Six Sum Product Equality

It's easy enough to find $1+1+1+1+2+6=1\times1\times1\times1\times2\times6=\boxed{12}$ .

To prove this is the only possible $n$ , let's try to find $x_1 \le x_2 \le \cdots \le x_6$ satisfying the condition. Let $S$ be their sum and $P$ their product.

Now, if $x_1\ge 2$ , then $P \ge 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot x_6 = 2^5 x_6$ but $S \le x_6+x_6+x_6+x_6+x_6+x_6 = 6x_6$ This is impossible, so $x_1=1$ . Now try $x_2\ge 2$ : $P\ge 2^4 x_6,\;\;\;S\le 6x_6$ so $x_2=1$ . The same thing works to show $x_3=1$ .

If we assume $x_4\ge2$ , though, the same idea doesn't work; we get $P\ge 2^2 x_6,\;\;\;S\le 6x_6$ which isn't a contradiction. However, we can show that $x_4 <3$ this way.

We now know the set is either $\left(1,1,1,1,x_5,x_6 \right)$ or $\left(1,1,1,2,x_5,x_6\right)$ . Put $x_5=a$ and $x_6=b$ (remember that $a \le b$ ).

Case $(1,1,1,2,a,b)$ : we have $\begin{aligned} 5+a+b &=2ab \\ 4a b - 2a - 2b &= 10 \\ (2a-1)(2b-1) &= 11 \end{aligned}$

Since $11$ is prime, one of the factors must be $1$ , which would mean $a=1$ or $b=1$ ; but this contradicts $x_4 \le x_5 \le x_6$ .

So the only possibility is case $(1,1,1,1,a,b)$ : now $\begin{aligned} 4+a+b &=ab \\ a b - a - b &= 4 \\ (a-1)(b-1) &= 5 \end{aligned}$

But this can only happen if $a=2$ and $b=6$ , proving the uniqueness of the solution.

Simple python code to find the sum.

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def find_six_sum(n):
    for a in range(1, n):
        for b in range(1, n):
            for c in range(1, n):
                for d in range(1, n):
                    for e in range(1, n):
                        for f in range(1, n):
                            s = a + b + c + d + e + f
                            p = a * b * c * d * e * f
                            if s == p:
                                print( a, b, c, d, e, f)
                                return

find_six_sum(10)