Arrange n tiles in a rectangle pattern












1














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.




  1. a 1 by 1 square, fairly simple.

  2. a 1 by 2 rectangle.

  3. a 1 by 3 rectangle.

  4. a 2 by 2 square.

  5. this is impossible because a 1 by 5 rectangle is thinner than the previous shape.

  6. a 2 by 3 rectangle.

  7. this is impossible.

  8. a 2 by 4 rectangle.

  9. 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







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    1














    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.




    1. a 1 by 1 square, fairly simple.

    2. a 1 by 2 rectangle.

    3. a 1 by 3 rectangle.

    4. a 2 by 2 square.

    5. this is impossible because a 1 by 5 rectangle is thinner than the previous shape.

    6. a 2 by 3 rectangle.

    7. this is impossible.

    8. a 2 by 4 rectangle.

    9. 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







    share|cite|improve this question







    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.























      1












      1








      1







      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.




      1. a 1 by 1 square, fairly simple.

      2. a 1 by 2 rectangle.

      3. a 1 by 3 rectangle.

      4. a 2 by 2 square.

      5. this is impossible because a 1 by 5 rectangle is thinner than the previous shape.

      6. a 2 by 3 rectangle.

      7. this is impossible.

      8. a 2 by 4 rectangle.

      9. 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







      share|cite|improve this question







      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.




      1. a 1 by 1 square, fairly simple.

      2. a 1 by 2 rectangle.

      3. a 1 by 3 rectangle.

      4. a 2 by 2 square.

      5. this is impossible because a 1 by 5 rectangle is thinner than the previous shape.

      6. a 2 by 3 rectangle.

      7. this is impossible.

      8. a 2 by 4 rectangle.

      9. 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






      share|cite|improve this question







      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.











      share|cite|improve this question







      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.









      share|cite|improve this question




      share|cite|improve this question






      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 3 hours ago









      Alex Laquerre

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      61




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      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.






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      Check out our Code of Conduct.






















          1 Answer
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          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.



          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.






          share|cite|improve this answer























          • 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 hours 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 hours 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 hours 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
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour ago











          Your Answer





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          4














          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.



          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.






          share|cite|improve this answer























          • 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 hours 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 hours 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 hours 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
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour ago
















          4














          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.



          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.






          share|cite|improve this answer























          • 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 hours 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 hours 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 hours 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
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour ago














          4












          4








          4






          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.



          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.






          share|cite|improve this answer














          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.



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 39 mins ago

























          answered 2 hours ago









          John Omielan

          72418




          72418












          • 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 hours 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 hours 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 hours 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
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour 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 hours 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 hours 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 hours 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
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour 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 hours 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 hours 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 hours 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 hours 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 hours 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 hours 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
          1 hour 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
          1 hour ago






          1




          1




          You are welcome to edit it into your answer, which will make it more accessible to readers.
          – Ross Millikan
          1 hour ago




          You are welcome to edit it into your answer, which will make it more accessible to readers.
          – Ross Millikan
          1 hour ago










          Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.










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          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.
















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          數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
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          When can things happen in Etherscan, such as the picture below?

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          2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi 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  |  ...
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          格利澤436b

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          格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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