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 = 22.214.171.124.126.96.36.199.188.8.131.52.5.5.5
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 + 184.108.40.206 + 220.127.116.11.2 = 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 + 18.104.22.168 + 22.214.171.124.2 = 316 + 125 + 120 + 150 = 711!
So the answer is $3.16, $1.50, $1.25 and $1.20