300000
English | Français | فارسی | 中文 | Українська | Azerbaijani | ខ្មែរ | Tiếng Việt | Bahasa Melayu | Deutsch | O'zbek | РусскийSegi Empat Ajaib
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300000
English | Français | فارسی | 中文 | Українська | Azerbaijani | ខ្មែរ | Tiếng Việt | Bahasa Melayu | Deutsch | O'zbek | Русский15 * 35 50 * ? 25 * *This is easy to solve. The sum in each line is 15+50+25=90. From the diagonal, it follows that the middle value is 90−25−35=30 and from the 2^{nd} row: 50+30+?=90 ⇒ ? = 10.
? * * * * 47 * 63 *What is the standard way of solving this problem? One could introduce variables
P Q R U V 47 X 63 Zand formulate all conditions as equations such as
P+Q+R = U+V+47 = X+63+Z = P+U+X = Q+V+63 = R+47+Z = P+V+Z = R+V+X,and solve this system of equations by eliminating all unknowns except P to get one equation for P and solve that. These are 7 conditions for 7 unknowns.
P Q R U M W X Y Zthen these quantities that do not change under rotations and mirrorings are
C = sum of corner values (= P+R+X+Z) E = sum of middle edge values (= Q+U+W+Y) M = middle value (= M) S = sum in each line (= P+Q+R = P+U+X = X+M+P = ...) .
2C + E = 4S (1) M + C + E = 3S (2) C + 2M = 2S (3) C − M = S (4) 3M = S (5)
The above relations result as follows:
(1) from summing the 4 edges
(2) from summing all numbers
(3) from summing the 2 diagonals
(4) from (1) − (2)
(5) from (3) − (4)
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Verify (1), (2) and (3) by replacing C, E and S by their sums.
3 2 7 * ? * * * *
Since all 3 numbers in the 1^{st} row are given, we know that S = 3+2+7 = 12. Thus by (5), ? = 12/3 = 4.
11 * 4 * 6 * * ? *
Let Q = the middle value of 1^{st} row. Then by (5), 11+Q+4 = 3×6 ⇒ Q = 18−4−11 = 3.
Thus, ? = 18−6−3 = 9.
* * 5 * 9 * ? * *
By (5), ?+9+5 = 3×9 = 27 ⇒ ? = 27−9−5 = 13.
* * 5 * ? * 9 * *
By (5), 5+?+9 = 3×? ⇒ 14 = 2×? ⇒ ? = 14/2 = 7.
P * * * * * * * 14
Without knowing that this Magic Square is filled with the whole numbers from 7 to 15 inclusive, the one given number would not have been enough. But with this information and by (5),
3S = (7+15) + (8+14) + (9+13) + (10+12) + 11 = 4×2 + 11 = 99 ⇒ S = 33 ⇒ M = 11 ⇒ 14 + 11 + P = 33 ⇒ P = 8.
? * * * * 47 * 63 *Here we need more relations.
Our strategy will be to start with the simplest possible Magic Square
0 0 0 0 0 0 (6) 0 0 0
and find 'deformations', i.e. methods to change the Magic Square into another Magic Square. The first such deformation is to add the same value 1 to all fields:
1 1 1 1 1 1 (7) 1 1 1
Which is also a Magic Square. What other generalizing deformations are there?
Proof.
Because addition is commutative: If 2 Magic Squares
a11 a12 a13 a21 a22 a23 (8) a31 a32 a33
and
b11 b12 b13 b21 b22 b23 (9) b31 b32 b33
are added together:
a11+b11 a12+b12 a13+b13 a21+b21 a22+b22 a23+b23 (10) a31+b31 a32+b32 a33+b33
then this is also a Magic Square because, for example, the first two rows of (10) have equal sums.
a11+b11 + a21+b21 + a31+b31 = a11+a21+a31 + b11+b21+b31 (commutativity of addition) = a12+a22+a32 + b12+b22+b32 (because (8) and (9) are Magic Squares) = a12+b12 + a22+b22 + a32+b32 (which we wanted to show)
Similarly, one can show the equality of sums of other lines of (10).
Because of the distributive law, if (8):
a11 a12 a13 a21 a22 a23 a31 a32 a33
is a Magic Square then
b×a11 b×a12 b×a13 b×a21 b×a22 b×a23 b×a31 b×a32 b×a33
is a Magic Square too because, for example, if
a11 + a12 + a13 = a21 + a22 + a23 then b×(a11 + a12 + a13) = b×(a21 + a22 + a23) and b×a11 + b×a12 + b×a13 = b×a21 + b×a22 + b×a23 .
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1 * * * 0 * * * *Because M=0, then S must also be zero, i.e.
1 * * * 0 * * * −1The upper-right corner can stay 0 but then all other changes are fixed:
+1 −1 0 −1 0 +1 (11) 0 +1 −1This is a Magic Square itself.
0 −1 +1 +1 0 −1 (12) −1 +1 0We now apply theorem 1 three times.
0 * 0 * 0 * * * *with an S value of zero (because the middle value is zero). This easily shows that the whole Magic Square is
0 0 0 0 0 0 0 0 0But, that means each Magic Square can be written as a sum of multiples of squares M×(7) + A×(11) + B×(12).
M+A M−A−B M +B M−A+B M M+A−B (13) M −B M+A+B M−A .
As shown above, each Magic Square can be reduced to the square consisting of only zeros by subtracting multiples of (7), (11) and (12).
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For example, in the upper-left corner in (13), M+A = ((M+A+B)+(M+A−B))/2, and likewise for the other 3 corners.
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? * * * * 47 * 63 *by reducing it to ? = (63+47)/2 = 55. Here is another relation:
By inspection of (13).
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* 80 * * * 56 * * *
First we can apply lemma 2 to get value in the lower-left corner the average (80+56)/2 = 136/2 = 68:
* 80 * * * 56 68 * *
Then, after adding 0 in the middle and mirroring numbers in the middle by switching signs:
* 80 −68 −56 0 56 68 −80 *
Finally since all sums are 0, the upper-left corner must be 68−80 = −12. Therefore, the simplest solution of this example is:
−12 80 −68 −56 0 56 68 −80 12
P * * U ? W X * *where P, U, X, W are known and we want to find ?.
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