# Set of numbers from $1-9$ multiplied together to get the smallest possible value

The numbers $$x_1,$$ $$x_2,$$ $$x_3,$$ $$y_1,$$ $$y_2,$$ $$y_3,$$ $$z_1,$$ $$z_2,$$ $$z_3$$ are equal to the numbers $$1,$$ $$2,$$ $$3,$$ $$\dots,$$ $$9$$ in some order. Find the smallest possible value of $$x_1 x_2 x_3 + y_1 y_2 y_3 + z_1 z_2 z_3.$$

I would assume the lowest number, $$1,$$ would have to be multiplied by $$9,$$ the highest. I do not know how to approach this with AM-GM, though.

• You in fact can have a proof by exhaustion... A dirty trick – user12986714 May 10 '20 at 1:22
• Sorry, is there a slick way to do this? I don't really want to brute force, though I can. – Frost Bite May 10 '20 at 1:23
• I don't want to brute force this problem, but rather to do it smartly. – Frost Bite May 10 '20 at 1:30
• $\sqrt[3]{9!}\approx 70.327$ and $3\times\sqrt[3]{9!}\approx 213.98$, so $72+72+70=214$ should be minimal – Daniel Mathias May 10 '20 at 1:34
• $9\cdot8\cdot 1 + 6\cdot 5 \cdot 2 +4\cdot3\cdot 7= 72+60+84=216$ is also pretty close. – mechanodroid May 10 '20 at 1:53

The following Mathematica script confirms that there is only one solution:

prod[p_] := p[[1]] p[[2]] p[[3]] + p[[4]] p[[5]] p[[6]] + p[[7]] p[[8]] p[[9]];
perms = Permutations[{1, 2, 3, 4, 5, 6, 7, 8, 9}];
unique = Select[perms, (#[[1]] < #[[2]] < #[[3]] && #[[4]] < #[[5]] < #[[6]] && #[[7]] < #[[8]] < #[[9]] && #[[1]] < #[[4]] < #[[7]]) &];
products = Map[prod, unique];
min = Min[products];
result = Select[unique, (prod[#] == min) &];
Print[min, " ", result]


The script prints:

214 {{1, 8, 9, 2, 5, 7, 3, 4, 6}}


The next best two solutions are:

215 {{1, 7, 9, 2, 5, 8, 3, 4, 6}}
216 {{1, 8, 9, 2, 5, 6, 3, 4, 7}, {1, 8, 9, 2, 6, 7, 3, 4, 5}}


This isn't as systematic or as linear as I'd like but:

$$\frac {M+N+P}3 \le \sqrt[3]{MNP}=\sqrt[3]{9!}\approx 71.3$$ and equality is closest when the $$w_1w_2w_3$$ clusters are each closet to the geometric mean of $$71.3$$.

So we must have one term contaning the $$9$$. That is, wolog, $$9x_1x_2 \approx 71.3$$ so $$x_1x_2 \approx 7.9$$. Clearly the best option is $$x_1x_2 = 8$$ and $$x_1,x_2 = 2,4$$ or $$1,8$$.

$$1,8$$ has the advantage as it simultaneously finds clusters for the far off values of $$1$$ and $$8$$ as well.

In any event we also have to find, wolog, $$7y_1y_2 \approx 71.3$$ so $$y_1x_2\approx 10.1$$ and the best solution for that is $$7,2,5$$.

So with $$9,1,8$$ and $$7,2,5$$ that leaves $$z_1, z_2, z_3 = 3,4,6$$ and $$3*4*6=72$$ with our clusters being $$70, 72, 72$$ around our geometric mean of $$71.3$$ the only possible smaller sum would be be a tighter cluster and the only tighter cluster that is smaller it $$71,71,71$$, which is not an option.

So we can be assured that is the least.

We were "lucky" though

Still if say we had gotten something pretty close first, say,

$$9*8*1 + 6*5*2 + 7*4*2 = 72 + 60 + 84$$ we could see if we can improve it be attempting to finding a tighter cluster of values closer to $$71.3$$, than $$60$$ and $$84$$ are. The contenders that can formed by trios less than or equal to $$9$$ are:

$$63=7*1*9; 64=8*2*4; 70=7*2*5; 72=9*8*1=9*2*4=6*3*4;80=8*2*5$$ ... and it would take brute force to choose among those. But even among that we want to get as close to $$71.3$$ as possible and we have $$3$$ ways to do $$72$$ and one way to do $$70$$.... well, that screams to uss to look if two of the ways to do $$72$$ are mutually compatible with the way to do $$70$$.