Reverse engineering a math magic trick involving matrices.
This problem was brought up by my mother from a corporate party along with a question on how that worked.
There was a showman who asked to tell him a number from $10$ to $99$ (If i'm not mistaken). The number $83$ was named after which he took a piece of paper and quickly put down a matrix (note he did that fast):
$$
begin{bmatrix}
8 & 11 & 63 & 1\
62 & 2 & 7 & 12\
3 & 65 & 9 & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
If you take a closer look every row, column and diagonal has the sum of $83$. Moreover consider the corners of the matrix also have the sum of $83$. For example:
$$
begin{bmatrix}
colorred{8} & colorred{11} & 63 & 1\
colorred{62} & colorred{2} & 7 & 12\
3 & 65 & 9 & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
Also the central square is $83$ in sum as well:
$$
begin{bmatrix}
8 & 11 & 63 & 1\
62 & colorred{2} & colorred{7} & 12\
3 & colorred{65} & colorred{9} & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
Clearly numbers $1,2,3,4,5,6,7,8,9,10,11,12$ are filled in in a circular manner. And then consequent $62, 63, 64, 65$ are as well. I'm not very familial with linear algebra so my question is:
What was that rule he used to build it? Can we construct a matrix with the same properties given a random number in some range? Is it possible to build a similar one but for $5times 5$, $6times 6$ or $Ntimes N$ matrix?
recreational-mathematics magic-square
edited Dec 22 at 21:20
anomaly
17.3k42663
asked Dec 22 at 12:48
roman
1,92921221
|
show 5 more comments
This problem was brought up by my mother from a corporate party along with a question on how that worked.
There was a showman who asked to tell him a number from $10$ to $99$ (If i'm not mistaken). The number $83$ was named after which he took a piece of paper and quickly put down a matrix (note he did that fast):
$$
begin{bmatrix}
8 & 11 & 63 & 1\
62 & 2 & 7 & 12\
3 & 65 & 9 & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
If you take a closer look every row, column and diagonal has the sum of $83$. Moreover consider the corners of the matrix also have the sum of $83$. For example:
$$
begin{bmatrix}
colorred{8} & colorred{11} & 63 & 1\
colorred{62} & colorred{2} & 7 & 12\
3 & 65 & 9 & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
Also the central square is $83$ in sum as well:
$$
begin{bmatrix}
8 & 11 & 63 & 1\
62 & colorred{2} & colorred{7} & 12\
3 & colorred{65} & colorred{9} & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
Clearly numbers $1,2,3,4,5,6,7,8,9,10,11,12$ are filled in in a circular manner. And then consequent $62, 63, 64, 65$ are as well. I'm not very familial with linear algebra so my question is:
What was that rule he used to build it? Can we construct a matrix with the same properties given a random number in some range? Is it possible to build a similar one but for $5times 5$, $6times 6$ or $Ntimes N$ matrix?
recreational-mathematics magic-square
edited Dec 22 at 21:20
anomaly
17.3k42663
asked Dec 22 at 12:48
roman
1,92921221
What kind of show was that?
– lcv
Dec 22 at 13:32
10
It’s hard to imagine a less geeky show :-)
– lcv
Dec 22 at 13:38
8
This feels more like recreational mathematics than linear algebra.
– Andreas Rejbrand
Dec 22 at 16:50
1
Reminds me of a paper I read before about constructing magic squares. Have a look at this: researchgate.net/publication/…
– Zushauque
Dec 22 at 20:00
1
To show that you can get any sum on the rows, columns, and two main diagonals for any $N times N$ magic square ($N ge 4$), see my Answer here: math.stackexchange.com/questions/1992336/…
– Bram28
Dec 23 at 15:40
|
show 5 more comments
This problem was brought up by my mother from a corporate party along with a question on how that worked.
There was a showman who asked to tell him a number from $10$ to $99$ (If i'm not mistaken). The number $83$ was named after which he took a piece of paper and quickly put down a matrix (note he did that fast):
$$
begin{bmatrix}
8 & 11 & 63 & 1\
62 & 2 & 7 & 12\
3 & 65 & 9 & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
If you take a closer look every row, column and diagonal has the sum of $83$. Moreover consider the corners of the matrix also have the sum of $83$. For example:
$$
begin{bmatrix}
colorred{8} & colorred{11} & 63 & 1\
colorred{62} & colorred{2} & 7 & 12\
3 & 65 & 9 & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
Also the central square is $83$ in sum as well:
$$
begin{bmatrix}
8 & 11 & 63 & 1\
62 & colorred{2} & colorred{7} & 12\
3 & colorred{65} & colorred{9} & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
Clearly numbers $1,2,3,4,5,6,7,8,9,10,11,12$ are filled in in a circular manner. And then consequent $62, 63, 64, 65$ are as well. I'm not very familial with linear algebra so my question is:
What was that rule he used to build it? Can we construct a matrix with the same properties given a random number in some range? Is it possible to build a similar one but for $5times 5$, $6times 6$ or $Ntimes N$ matrix?
recreational-mathematics magic-square
edited Dec 22 at 21:20
anomaly
17.3k42663
asked Dec 22 at 12:48
roman
1,92921221
This problem was brought up by my mother from a corporate party along with a question on how that worked.
There was a showman who asked to tell him a number from $10$ to $99$ (If i'm not mistaken). The number $83$ was named after which he took a piece of paper and quickly put down a matrix (note he did that fast):
$$
begin{bmatrix}
8 & 11 & 63 & 1\
62 & 2 & 7 & 12\
3 & 65 & 9 & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
If you take a closer look every row, column and diagonal has the sum of $83$. Moreover consider the corners of the matrix also have the sum of $83$. For example:
$$
begin{bmatrix}
colorred{8} & colorred{11} & 63 & 1\
colorred{62} & colorred{2} & 7 & 12\
3 & 65 & 9 & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
Also the central square is $83$ in sum as well:
$$
begin{bmatrix}
8 & 11 & 63 & 1\
62 & colorred{2} & colorred{7} & 12\
3 & colorred{65} & colorred{9} & 6 \
10 & 5 & 4 & 64
end{bmatrix}
$$
Clearly numbers $1,2,3,4,5,6,7,8,9,10,11,12$ are filled in in a circular manner. And then consequent $62, 63, 64, 65$ are as well. I'm not very familial with linear algebra so my question is:
What was that rule he used to build it? Can we construct a matrix with the same properties given a random number in some range? Is it possible to build a similar one but for $5times 5$, $6times 6$ or $Ntimes N$ matrix?
recreational-mathematics magic-square
recreational-mathematics magic-square
edited Dec 22 at 21:20
anomaly
17.3k42663
asked Dec 22 at 12:48
roman
1,92921221
edited Dec 22 at 21:20
anomaly
17.3k42663
asked Dec 22 at 12:48
roman
1,92921221
edited Dec 22 at 21:20
anomaly
17.3k42663
edited Dec 22 at 21:20
anomaly
17.3k42663
edited Dec 22 at 21:20
anomaly
17.3k42663
17.3k42663
asked Dec 22 at 12:48
roman
1,92921221
asked Dec 22 at 12:48
roman
1,92921221
asked Dec 22 at 12:48
roman
1,92921221
1,92921221
What kind of show was that?
– lcv
Dec 22 at 13:32
10
It’s hard to imagine a less geeky show :-)
– lcv
Dec 22 at 13:38
8
This feels more like recreational mathematics than linear algebra.
– Andreas Rejbrand
Dec 22 at 16:50
1
Reminds me of a paper I read before about constructing magic squares. Have a look at this: researchgate.net/publication/…
– Zushauque
Dec 22 at 20:00
1
To show that you can get any sum on the rows, columns, and two main diagonals for any $N times N$ magic square ($N ge 4$), see my Answer here: math.stackexchange.com/questions/1992336/…
– Bram28
Dec 23 at 15:40
|
show 5 more comments
What kind of show was that?
– lcv
Dec 22 at 13:32
10
It’s hard to imagine a less geeky show :-)
– lcv
Dec 22 at 13:38
8
This feels more like recreational mathematics than linear algebra.
– Andreas Rejbrand
Dec 22 at 16:50
1
Reminds me of a paper I read before about constructing magic squares. Have a look at this: researchgate.net/publication/…
– Zushauque
Dec 22 at 20:00
1
To show that you can get any sum on the rows, columns, and two main diagonals for any $N times N$ magic square ($N ge 4$), see my Answer here: math.stackexchange.com/questions/1992336/…
– Bram28
Dec 23 at 15:40
What kind of show was that?
– lcv
Dec 22 at 13:32
What kind of show was that?
– lcv
Dec 22 at 13:32
10
10
It’s hard to imagine a less geeky show :-)
– lcv
Dec 22 at 13:38
It’s hard to imagine a less geeky show :-)
– lcv
Dec 22 at 13:38
8
8
This feels more like recreational mathematics than linear algebra.
– Andreas Rejbrand
Dec 22 at 16:50
This feels more like recreational mathematics than linear algebra.
– Andreas Rejbrand
Dec 22 at 16:50
1
1
Reminds me of a paper I read before about constructing magic squares. Have a look at this: researchgate.net/publication/…
– Zushauque
Dec 22 at 20:00
Reminds me of a paper I read before about constructing magic squares. Have a look at this: researchgate.net/publication/…
– Zushauque
Dec 22 at 20:00
1
1
To show that you can get any sum on the rows, columns, and two main diagonals for any $N times N$ magic square ($N ge 4$), see my Answer here: math.stackexchange.com/questions/1992336/…
– Bram28
Dec 23 at 15:40
To show that you can get any sum on the rows, columns, and two main diagonals for any $N times N$ magic square ($N ge 4$), see my Answer here: math.stackexchange.com/questions/1992336/…
– Bram28
Dec 23 at 15:40
|
show 5 more comments
3 Answers
3
active
oldest
votes
This is all based on the amazing matrix $$pmatrix{8&11&14&1\13&2&7&12\3&16&9&6\10&5&4&15}$$
Note that this works for $34$. For $34+n$ you just replace $13,14,15,16$ with $13+n,14+n,15+n,16+n$. I do not know what the showman's backup plan for numbers below $34$ is.
EDIT: There is a Numberphile video on YouTube on this exact trick.
edited Dec 23 at 21:43
Rodrigo de Azevedo
12.8k41854
answered Dec 22 at 13:05
SmileyCraft
2,424413
5
I also believe it was not 10, but rather some larger two digit number
– roman
Dec 22 at 13:19
2
Thanks for sharing a link to the video! Also it looks like magic does not exist after all :)
– roman
Dec 22 at 15:30
1
It's possible that his backup plan was to subtract some number from those four.
– Jarred Allen
Dec 23 at 18:36
@JarredAllen Or perhaps use a smaller matrix? Subtract all numbers from 100, effectively inverting the trick? No clue, but there are possibilities indeed.
– Mast
Dec 24 at 6:45
add a comment |
He did not write down 'a matrix'. He wrote down a magic square.
There are various algorithms for constructing them, the easiest to do mentally is to start with a generic starter matrix you memoize:
A 1 12 7
11 8 B 2
5 10 3 C
4 D 6 9
And solve for A, B, C, D for your target number. With some practice you can do this super fast.
answered Dec 22 at 13:06
orlp
7,3411230
8
Of course it is a matrix as well.
– Jannik Pitt
Dec 22 at 13:27
4
@JannikPitt Of course. I just remarked on it as if someone arriving by private helicopter to do grocery shopping is described as "arriving by vehicle".
– orlp
Dec 22 at 17:16
add a comment |
Observe that the numbers in the sixties lie on different rows, different columns and different quadrants. If you increase them simultaneously, all the twelve sums increase by the same amount.
In the pure magic square (numbers $1$ to $16$), the sum is $34$, hence if suffices to add $83-34=49$ to these four elements.
It would be more logical to request a sum in the range $[34,83]$ so that all elements are at most two-digit naturals.
For stronger mystification, you can split the required increment in two or more and adjust on the other possible groups.
answered Dec 22 at 13:49
Yves Daoust
124k671221
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049428%2freverse-engineering-a-math-magic-trick-involving-matrices%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is all based on the amazing matrix $$pmatrix{8&11&14&1\13&2&7&12\3&16&9&6\10&5&4&15}$$
Note that this works for $34$. For $34+n$ you just replace $13,14,15,16$ with $13+n,14+n,15+n,16+n$. I do not know what the showman's backup plan for numbers below $34$ is.
EDIT: There is a Numberphile video on YouTube on this exact trick.
edited Dec 23 at 21:43
Rodrigo de Azevedo
12.8k41854
answered Dec 22 at 13:05
SmileyCraft
2,424413
5
I also believe it was not 10, but rather some larger two digit number
– roman
Dec 22 at 13:19
2
Thanks for sharing a link to the video! Also it looks like magic does not exist after all :)
– roman
Dec 22 at 15:30
1
It's possible that his backup plan was to subtract some number from those four.
– Jarred Allen
Dec 23 at 18:36
@JarredAllen Or perhaps use a smaller matrix? Subtract all numbers from 100, effectively inverting the trick? No clue, but there are possibilities indeed.
– Mast
Dec 24 at 6:45
add a comment |
This is all based on the amazing matrix $$pmatrix{8&11&14&1\13&2&7&12\3&16&9&6\10&5&4&15}$$
Note that this works for $34$. For $34+n$ you just replace $13,14,15,16$ with $13+n,14+n,15+n,16+n$. I do not know what the showman's backup plan for numbers below $34$ is.
EDIT: There is a Numberphile video on YouTube on this exact trick.
edited Dec 23 at 21:43
Rodrigo de Azevedo
12.8k41854
answered Dec 22 at 13:05
SmileyCraft
2,424413
5
I also believe it was not 10, but rather some larger two digit number
– roman
Dec 22 at 13:19
2
Thanks for sharing a link to the video! Also it looks like magic does not exist after all :)
– roman
Dec 22 at 15:30
1
It's possible that his backup plan was to subtract some number from those four.
– Jarred Allen
Dec 23 at 18:36
@JarredAllen Or perhaps use a smaller matrix? Subtract all numbers from 100, effectively inverting the trick? No clue, but there are possibilities indeed.
– Mast
Dec 24 at 6:45
add a comment |
This is all based on the amazing matrix $$pmatrix{8&11&14&1\13&2&7&12\3&16&9&6\10&5&4&15}$$
Note that this works for $34$. For $34+n$ you just replace $13,14,15,16$ with $13+n,14+n,15+n,16+n$. I do not know what the showman's backup plan for numbers below $34$ is.
EDIT: There is a Numberphile video on YouTube on this exact trick.
edited Dec 23 at 21:43
Rodrigo de Azevedo
12.8k41854
answered Dec 22 at 13:05
SmileyCraft
2,424413
This is all based on the amazing matrix $$pmatrix{8&11&14&1\13&2&7&12\3&16&9&6\10&5&4&15}$$
Note that this works for $34$. For $34+n$ you just replace $13,14,15,16$ with $13+n,14+n,15+n,16+n$. I do not know what the showman's backup plan for numbers below $34$ is.
EDIT: There is a Numberphile video on YouTube on this exact trick.
edited Dec 23 at 21:43
Rodrigo de Azevedo
12.8k41854
answered Dec 22 at 13:05
SmileyCraft
2,424413
edited Dec 23 at 21:43
Rodrigo de Azevedo
12.8k41854
edited Dec 23 at 21:43
Rodrigo de Azevedo
12.8k41854
edited Dec 23 at 21:43
Rodrigo de Azevedo
12.8k41854
12.8k41854
answered Dec 22 at 13:05
SmileyCraft
2,424413
answered Dec 22 at 13:05
SmileyCraft
2,424413
answered Dec 22 at 13:05
SmileyCraft
2,424413
2,424413
5
I also believe it was not 10, but rather some larger two digit number
– roman
Dec 22 at 13:19
2
Thanks for sharing a link to the video! Also it looks like magic does not exist after all :)
– roman
Dec 22 at 15:30
1
It's possible that his backup plan was to subtract some number from those four.
– Jarred Allen
Dec 23 at 18:36
@JarredAllen Or perhaps use a smaller matrix? Subtract all numbers from 100, effectively inverting the trick? No clue, but there are possibilities indeed.
– Mast
Dec 24 at 6:45
add a comment |
5
I also believe it was not 10, but rather some larger two digit number
– roman
Dec 22 at 13:19
2
Thanks for sharing a link to the video! Also it looks like magic does not exist after all :)
– roman
Dec 22 at 15:30
1
It's possible that his backup plan was to subtract some number from those four.
– Jarred Allen
Dec 23 at 18:36
@JarredAllen Or perhaps use a smaller matrix? Subtract all numbers from 100, effectively inverting the trick? No clue, but there are possibilities indeed.
– Mast
Dec 24 at 6:45
5
5
I also believe it was not 10, but rather some larger two digit number
– roman
Dec 22 at 13:19
I also believe it was not 10, but rather some larger two digit number
– roman
Dec 22 at 13:19
2
2
Thanks for sharing a link to the video! Also it looks like magic does not exist after all :)
– roman
Dec 22 at 15:30
Thanks for sharing a link to the video! Also it looks like magic does not exist after all :)
– roman
Dec 22 at 15:30
1
1
It's possible that his backup plan was to subtract some number from those four.
– Jarred Allen
Dec 23 at 18:36
It's possible that his backup plan was to subtract some number from those four.
– Jarred Allen
Dec 23 at 18:36
@JarredAllen Or perhaps use a smaller matrix? Subtract all numbers from 100, effectively inverting the trick? No clue, but there are possibilities indeed.
– Mast
Dec 24 at 6:45
@JarredAllen Or perhaps use a smaller matrix? Subtract all numbers from 100, effectively inverting the trick? No clue, but there are possibilities indeed.
– Mast
Dec 24 at 6:45
add a comment |
He did not write down 'a matrix'. He wrote down a magic square.
There are various algorithms for constructing them, the easiest to do mentally is to start with a generic starter matrix you memoize:
A 1 12 7
11 8 B 2
5 10 3 C
4 D 6 9
And solve for A, B, C, D for your target number. With some practice you can do this super fast.
answered Dec 22 at 13:06
orlp
7,3411230
8
Of course it is a matrix as well.
– Jannik Pitt
Dec 22 at 13:27
4
@JannikPitt Of course. I just remarked on it as if someone arriving by private helicopter to do grocery shopping is described as "arriving by vehicle".
– orlp
Dec 22 at 17:16
add a comment |
He did not write down 'a matrix'. He wrote down a magic square.
There are various algorithms for constructing them, the easiest to do mentally is to start with a generic starter matrix you memoize:
A 1 12 7
11 8 B 2
5 10 3 C
4 D 6 9
And solve for A, B, C, D for your target number. With some practice you can do this super fast.
answered Dec 22 at 13:06
orlp
7,3411230
8
Of course it is a matrix as well.
– Jannik Pitt
Dec 22 at 13:27
4
@JannikPitt Of course. I just remarked on it as if someone arriving by private helicopter to do grocery shopping is described as "arriving by vehicle".
– orlp
Dec 22 at 17:16
add a comment |
He did not write down 'a matrix'. He wrote down a magic square.
There are various algorithms for constructing them, the easiest to do mentally is to start with a generic starter matrix you memoize:
A 1 12 7
11 8 B 2
5 10 3 C
4 D 6 9
And solve for A, B, C, D for your target number. With some practice you can do this super fast.
answered Dec 22 at 13:06
orlp
7,3411230
He did not write down 'a matrix'. He wrote down a magic square.
There are various algorithms for constructing them, the easiest to do mentally is to start with a generic starter matrix you memoize:
A 1 12 7
11 8 B 2
5 10 3 C
4 D 6 9
And solve for A, B, C, D for your target number. With some practice you can do this super fast.
answered Dec 22 at 13:06
orlp
7,3411230
edited Dec 22 at 17:16
answered Dec 22 at 13:06
orlp
7,3411230
answered Dec 22 at 13:06
orlp
7,3411230
answered Dec 22 at 13:06
orlp
7,3411230
7,3411230
8
Of course it is a matrix as well.
– Jannik Pitt
Dec 22 at 13:27
4
@JannikPitt Of course. I just remarked on it as if someone arriving by private helicopter to do grocery shopping is described as "arriving by vehicle".
– orlp
Dec 22 at 17:16
add a comment |
8
Of course it is a matrix as well.
– Jannik Pitt
Dec 22 at 13:27
4
@JannikPitt Of course. I just remarked on it as if someone arriving by private helicopter to do grocery shopping is described as "arriving by vehicle".
– orlp
Dec 22 at 17:16
8
8
Of course it is a matrix as well.
– Jannik Pitt
Dec 22 at 13:27
Of course it is a matrix as well.
– Jannik Pitt
Dec 22 at 13:27
4
4
@JannikPitt Of course. I just remarked on it as if someone arriving by private helicopter to do grocery shopping is described as "arriving by vehicle".
– orlp
Dec 22 at 17:16
@JannikPitt Of course. I just remarked on it as if someone arriving by private helicopter to do grocery shopping is described as "arriving by vehicle".
– orlp
Dec 22 at 17:16
add a comment |
Observe that the numbers in the sixties lie on different rows, different columns and different quadrants. If you increase them simultaneously, all the twelve sums increase by the same amount.
In the pure magic square (numbers $1$ to $16$), the sum is $34$, hence if suffices to add $83-34=49$ to these four elements.
It would be more logical to request a sum in the range $[34,83]$ so that all elements are at most two-digit naturals.
For stronger mystification, you can split the required increment in two or more and adjust on the other possible groups.
answered Dec 22 at 13:49
Yves Daoust
124k671221
add a comment |
Observe that the numbers in the sixties lie on different rows, different columns and different quadrants. If you increase them simultaneously, all the twelve sums increase by the same amount.
In the pure magic square (numbers $1$ to $16$), the sum is $34$, hence if suffices to add $83-34=49$ to these four elements.
It would be more logical to request a sum in the range $[34,83]$ so that all elements are at most two-digit naturals.
For stronger mystification, you can split the required increment in two or more and adjust on the other possible groups.
answered Dec 22 at 13:49
Yves Daoust
124k671221
add a comment |
Observe that the numbers in the sixties lie on different rows, different columns and different quadrants. If you increase them simultaneously, all the twelve sums increase by the same amount.
In the pure magic square (numbers $1$ to $16$), the sum is $34$, hence if suffices to add $83-34=49$ to these four elements.
It would be more logical to request a sum in the range $[34,83]$ so that all elements are at most two-digit naturals.
For stronger mystification, you can split the required increment in two or more and adjust on the other possible groups.
answered Dec 22 at 13:49
Yves Daoust
124k671221
Observe that the numbers in the sixties lie on different rows, different columns and different quadrants. If you increase them simultaneously, all the twelve sums increase by the same amount.
In the pure magic square (numbers $1$ to $16$), the sum is $34$, hence if suffices to add $83-34=49$ to these four elements.
It would be more logical to request a sum in the range $[34,83]$ so that all elements are at most two-digit naturals.
For stronger mystification, you can split the required increment in two or more and adjust on the other possible groups.
answered Dec 22 at 13:49
Yves Daoust
124k671221
edited Dec 22 at 13:55
answered Dec 22 at 13:49
Yves Daoust
124k671221
answered Dec 22 at 13:49
Yves Daoust
124k671221
answered Dec 22 at 13:49
Yves Daoust
124k671221
124k671221
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049428%2freverse-engineering-a-math-magic-trick-involving-matrices%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
What kind of show was that?
– lcv
Dec 22 at 13:32
10
It’s hard to imagine a less geeky show :-)
– lcv
Dec 22 at 13:38
8
This feels more like recreational mathematics than linear algebra.
– Andreas Rejbrand
Dec 22 at 16:50
1
Reminds me of a paper I read before about constructing magic squares. Have a look at this: researchgate.net/publication/…
– Zushauque
Dec 22 at 20:00
1
To show that you can get any sum on the rows, columns, and two main diagonals for any $N times N$ magic square ($N ge 4$), see my Answer here: math.stackexchange.com/questions/1992336/…
– Bram28
Dec 23 at 15:40