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Is there an $n \times n$ multiplication table such that if you form a border of width $k$ ("the frame") and sum its elements, the total will equal the sum of the remaining elements ("the picture")?

The diagram on the left is intended as a general representation. On the right, I created a $5 \times 5$ multiplication table (entries omitted) and summed them in their corresponding color. The width of the border was $k = 1$, and I found $\sum$frame $= 144$ but $\sum$picture $= 81$. So, in this case, the sums were not equal.

(Yes, there is a quick parity argument to show the sums here aren't equal, but I wanted to carry out the computations explicitly in an example to ensure that my question is understood.)

enter image description here

If the answer is no such table exists: how do you prove this?

If the answer is yes: what is the minimal $n$ (and its minimal $k$) for which this is possible?

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Hey, B.D. Don't forget to award that bounty to Matthew. It doesn't happen automatically, and since you've got your answer, you should take care of that so you aren't on the "featured" page anymore. –  Cameron Buie Nov 7 '12 at 21:27
    
@CameronBuie I bountied it because I thought his answer should get more recognition! –  Benjamin Dickman Nov 8 '12 at 0:38
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I see. Fair enough! (I upvoted it.) –  Cameron Buie Nov 8 '12 at 0:50
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1 Answer 1

up vote 4 down vote accepted
+50

I believe there is no such table.

Consider an $n \times n$ table with a border of $k$.

The sum of all entries in the table is $$ \left(\frac{1}{2}n(n+1) \right) ^2 $$ while the sum of all entries in the "picture" is $$ \sum_{i=k+1}^{n-k} \sum_{j=k+1}^{n-k} ij = \left( \sum_{i=k+1}^{n-k} i \right)^2 = (n+1)^2 \left(\frac{1}{2}n-k \right)^2$$

Thus, we seek a pair $n,k$ with $$ \frac{1}{2}\left(\frac{1}{2}n(n+1) \right) ^2 = (n+1)^2 \left(\frac{1}{2}n-k \right)^2.$$ This last equation simplifies to $$0=n^2-8kn+8k^2$$ from which we may conclude that $$n=(4 \pm \sqrt{8}) k.$$ Since $k$ and $n$ must be integers, we may conclude that there is no such pair.

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Excellent; thanks! –  Benjamin Dickman Nov 5 '12 at 1:59
    
@B.D Thanks for the extra points! Cheers! –  Matthew Conroy Nov 8 '12 at 16:43
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