples of 5.

Sub-option A

One of b and c (let’s say b) is a multiple of 625. But a + b + c = 561, so b < 561. Therefore this is impossible.

Sub-option B

Both of b and c are multiples of 5.

Then to have a + b + c = 561, the last digit of a can only be 1 or 6, meaning that a must be 316 (or 2² x 79)

So we’d have prices of 1.50 and 3.16, leaving b + c = 245 and bc = 1.5 x 100 x 100 = 15000 = 2³ x 3 x 5⁴

Since b + c is not a multiple of 25, one of b and c (let’s say b) must be a multiple of 125. But this leaves b = 125 as the only option, as the next multiple of 125 would give b = 250, but b < 245.

This leads to c = 245 – 125 = 120.

Checking the product of the prices, abc= 3.16 x 1.25 x 1.2 = 4.74 as required!

Hence the prices must have been £1.20, £1.25, £1.50 and £3.16.

Challenge 298 : A convenient sum and product?

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