Arrange $n$ tiles in a rectangle pattern
Is there a way to find how to arrange $n$ tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.
For example, here are the results I expect from $1$ to $9$.
- a 1 by 1 square, fairly simple.
- a 1 by 2 rectangle.
- a 1 by 3 rectangle.
- a 2 by 2 square.
- this is impossible because a 1 by 5 rectangle is thinner than the previous shape.
- a 2 by 3 rectangle.
- this is impossible.
- a 2 by 4 rectangle.
- a 3 by 3 square.
I am only looking at $n$ smaller than $100$ but would rather not have to hardcode everything.
geometry square-numbers rectangles
edited 2 days ago
dmtri
1,4201521
asked Dec 31 '18 at 2:18
Alex Laquerre
161
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Is there a way to find how to arrange $n$ tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.
For example, here are the results I expect from $1$ to $9$.
- a 1 by 1 square, fairly simple.
- a 1 by 2 rectangle.
- a 1 by 3 rectangle.
- a 2 by 2 square.
- this is impossible because a 1 by 5 rectangle is thinner than the previous shape.
- a 2 by 3 rectangle.
- this is impossible.
- a 2 by 4 rectangle.
- a 3 by 3 square.
I am only looking at $n$ smaller than $100$ but would rather not have to hardcode everything.
geometry square-numbers rectangles
edited 2 days ago
dmtri
1,4201521
asked Dec 31 '18 at 2:18
Alex Laquerre
161
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Is there a way to find how to arrange $n$ tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.
For example, here are the results I expect from $1$ to $9$.
- a 1 by 1 square, fairly simple.
- a 1 by 2 rectangle.
- a 1 by 3 rectangle.
- a 2 by 2 square.
- this is impossible because a 1 by 5 rectangle is thinner than the previous shape.
- a 2 by 3 rectangle.
- this is impossible.
- a 2 by 4 rectangle.
- a 3 by 3 square.
I am only looking at $n$ smaller than $100$ but would rather not have to hardcode everything.
geometry square-numbers rectangles
edited 2 days ago
dmtri
1,4201521
asked Dec 31 '18 at 2:18
Alex Laquerre
161
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there a way to find how to arrange $n$ tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.
For example, here are the results I expect from $1$ to $9$.
- a 1 by 1 square, fairly simple.
- a 1 by 2 rectangle.
- a 1 by 3 rectangle.
- a 2 by 2 square.
- this is impossible because a 1 by 5 rectangle is thinner than the previous shape.
- a 2 by 3 rectangle.
- this is impossible.
- a 2 by 4 rectangle.
- a 3 by 3 square.
I am only looking at $n$ smaller than $100$ but would rather not have to hardcode everything.
geometry square-numbers rectangles
geometry square-numbers rectangles
edited 2 days ago
dmtri
1,4201521
asked Dec 31 '18 at 2:18
Alex Laquerre
161
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
dmtri
1,4201521
asked Dec 31 '18 at 2:18
Alex Laquerre
161
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
dmtri
1,4201521
edited 2 days ago
dmtri
1,4201521
edited 2 days ago
dmtri
1,4201521
1,4201521
asked Dec 31 '18 at 2:18
Alex Laquerre
161
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 31 '18 at 2:18
Alex Laquerre
161
asked Dec 31 '18 at 2:18
Alex Laquerre
161
161
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
As previously written by Ross Millikan in the comments below, stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 days ago
John Omielan
99418
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 days ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 days ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 days ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
2 days ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
2 days ago
|
show 7 more comments
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
});
}
});
Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3057384%2farrange-n-tiles-in-a-rectangle-pattern%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
As previously written by Ross Millikan in the comments below, stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 days ago
John Omielan
99418
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 days ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 days ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 days ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
2 days ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
2 days ago
|
show 7 more comments
Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
As previously written by Ross Millikan in the comments below, stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 days ago
John Omielan
99418
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 days ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 days ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 days ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
2 days ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
2 days ago
|
show 7 more comments
Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
As previously written by Ross Millikan in the comments below, stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 days ago
John Omielan
99418
Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
As previously written by Ross Millikan in the comments below, stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 days ago
John Omielan
99418
edited 2 days ago
answered 2 days ago
John Omielan
99418
answered 2 days ago
John Omielan
99418
answered 2 days ago
John Omielan
99418
99418
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 days ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 days ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 days ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
2 days ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
2 days ago
|
show 7 more comments
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 days ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 days ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 days ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
2 days ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
2 days ago
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 days ago
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 days ago
2
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 days ago
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 days ago
1
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 days ago
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 days ago
1
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
2 days ago
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
2 days ago
1
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
2 days ago
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
2 days ago
|
show 7 more comments
Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3057384%2farrange-n-tiles-in-a-rectangle-pattern%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
