# 4 items add up to and multiply to 7.11 what are the value of the items?

A man walks into 711 and buys four items, the items add up to 7.11 and multiply to 7.11. What are the prices for the 4 items?

During a talk about truth in mathematics, the presenter asked this question to the audience. I'm just curious as to what the answer is.

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Hint: This is a Diophantine problem. Besides the sum and product, you also have that the answers (expressed in cents) are integers. –  Codie CodeMonkey Sep 21 '11 at 6:47
Wait, the man bought numbers? Where can I buy numbers? I call dibs on Euler-Mascheroni. –  anon Sep 21 '11 at 6:59
Hmmm, the prime factors of 711 are 3*3*79. There aren't many choices for the prices here, and none of them add up to 711. For example, 1*1*9*79 = 711, but the sum isn't even close. –  Codie CodeMonkey Sep 21 '11 at 7:01
@anon: if $\pi$ is too expensive, maybe I can just buy a piece? –  Codie CodeMonkey Sep 21 '11 at 7:05
@DeepYellow: $$\frac{abcd}{100^4}=\frac{711}{100}$$ $$\implies abcd=2^6 3^2 5^6 79$$ –  anon Sep 21 '11 at 7:05

$$\text{dollars}:\quad 3.16,\quad 1.25,\quad 1.50,\quad 1.20$$ $$\text{cents}:\quad 2^2\cdot 79,\quad 5^3,\quad 2\cdot3\cdot5^2,\quad 2^3\cdot 3 \cdot5$$
I cheated; I just googled the problem and found that someone brute forced a solution. It's possible that one could do some form of case-by-case modular analysis, but even if that's feasible it sounds exhausting. (One pattern that jumps out to me, though, is how the factors of $2$ and $5$ are spread; perhaps one could prove the distribution of them among the solutions totally a priori?) You can also get pretty damn close by making $a=0.79, b=2.00$, $c=1.75$-$1.76$ and $d=2.57$-$2.56$.