# Max product of $20$ numbers with mean $12$ if $n$-th number is $\frac{n}2$ or $2n$?

A list contains $$20$$ numbers. For each positive integer $$n$$, from $$1$$ to $$20$$, the $$n$$th number in the list is either $$\frac{1}{2}n$$ or $$2n$$. If the mean of the number in the list is exactly $$12$$ and the product of the numbers is $$P$$, what is the greatest possible value of $$\frac{P}{20!}$$?

I have tried some examples, but they bring me nowhere. I then tried creating equations: I can call the sum of the numbers that you are halving by $$a$$ and the sum of the numbers that you are doubling by $$b$$. However, after trying repeatedly, I don't see any way to create an equation. Can I have a hint?

Also, if you are nice, may you please help me on this question($N$'s base-5 and base-6 representations, treated as base-10, yield sum $S$. For which $N$ are $S$'s rightmost two digits the same as $2N$'s?)?

Thanks!

Max0815

This is an extension to the answer provided by D.B. to confirm the stated result.

Since the mean is $$12$$, with $$20$$ numbers, the sum $$S$$ is such that $$S / 20 = 12$$, so $$S = 240$$. Now, for comparison, consider as an upper range that all $$20$$ values use $$2n$$, to then get a sum of

$$2\left(1 + 2 + \ldots + 19 + 20\right) = 2\frac{\left(20\right)\left(21\right)}{2} = 420 \tag{1}\label{eq1}$$

As such, this is $$420 - 240 = 180$$ too high. Now, for each term which used the $$n/2$$ value instead, it would reduce the sum by $$3n/2$$. As such, if the sum of these $$n$$ values is $$S_1$$, we have that that the total reduction is $$3S_1/2 = 180$$ giving that $$S_1 = 120$$. As mentioned, we want the number of these terms to be the minimum, so we start using the largest ones. Using the $$7$$ values $$14, 15, \ldots 20$$ gives a sum of $$119$$, meaning just need to use $$1$$ as well to get the sum to $$120$$, for a total of $$8$$ terms, as conjectured. Note you cannot use $$7$$ or less since, as just shown, the $$7$$ largest values add to $$119$$, so any other $$7$$ or fewer terms will add up to less than $$119$$, so $$8$$ is the minimum # of terms required.

One small thing to note is that which $$8$$ terms are to use $$n/2$$ is not uniquely defined. For example, the $$1$$ and $$14$$ terms can be replaced by any other $$2$$ which add to $$15$$, such as $$2$$ and $$13$$, $$3$$ and $$12$$, $$\ldots$$, $$7$$ and $$8$$.

• That you for the extension explaination! – Max0815 Feb 5 at 23:35
• @Max0815 You are welcome. I'm glad that I could help explain things. – John Omielan Feb 6 at 0:30
• :) yep clarified a lot – Max0815 Feb 6 at 0:37

Suppose, for $$m \leq n$$ that $$m$$ is the number of terms in the sequence with the factor of $$1/2$$. Then,

$$P = \prod_{n = 1}^{20} n = (\frac{1}{2})^m(2)^{n-m} 20!.$$ Then, $$\frac{P}{20!} = 2^{n-2m}.$$

To find $$m$$ and $$n$$, note that the mean of the sequence is $$12$$. We want the mean to be $$12$$ with $$m$$ as small as possible. So, we start by tacking on the factor of $$1/2$$ onto the largest terms in the sequence. $$\frac{1}{20}(\sum_{n=1}^{13} 2n + (1/2)*\sum_{n=14}^{20} n) = 9.1+2.975 \approx 12,$$ which is not quite there. We will need to increase $$m$$ by one and decrease $$n$$ by one. So, I think the answer is $$2^{20-2*8} = 2^{4} = 16.$$ But I'm not sure how to make the mean exactly $$12$$.

Anyway, this is how I would go about the problem.

• Your use of n is inconsistent and confusing. If 8 terms use the 1/2 factor, then the other 12 terms use the 2 factor, and P is 16, not 1/16. – Grimy Feb 5 at 11:57
• There's some confusion of variables in your answer, so it's one reason I believe you get the incorrect answer. You use $n$ for the number of items of the list, which is a fixed $20$, such as at the start where you say "for $m \le n$", but you also use it as the dummy variables in your product and sums. In addition, you make the statement "We will need to increase $m$ by one and decrease $n$ by one". I don't know what you mean by the latter part. Finally, you use $12$ for $n$ in your calculation, instead of $20$. As such, as Grimy says, the proper answer is $2^{20 - 2 \times 8} = 2^4 = 16$. – John Omielan Feb 5 at 11:58
• Ok. Thanks for the correction. – D.B. Feb 5 at 21:42