# If the product of $n$ distinct positive integers is $n^n$. What is the minimum value of their average if $n = 6$?

As $$n=6$$, then $$n^n = 6^6 = 46656$$ and I was able to break it into the product of lowest positive integers in the form as below :-

$$n^n = 6^6 = 46656 = 1 \times 4 \times 6 \times 9 \times 12 \times18$$

and I am not able to simplify it to any other form of product of lowest positive integers and for this case I am able to get to the average of $$8.\overline{3}$$ but this is not the correct answer that has been provided.

Can someone please tell me if this can be simplified more than this? What is the way to tackle these types of question in a much efficient way?

Thanks in advance !

## 1 Answer

A good way to start is to use AM-GM. Call the numbers $$x_1, x_2, \dots, x_6.$$ Then by AM-GM, $$\frac{x_1+x_2+\dots+x_6}{6} \ge \sqrt[6]{x_1x_2\dots x_6}.$$

We know the RHS to be $$\sqrt[6]{46656} = 6.$$ So the mean, which is the stuff on the LHS, is at least $$6.$$ So whatever answer we get must be at least $$6$$.

From here it's a good idea on these sort of "mean/set" contest problems to start with one solution like yours, $$1\times 4\times 6\times 9\times 12\times 18,$$ and narrow down. Let's look for any factor pairs that we can make smaller when the two are summed together.

We see that $$1\times 4 = 2\times 2,$$ with the RHS producing a smaller mean. But the problem is that we have a repeat of the number $$2.$$ Similarly for something like $$6\times 18 = 9 \times 12,$$ we have a repeat of the numbers $$9$$ and $$12$$, since in the original solution we have $$1\times 4\times 6\times \boxed{9\times 12}\times 18.$$ So we need to find factor pairs that have a smaller sum and don't repeat any numbers.

Starting from the smallest and going up, we find that $$1\times 4 = 2\times 2,$$ which doesn't work, but $$1\times 6 = 2\times 3$$ works since there is no other $$2$$ or $$3$$. So we replace $$1\times 6$$ with $$2\times 3$$ to get $$2\times 3 \times 4 \times 9 \times 12 \times 18.$$

But let's see if we can narrow down even more. We see $$2\times 9=3\times 6$$ but that doesn't work because there'd be two $$3$$s. Working our way, we eventually find that $$4\times 12 = 6\times 8,$$ a working pair that has a smaller mean! So we replace it and get $$2\times 3\times 6\times 8\times 9\times 18.$$

Continuing, we will eventually see that there are no other ways to make the mean smaller. So the mean is $$\frac{2+3+6+8+9+18}{6} = \boxed{\frac{23}{3}.}$$

Note: Problems like these don't really have a formula or something like a cheat sheet, it's more just experimentation and a right set strategy can save time and be accurate.

• Note: Simply optimizing factor pairs each time might not lead to the optimal solution. EG we might need to optimize triples, or might need to take a suboptimal step first. $\quad$ I think that a rigorous answer would pretty much have to list out all possible values. Commented May 18, 2021 at 19:16
• I can confirm your result, after an exhaustive search with Mathematica. The second best is $2\times3\times4\times9\times12\times18$, with sum $48$. Commented May 18, 2021 at 20:52
• @CalvinLin You're completely right. In a contest math setting though (like this problem), not full rigor is needed Commented May 18, 2021 at 23:54
• Whoa...this is very well explained and thanks every one for your answers, comments and suggestions. Commented May 19, 2021 at 11:10