Rolling a die 1000 times , a string of the product will be a cube












3












$begingroup$


Each of the six faces of a die is marked with an integer, not necessarily positive. The
die is rolled 1000 times. Show that there is a time interval such that the product of all rolls
in this interval is a cube of an integer. (For example, it could happen that the product of
all outcomes between 5th and 20th throws is a cube; obviously, the interval has to include
at least one throw!)
i looked at abcdef as a possible product , somehow 1000 of these 'terms' no matter how we shuffle them , a product like ababab or adadad or acdacdacd will be in the string but i cant prove how :(






combinatorics







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$endgroup$

















    3












    $begingroup$


    Each of the six faces of a die is marked with an integer, not necessarily positive. The
    die is rolled 1000 times. Show that there is a time interval such that the product of all rolls
    in this interval is a cube of an integer. (For example, it could happen that the product of
    all outcomes between 5th and 20th throws is a cube; obviously, the interval has to include
    at least one throw!)
    i looked at abcdef as a possible product , somehow 1000 of these 'terms' no matter how we shuffle them , a product like ababab or adadad or acdacdacd will be in the string but i cant prove how :(






    combinatorics







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      2



      $begingroup$


      Each of the six faces of a die is marked with an integer, not necessarily positive. The
      die is rolled 1000 times. Show that there is a time interval such that the product of all rolls
      in this interval is a cube of an integer. (For example, it could happen that the product of
      all outcomes between 5th and 20th throws is a cube; obviously, the interval has to include
      at least one throw!)
      i looked at abcdef as a possible product , somehow 1000 of these 'terms' no matter how we shuffle them , a product like ababab or adadad or acdacdacd will be in the string but i cant prove how :(






      combinatorics







      share|cite|improve this question









      $endgroup$




      Each of the six faces of a die is marked with an integer, not necessarily positive. The
      die is rolled 1000 times. Show that there is a time interval such that the product of all rolls
      in this interval is a cube of an integer. (For example, it could happen that the product of
      all outcomes between 5th and 20th throws is a cube; obviously, the interval has to include
      at least one throw!)
      i looked at abcdef as a possible product , somehow 1000 of these 'terms' no matter how we shuffle them , a product like ababab or adadad or acdacdacd will be in the string but i cant prove how :(






      combinatorics




      combinatorics






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      Randin DRandin D

      826




      826






















          1 Answer
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          $begingroup$

          We can show something stronger; there exists an interval of rolls where each side has appeared a number of times which is a multiple of $3$.



          For each $i=1,2,dots,1000$, let $a_i$ be number of times that the first face has appeared in rolls numbered $1,2,dots,i$. Same for $b_i,c_i,d_i,e_i,f_i$. Consider the $1000$ ordered six-tuples of values $T_i:=(a_i,b_i,dots,f_i)$, where each coordinate is only recorded modulo $3$. There are $3^6=729$ possible six-tuples, and there are $1000$ rolls, so by the pigeonhole principle, there are two indices $i<j$ for which $T_i=T_j$. This implies that among rolls numbered $i+1,i+2,dots,j$, each face appears a number of times which is a multiple of three.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            (a) There are $1000$ numbers $iin{1,2,dots,1000}$, and $729$ possible tuples. $1000$ pigeons in $729$ holes. (b) If each face appears a multiple of three times, and the numbers of the faces are $x,y,z,w,s,t$, then the product of the numbers in that interval $[i+1,j]$ is $x^{3i}y^{3j}cdots t^{3k}$, which is the cube of $x^iy^jcdots t^k$.
            $endgroup$
            – Mike Earnest
            2 hours ago










          • $begingroup$
            i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            how do u get the 729 again?
            $endgroup$
            – Randin D
            2 hours ago












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          1 Answer
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          active

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          active

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          active

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          5












          $begingroup$

          We can show something stronger; there exists an interval of rolls where each side has appeared a number of times which is a multiple of $3$.



          For each $i=1,2,dots,1000$, let $a_i$ be number of times that the first face has appeared in rolls numbered $1,2,dots,i$. Same for $b_i,c_i,d_i,e_i,f_i$. Consider the $1000$ ordered six-tuples of values $T_i:=(a_i,b_i,dots,f_i)$, where each coordinate is only recorded modulo $3$. There are $3^6=729$ possible six-tuples, and there are $1000$ rolls, so by the pigeonhole principle, there are two indices $i<j$ for which $T_i=T_j$. This implies that among rolls numbered $i+1,i+2,dots,j$, each face appears a number of times which is a multiple of three.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            (a) There are $1000$ numbers $iin{1,2,dots,1000}$, and $729$ possible tuples. $1000$ pigeons in $729$ holes. (b) If each face appears a multiple of three times, and the numbers of the faces are $x,y,z,w,s,t$, then the product of the numbers in that interval $[i+1,j]$ is $x^{3i}y^{3j}cdots t^{3k}$, which is the cube of $x^iy^jcdots t^k$.
            $endgroup$
            – Mike Earnest
            2 hours ago










          • $begingroup$
            i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            how do u get the 729 again?
            $endgroup$
            – Randin D
            2 hours ago
















          5












          $begingroup$

          We can show something stronger; there exists an interval of rolls where each side has appeared a number of times which is a multiple of $3$.



          For each $i=1,2,dots,1000$, let $a_i$ be number of times that the first face has appeared in rolls numbered $1,2,dots,i$. Same for $b_i,c_i,d_i,e_i,f_i$. Consider the $1000$ ordered six-tuples of values $T_i:=(a_i,b_i,dots,f_i)$, where each coordinate is only recorded modulo $3$. There are $3^6=729$ possible six-tuples, and there are $1000$ rolls, so by the pigeonhole principle, there are two indices $i<j$ for which $T_i=T_j$. This implies that among rolls numbered $i+1,i+2,dots,j$, each face appears a number of times which is a multiple of three.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            (a) There are $1000$ numbers $iin{1,2,dots,1000}$, and $729$ possible tuples. $1000$ pigeons in $729$ holes. (b) If each face appears a multiple of three times, and the numbers of the faces are $x,y,z,w,s,t$, then the product of the numbers in that interval $[i+1,j]$ is $x^{3i}y^{3j}cdots t^{3k}$, which is the cube of $x^iy^jcdots t^k$.
            $endgroup$
            – Mike Earnest
            2 hours ago










          • $begingroup$
            i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            how do u get the 729 again?
            $endgroup$
            – Randin D
            2 hours ago














          5












          5








          5





          $begingroup$

          We can show something stronger; there exists an interval of rolls where each side has appeared a number of times which is a multiple of $3$.



          For each $i=1,2,dots,1000$, let $a_i$ be number of times that the first face has appeared in rolls numbered $1,2,dots,i$. Same for $b_i,c_i,d_i,e_i,f_i$. Consider the $1000$ ordered six-tuples of values $T_i:=(a_i,b_i,dots,f_i)$, where each coordinate is only recorded modulo $3$. There are $3^6=729$ possible six-tuples, and there are $1000$ rolls, so by the pigeonhole principle, there are two indices $i<j$ for which $T_i=T_j$. This implies that among rolls numbered $i+1,i+2,dots,j$, each face appears a number of times which is a multiple of three.






          share|cite|improve this answer









          $endgroup$



          We can show something stronger; there exists an interval of rolls where each side has appeared a number of times which is a multiple of $3$.



          For each $i=1,2,dots,1000$, let $a_i$ be number of times that the first face has appeared in rolls numbered $1,2,dots,i$. Same for $b_i,c_i,d_i,e_i,f_i$. Consider the $1000$ ordered six-tuples of values $T_i:=(a_i,b_i,dots,f_i)$, where each coordinate is only recorded modulo $3$. There are $3^6=729$ possible six-tuples, and there are $1000$ rolls, so by the pigeonhole principle, there are two indices $i<j$ for which $T_i=T_j$. This implies that among rolls numbered $i+1,i+2,dots,j$, each face appears a number of times which is a multiple of three.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Mike EarnestMike Earnest

          28.5k22153




          28.5k22153












          • $begingroup$
            i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            (a) There are $1000$ numbers $iin{1,2,dots,1000}$, and $729$ possible tuples. $1000$ pigeons in $729$ holes. (b) If each face appears a multiple of three times, and the numbers of the faces are $x,y,z,w,s,t$, then the product of the numbers in that interval $[i+1,j]$ is $x^{3i}y^{3j}cdots t^{3k}$, which is the cube of $x^iy^jcdots t^k$.
            $endgroup$
            – Mike Earnest
            2 hours ago










          • $begingroup$
            i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            how do u get the 729 again?
            $endgroup$
            – Randin D
            2 hours ago


















          • $begingroup$
            i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            (a) There are $1000$ numbers $iin{1,2,dots,1000}$, and $729$ possible tuples. $1000$ pigeons in $729$ holes. (b) If each face appears a multiple of three times, and the numbers of the faces are $x,y,z,w,s,t$, then the product of the numbers in that interval $[i+1,j]$ is $x^{3i}y^{3j}cdots t^{3k}$, which is the cube of $x^iy^jcdots t^k$.
            $endgroup$
            – Mike Earnest
            2 hours ago










          • $begingroup$
            i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
            $endgroup$
            – Randin D
            2 hours ago










          • $begingroup$
            how do u get the 729 again?
            $endgroup$
            – Randin D
            2 hours ago
















          $begingroup$
          i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
          $endgroup$
          – Randin D
          2 hours ago




          $begingroup$
          i dont get it, how again can u use pigeonhole? and how does this work for cubes not 3k
          $endgroup$
          – Randin D
          2 hours ago












          $begingroup$
          (a) There are $1000$ numbers $iin{1,2,dots,1000}$, and $729$ possible tuples. $1000$ pigeons in $729$ holes. (b) If each face appears a multiple of three times, and the numbers of the faces are $x,y,z,w,s,t$, then the product of the numbers in that interval $[i+1,j]$ is $x^{3i}y^{3j}cdots t^{3k}$, which is the cube of $x^iy^jcdots t^k$.
          $endgroup$
          – Mike Earnest
          2 hours ago




          $begingroup$
          (a) There are $1000$ numbers $iin{1,2,dots,1000}$, and $729$ possible tuples. $1000$ pigeons in $729$ holes. (b) If each face appears a multiple of three times, and the numbers of the faces are $x,y,z,w,s,t$, then the product of the numbers in that interval $[i+1,j]$ is $x^{3i}y^{3j}cdots t^{3k}$, which is the cube of $x^iy^jcdots t^k$.
          $endgroup$
          – Mike Earnest
          2 hours ago












          $begingroup$
          i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
          $endgroup$
          – Randin D
          2 hours ago




          $begingroup$
          i get pigeonhole ..1000 into 729 means 3 will be in one pigeonhole right? and thats the cube?
          $endgroup$
          – Randin D
          2 hours ago












          $begingroup$
          how do u get the 729 again?
          $endgroup$
          – Randin D
          2 hours ago




          $begingroup$
          how do u get the 729 again?
          $endgroup$
          – Randin D
          2 hours ago


















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