2

EDIT. Put another way, we know from the summation that each line must equal 15. 4 lines should equal 60, but there is some over-counting. How can we know that this over-counting is by 15, without knowing in advance that the middle square is worth 5?

I know there are answers on here proving why the centre square of a 3x3 magic square must be 5. However, I would like help understanding a specific explanation because it bothers me that something explained casually on a video as an "of course" does not yet make sense to me.

Based on this image:

enter image description here

the instructor claims that $4S = total sum + 3.centre$.

I think variable names may be causing me some confusion here. What is $totalsum$? I thought originality it meant the sum of the 4 lines, but then the instructor goes on to say that we know $totalsum$ is $3S$ which we know is $45$.

I know that $4S = 60$, and I can see that the centre square is counted 4 times. Beyond this though, the penny is stuck in the slot..

I would be grateful if someone could provide an explanation based on the approach given, but filling in the gaps and clarifying the argument.

The transcript from the lesson is below. Please note that the transcription software has made things sound worse than they actually are, even though for me the explanation was not sufficiently clear to understand in spite of a sincere effort to do so.

And the trick is that we can predict what is in the centre. So look, we have sum up four lines. This is the one line, this is the second line, third line and fourth line. First we know that if we sum all these numbers along these lines, we get four times the constant, this s, which is the sum of each line and each diagonal. So we have 4S. On the other hand, we can see that everything is counted exactly once, except for the centre which is count for four lines, it's count four times. So the centre, there is additional to 3×center. So 4S is the total sum which we know is 3S and 3×center. So the centre is one third, so this four and three, so the difference is S. And the centre is one third of the sum on every line. In our case is five. So, here is the equation and here is the solution in general, and in our case it's five.

Robin Andrews
  • 957

3 Answers3

2

This should be a cleaned-up version of the given explanation. Let the square be $$\begin{bmatrix} a&b&c\\d&e&f\\g&h&i\end{bmatrix}$$ Then $S$ is the sum of any row or column or diagonal. We are using $1$ to $9$ exactly once, and the three rows use each cell exactly once, so $$(a+b+c)+(d+e+f)+(g+h+i)=3S=1+\cdots+9=45\implies S=15\tag1$$ Now we take the sum of all lines passing through the centre: $$(\color{blue}a+\color{blue}e+\color{blue}i)+(\color{blue}b+e+\color{blue}h)+\color{blue}c+e+\color{blue}g)+(\color{blue}d+e+\color{blue}f)=4S\tag2$$ The blue variables are precisely the three-line sum $(1)$ we have already worked out. Thus $(2)=4S$ equals $3S+3e$, i.e. $S=3e$ and $e=5$.

Parcly Taxel
  • 103,344
  • I have a suspicion that the meaning of "total sum" switches mid explanation. It makes sense to me that $4S$ is the "total sum" of the 4 lines when the term $4S$ is first introduced. Does that seem like a reasonable interpretation? In this case am I right that there is an unannounced switch in meaning, or am I just making things complicated in my head? – Robin Andrews Sep 24 '22 at 09:31
  • If not, I think there is a step missing. Why would $4S$ equal the total sum of the whole grid, adjusted by compensating for overcounting the centre? I mean there are 8 lines in total... – Robin Andrews Sep 24 '22 at 09:35
  • Well that's radically confusing, IMO. And my poor foggy brain is not able to disambiguate yet. – Robin Andrews Sep 24 '22 at 09:38
  • Is insufficient disambiguation not a type of "accidental misdirection?" A computer program would fail on such an error. – Robin Andrews Sep 24 '22 at 09:41
  • @RobinAndrews $4S$ is the total sum, and (in my opinion) there is no switch in what $S$ represents. My take on the posting is that $S$ is consistently used to represent the constant value that is the sum of 3 cells that are in the same row, or same column, or along either of the two main diagonals. The first thing to realize is that since $$\sum_{i=1}^9 (i) = 45,$$ that $S$ must equal $(15)$. The posting consistently (in effect) assigns this value to $S$. This explains why $$60 = 4S = \left[\sum_{i=1}^9 (i)\right] + 3M,$$ where $M$ is the value of the middle cell. ...see next comment. – user2661923 Sep 24 '22 at 09:42
  • The posting is not particularly well presented, and I found the explanation to be somewhat convoluted in parts, but the presentation is consistently trying to communicate the message in my previous comment. – user2661923 Sep 24 '22 at 09:44
  • $4S$ initially is the sum of the 4 lines, when first introduced, I'm assuming, perhaps incorrectly? I'm still having trouble seeing why the sum of 4 arbitrary lines can be considered to be related to the sum of the numbers in the whole grid in the precisely defined way of being equal to that sum plus 3 times the middle square. I probably need to ponder a bit more. It would be interesting to see the explanation with distinct variables used: one for the sum of a single line, maybe 1 for 4 times that number, and one for the sum of the whole grid. I'm hoping for an "aha moment." – Robin Andrews Sep 24 '22 at 09:52
  • I've just seen your edit. I've also just realised that the 4 lines represent a complete covering of the grid, with some over-counting of the middle square. This hadn't occurred to me before an so I couldn't see why we could relate the sum of the 4 lines to the summation of the grid. I think the fog is clearing... Thanks for your patience. – Robin Andrews Sep 24 '22 at 13:32
2

The "total sum" is $T=1+2+ ... +9=45.$

Each of the 3 rows has the same sum, and the sum of all the members of all 3 rows is $T.$ So the sum of each row is $T/3=15.$ Each diagonal and each column also has this sum $15.$

Let $C$ be the number in the center. Each of the 4 green lines sums to $15.$ So adding these sums gives a value $\quad 4$x$15=60.$ But this adds each non-center cell once, plus $4$x$C.$

This is the same as adding $every$ cell once (giving $T$), plus $3$x$C.$

That is, the sum of the sums of the 4 green lines is $T+3$x$C.$

Therefore $60=T+3$x$C=45+3$x$C.$

DanielWainfleet
  • 57,985
  • Thanks. If the question were phrased "some numbers add up to 60. They should add up to 45. Exactly one number has been counted 4.times instead of once. What is the value of that number?," would that capture the essence of this problem and be solvable please? – Robin Andrews Sep 24 '22 at 13:41
  • @RobinAndrews . I don't understand your comment. – DanielWainfleet Sep 24 '22 at 13:47
  • By considering analogous expressions of a problem we can gain insight that helps us understand the original expression. I'm confident now that the answer to the two questions in the above comment is yes. So we can write 60=45+3x and easily solve. My obstacle was I didn't initially realise the 4 lines represented a covering of the whole grid. Very obvious in hindsight, but explicitly drawing attention to this would have made the initial explanation far clearer, IMO. – Robin Andrews Sep 24 '22 at 13:53
2

I appreciate everyone's input. It has helped me to identify the crux of the issue I had understanding the explanation. This was that it was not explicitly pointed out that the 4 lines form a covering of the whole grid with some over-counting, and thus knowing the actual sum for the whole grid makes it easy to represent the problem as $60 = 45 + 3x$.

This point about the 4.lines covering the whole grid is probably so obvious to someone who already understands the explanation that it doesn't seem to be worth mentioning, yet for me the entire explanation hinged on it. One thing that would have helped is a better variable name: "total sum" seemed to me to mean the same as $4S$. I think "grid sum" or similar would have made the situation much clearer.

Robin Andrews
  • 957