- 29 Jul '04 18:23A customer at a 7-11 store selected four items to buy, and was told that the cost was $7.11. He was curious that the cost was the same as the store name, so he enquired as to how the figure was derived. The clerk said that he had simply multiplied the prices of the four individual items. The customer protested that the four prices should have been ADDED, not MULTIPLIED. The clerk said that that was OK with him, but, the result was still the same: exactly $7.11. What were the four prices?

-Ray. - 04 Aug '04 18:47

Very small, i just happened to remember that post, i do wish to find out a way to calculate it though.*Originally posted by rgoudie***What were the odds of that?**

-Ray.

The best way i can think of is that you write the numbers as sum of powers of 10, then multiply is out, add it up, get conditions for the 15 parameters, then solve it... - 04 Aug '04 19:16

Sounds like a feasible plan.*Originally posted by TheMaster37***Very small, i just happened to remember that post, i do wish to find out a way to calculate it though.**

The best way i can think of is that you write the numbers as sum of powers of 10, then multiply is out, add it up, get conditions for the 15 parameters, then solve it...

Given that the task was simply to discover the values that work, I would have come up some small program to find the values using brute force.

-Ray. - 07 Aug '04 22:57

Managed It (without looking at the answer)!*Originally posted by iamatiger***Challenge - first person to find a way to derive the answer with pencil and paper only wins.**

a*b*c*d = 7.11

a+b+c+d = 7.11

We know that a,b,c and d can have at most 2 decimal places

let A=100*a, B=100*b, C=100*c, D=100*d

then A,B,C and D are integers, and

A+B+C+D = 711 (we've multiplied both sides by 100)

A*B*C*D = 711*10^6 (we've multiplied both sides by 10^8)

next factorise 711*10^6

711*10^6 = 79.2.2.2.2.2.2.3.3.5.5.5.5.5.5

We then know that A, B C and D must be made up from some combination of those factors. And A + B + C + D = 711

Its pretty easy to then try various possible combinations on paper, considering i) what we can multiply the largest factor of 79 by, without "busting" and ii) How a total ending in 1 can be obtained (this is quite a limiting consideration as numbers with a factor of 5 can only end in 0 or 5)

After some fiddling I tried:

79.2.2 + 5.5.5 + 5.5.2.2 + 5.3.3.2.2 = 316+125+100+180 = 721

Hmm, close - Is there any way to reduce this total by 10? - well, if we swap a factor of 2 and a factor of 3 round in the last two numbers that should make the total slightly smaller....

79.2.2 + 5.5.5 + 5.5.3.2 + 5.3.2.2.2 = 316 + 125 + 120 + 150 = 711!

So the answer is $3.16, $1.50, $1.25 and $1.20