*Originally posted by iamatiger*

**Challenge - first person to find a way to derive the answer with pencil and paper only wins.**

Managed It (without looking at the answer)!

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