Sum of freely chosen subset of $n$-tuple is divisible by $n$.












2












$begingroup$


I am struggling with the following task:



Be $n in N$ and $(a_1, a_2, ldots, a_n) in mathbb{Z}^n$. Prove that there is always $i, j in underline{n}$, with $i≤j$, so that $sum_{k=i}^j a_k$ is divisible by n.



My first idea was to use a distinction of cases. In the first case a multiple of n is an element of $mathbb{Z}^n$ so you could select the sum to be just this element so it would obviously be divisible.



In the second case, all elements of the tuple are the same, so that adding them all up will result in a multiple of $n$ once more, making it divisible.



I am lost however on what to do in the third case, where the two above are not true. I would truly appreciate your input.






elementary-number-theory discrete-mathematics divisibility







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I am struggling with the following task:



    Be $n in N$ and $(a_1, a_2, ldots, a_n) in mathbb{Z}^n$. Prove that there is always $i, j in underline{n}$, with $i≤j$, so that $sum_{k=i}^j a_k$ is divisible by n.



    My first idea was to use a distinction of cases. In the first case a multiple of n is an element of $mathbb{Z}^n$ so you could select the sum to be just this element so it would obviously be divisible.



    In the second case, all elements of the tuple are the same, so that adding them all up will result in a multiple of $n$ once more, making it divisible.



    I am lost however on what to do in the third case, where the two above are not true. I would truly appreciate your input.






    elementary-number-theory discrete-mathematics divisibility







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      I am struggling with the following task:



      Be $n in N$ and $(a_1, a_2, ldots, a_n) in mathbb{Z}^n$. Prove that there is always $i, j in underline{n}$, with $i≤j$, so that $sum_{k=i}^j a_k$ is divisible by n.



      My first idea was to use a distinction of cases. In the first case a multiple of n is an element of $mathbb{Z}^n$ so you could select the sum to be just this element so it would obviously be divisible.



      In the second case, all elements of the tuple are the same, so that adding them all up will result in a multiple of $n$ once more, making it divisible.



      I am lost however on what to do in the third case, where the two above are not true. I would truly appreciate your input.






      elementary-number-theory discrete-mathematics divisibility







      share|cite|improve this question











      $endgroup$




      I am struggling with the following task:



      Be $n in N$ and $(a_1, a_2, ldots, a_n) in mathbb{Z}^n$. Prove that there is always $i, j in underline{n}$, with $i≤j$, so that $sum_{k=i}^j a_k$ is divisible by n.



      My first idea was to use a distinction of cases. In the first case a multiple of n is an element of $mathbb{Z}^n$ so you could select the sum to be just this element so it would obviously be divisible.



      In the second case, all elements of the tuple are the same, so that adding them all up will result in a multiple of $n$ once more, making it divisible.



      I am lost however on what to do in the third case, where the two above are not true. I would truly appreciate your input.






      elementary-number-theory discrete-mathematics divisibility




      elementary-number-theory discrete-mathematics divisibility






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 14:12









      rtybase

      11.1k21533




      11.1k21533










      asked Dec 13 '18 at 12:11









      JulianGiJulianGi

      111




      111






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Some elements of $(a_1, a_2, ldots, a_n)$ may be different and it need not contain a multiple of $n$ of course. (it is chosen just randomly from $mathbb{Z}^n$.)



          You need a little trick to solve this problem in an easy way. Consider $n$ values
          $$
          a_1, a_1+a_2, a_1+a_2+a_3, ldots , a_1+a_2 +cdots a_n.
          $$           If at least one of them is a multiple of $n$, it's okay. If not, what can be said about their remainders divided by $n$?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I think we have almost identical answers (+1).
            $endgroup$
            – rtybase
            Dec 13 '18 at 12:31



















          2












          $begingroup$

          Using pigeonhole principle, let's look at all the remainders of
          $$a_1 pmod{n}$$
          $$a_1+a_2pmod{n}$$
          $$a_1+a_2+a_3pmod{n}$$
          $$...$$
          $$a_1+a_2+a_3+...+a_npmod{n}$$
          If at least one is $0$, we are done.



          If none is $0$, then at least $2$ will have the same value (there will be $n$ remainders, taking $n-1$ possible values within ${1,...,n-1}$, pigeonhole principle). E.g.
          $$a_1+a_2+a_3+...+a_j equiv a_1+a_2+a_3+...+a_i pmod{n}$$
          The remaining part is to subtract, assume $i<j$



          $$(a_1+a_2+a_3+...+a_j)- (a_1+a_2+a_3+...+a_i)equiv 0 pmod{n}$$
          or
          $$a_{j+1}+a_{j+2}+a_{j+3}+...+a_iequiv 0 pmod{n}$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            This is very extensive and well written. Thanks!
            $endgroup$
            – Empty2k12
            Dec 13 '18 at 12:59










          • $begingroup$
            Thanks for your answer. Now I understand how this works. However I have trouble to understand how you reach the conclusion, that "If none is 0, then at least 2 will have the same value". If this would be given the rest of your calculations fall into place.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 13:46








          • 1




            $begingroup$
            @JulianGi recall divisibility, remainder is $0leq r <n$. We excluded $0$ so $1leq r <n$ or we have $n-1$ possible values from $1$ to $n-1$. The sequence constructed $a_1, a_1+a_2,...,a_1+a_2+...+a_n$ contains $n$ elements, taking the remainder of $pmod{n}$ yields $n$ remainders, taking values from $1$ to $n-1$. There is no one to one mapping between $n$ remainders taking values from ${1,2,...n-1}$ and ${1,2,...n-1}$, so "at least 2 (remainders) will have the same value ...".
            $endgroup$
            – rtybase
            Dec 13 '18 at 14:01








          • 1




            $begingroup$
            @rtybase That makes a lot of sense. Thank you very much.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 14:07













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          2 Answers
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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          3












          $begingroup$

          Some elements of $(a_1, a_2, ldots, a_n)$ may be different and it need not contain a multiple of $n$ of course. (it is chosen just randomly from $mathbb{Z}^n$.)



          You need a little trick to solve this problem in an easy way. Consider $n$ values
          $$
          a_1, a_1+a_2, a_1+a_2+a_3, ldots , a_1+a_2 +cdots a_n.
          $$           If at least one of them is a multiple of $n$, it's okay. If not, what can be said about their remainders divided by $n$?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I think we have almost identical answers (+1).
            $endgroup$
            – rtybase
            Dec 13 '18 at 12:31
















          3












          $begingroup$

          Some elements of $(a_1, a_2, ldots, a_n)$ may be different and it need not contain a multiple of $n$ of course. (it is chosen just randomly from $mathbb{Z}^n$.)



          You need a little trick to solve this problem in an easy way. Consider $n$ values
          $$
          a_1, a_1+a_2, a_1+a_2+a_3, ldots , a_1+a_2 +cdots a_n.
          $$           If at least one of them is a multiple of $n$, it's okay. If not, what can be said about their remainders divided by $n$?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I think we have almost identical answers (+1).
            $endgroup$
            – rtybase
            Dec 13 '18 at 12:31














          3












          3








          3





          $begingroup$

          Some elements of $(a_1, a_2, ldots, a_n)$ may be different and it need not contain a multiple of $n$ of course. (it is chosen just randomly from $mathbb{Z}^n$.)



          You need a little trick to solve this problem in an easy way. Consider $n$ values
          $$
          a_1, a_1+a_2, a_1+a_2+a_3, ldots , a_1+a_2 +cdots a_n.
          $$           If at least one of them is a multiple of $n$, it's okay. If not, what can be said about their remainders divided by $n$?






          share|cite|improve this answer











          $endgroup$



          Some elements of $(a_1, a_2, ldots, a_n)$ may be different and it need not contain a multiple of $n$ of course. (it is chosen just randomly from $mathbb{Z}^n$.)



          You need a little trick to solve this problem in an easy way. Consider $n$ values
          $$
          a_1, a_1+a_2, a_1+a_2+a_3, ldots , a_1+a_2 +cdots a_n.
          $$           If at least one of them is a multiple of $n$, it's okay. If not, what can be said about their remainders divided by $n$?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 12:24

























          answered Dec 13 '18 at 12:16









          SongSong

          16.2k1739




          16.2k1739








          • 1




            $begingroup$
            I think we have almost identical answers (+1).
            $endgroup$
            – rtybase
            Dec 13 '18 at 12:31














          • 1




            $begingroup$
            I think we have almost identical answers (+1).
            $endgroup$
            – rtybase
            Dec 13 '18 at 12:31








          1




          1




          $begingroup$
          I think we have almost identical answers (+1).
          $endgroup$
          – rtybase
          Dec 13 '18 at 12:31




          $begingroup$
          I think we have almost identical answers (+1).
          $endgroup$
          – rtybase
          Dec 13 '18 at 12:31











          2












          $begingroup$

          Using pigeonhole principle, let's look at all the remainders of
          $$a_1 pmod{n}$$
          $$a_1+a_2pmod{n}$$
          $$a_1+a_2+a_3pmod{n}$$
          $$...$$
          $$a_1+a_2+a_3+...+a_npmod{n}$$
          If at least one is $0$, we are done.



          If none is $0$, then at least $2$ will have the same value (there will be $n$ remainders, taking $n-1$ possible values within ${1,...,n-1}$, pigeonhole principle). E.g.
          $$a_1+a_2+a_3+...+a_j equiv a_1+a_2+a_3+...+a_i pmod{n}$$
          The remaining part is to subtract, assume $i<j$



          $$(a_1+a_2+a_3+...+a_j)- (a_1+a_2+a_3+...+a_i)equiv 0 pmod{n}$$
          or
          $$a_{j+1}+a_{j+2}+a_{j+3}+...+a_iequiv 0 pmod{n}$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            This is very extensive and well written. Thanks!
            $endgroup$
            – Empty2k12
            Dec 13 '18 at 12:59










          • $begingroup$
            Thanks for your answer. Now I understand how this works. However I have trouble to understand how you reach the conclusion, that "If none is 0, then at least 2 will have the same value". If this would be given the rest of your calculations fall into place.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 13:46








          • 1




            $begingroup$
            @JulianGi recall divisibility, remainder is $0leq r <n$. We excluded $0$ so $1leq r <n$ or we have $n-1$ possible values from $1$ to $n-1$. The sequence constructed $a_1, a_1+a_2,...,a_1+a_2+...+a_n$ contains $n$ elements, taking the remainder of $pmod{n}$ yields $n$ remainders, taking values from $1$ to $n-1$. There is no one to one mapping between $n$ remainders taking values from ${1,2,...n-1}$ and ${1,2,...n-1}$, so "at least 2 (remainders) will have the same value ...".
            $endgroup$
            – rtybase
            Dec 13 '18 at 14:01








          • 1




            $begingroup$
            @rtybase That makes a lot of sense. Thank you very much.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 14:07


















          2












          $begingroup$

          Using pigeonhole principle, let's look at all the remainders of
          $$a_1 pmod{n}$$
          $$a_1+a_2pmod{n}$$
          $$a_1+a_2+a_3pmod{n}$$
          $$...$$
          $$a_1+a_2+a_3+...+a_npmod{n}$$
          If at least one is $0$, we are done.



          If none is $0$, then at least $2$ will have the same value (there will be $n$ remainders, taking $n-1$ possible values within ${1,...,n-1}$, pigeonhole principle). E.g.
          $$a_1+a_2+a_3+...+a_j equiv a_1+a_2+a_3+...+a_i pmod{n}$$
          The remaining part is to subtract, assume $i<j$



          $$(a_1+a_2+a_3+...+a_j)- (a_1+a_2+a_3+...+a_i)equiv 0 pmod{n}$$
          or
          $$a_{j+1}+a_{j+2}+a_{j+3}+...+a_iequiv 0 pmod{n}$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            This is very extensive and well written. Thanks!
            $endgroup$
            – Empty2k12
            Dec 13 '18 at 12:59










          • $begingroup$
            Thanks for your answer. Now I understand how this works. However I have trouble to understand how you reach the conclusion, that "If none is 0, then at least 2 will have the same value". If this would be given the rest of your calculations fall into place.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 13:46








          • 1




            $begingroup$
            @JulianGi recall divisibility, remainder is $0leq r <n$. We excluded $0$ so $1leq r <n$ or we have $n-1$ possible values from $1$ to $n-1$. The sequence constructed $a_1, a_1+a_2,...,a_1+a_2+...+a_n$ contains $n$ elements, taking the remainder of $pmod{n}$ yields $n$ remainders, taking values from $1$ to $n-1$. There is no one to one mapping between $n$ remainders taking values from ${1,2,...n-1}$ and ${1,2,...n-1}$, so "at least 2 (remainders) will have the same value ...".
            $endgroup$
            – rtybase
            Dec 13 '18 at 14:01








          • 1




            $begingroup$
            @rtybase That makes a lot of sense. Thank you very much.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 14:07
















          2












          2








          2





          $begingroup$

          Using pigeonhole principle, let's look at all the remainders of
          $$a_1 pmod{n}$$
          $$a_1+a_2pmod{n}$$
          $$a_1+a_2+a_3pmod{n}$$
          $$...$$
          $$a_1+a_2+a_3+...+a_npmod{n}$$
          If at least one is $0$, we are done.



          If none is $0$, then at least $2$ will have the same value (there will be $n$ remainders, taking $n-1$ possible values within ${1,...,n-1}$, pigeonhole principle). E.g.
          $$a_1+a_2+a_3+...+a_j equiv a_1+a_2+a_3+...+a_i pmod{n}$$
          The remaining part is to subtract, assume $i<j$



          $$(a_1+a_2+a_3+...+a_j)- (a_1+a_2+a_3+...+a_i)equiv 0 pmod{n}$$
          or
          $$a_{j+1}+a_{j+2}+a_{j+3}+...+a_iequiv 0 pmod{n}$$






          share|cite|improve this answer











          $endgroup$



          Using pigeonhole principle, let's look at all the remainders of
          $$a_1 pmod{n}$$
          $$a_1+a_2pmod{n}$$
          $$a_1+a_2+a_3pmod{n}$$
          $$...$$
          $$a_1+a_2+a_3+...+a_npmod{n}$$
          If at least one is $0$, we are done.



          If none is $0$, then at least $2$ will have the same value (there will be $n$ remainders, taking $n-1$ possible values within ${1,...,n-1}$, pigeonhole principle). E.g.
          $$a_1+a_2+a_3+...+a_j equiv a_1+a_2+a_3+...+a_i pmod{n}$$
          The remaining part is to subtract, assume $i<j$



          $$(a_1+a_2+a_3+...+a_j)- (a_1+a_2+a_3+...+a_i)equiv 0 pmod{n}$$
          or
          $$a_{j+1}+a_{j+2}+a_{j+3}+...+a_iequiv 0 pmod{n}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 14:07

























          answered Dec 13 '18 at 12:21









          rtybasertybase

          11.1k21533




          11.1k21533








          • 1




            $begingroup$
            This is very extensive and well written. Thanks!
            $endgroup$
            – Empty2k12
            Dec 13 '18 at 12:59










          • $begingroup$
            Thanks for your answer. Now I understand how this works. However I have trouble to understand how you reach the conclusion, that "If none is 0, then at least 2 will have the same value". If this would be given the rest of your calculations fall into place.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 13:46








          • 1




            $begingroup$
            @JulianGi recall divisibility, remainder is $0leq r <n$. We excluded $0$ so $1leq r <n$ or we have $n-1$ possible values from $1$ to $n-1$. The sequence constructed $a_1, a_1+a_2,...,a_1+a_2+...+a_n$ contains $n$ elements, taking the remainder of $pmod{n}$ yields $n$ remainders, taking values from $1$ to $n-1$. There is no one to one mapping between $n$ remainders taking values from ${1,2,...n-1}$ and ${1,2,...n-1}$, so "at least 2 (remainders) will have the same value ...".
            $endgroup$
            – rtybase
            Dec 13 '18 at 14:01








          • 1




            $begingroup$
            @rtybase That makes a lot of sense. Thank you very much.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 14:07
















          • 1




            $begingroup$
            This is very extensive and well written. Thanks!
            $endgroup$
            – Empty2k12
            Dec 13 '18 at 12:59










          • $begingroup$
            Thanks for your answer. Now I understand how this works. However I have trouble to understand how you reach the conclusion, that "If none is 0, then at least 2 will have the same value". If this would be given the rest of your calculations fall into place.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 13:46








          • 1




            $begingroup$
            @JulianGi recall divisibility, remainder is $0leq r <n$. We excluded $0$ so $1leq r <n$ or we have $n-1$ possible values from $1$ to $n-1$. The sequence constructed $a_1, a_1+a_2,...,a_1+a_2+...+a_n$ contains $n$ elements, taking the remainder of $pmod{n}$ yields $n$ remainders, taking values from $1$ to $n-1$. There is no one to one mapping between $n$ remainders taking values from ${1,2,...n-1}$ and ${1,2,...n-1}$, so "at least 2 (remainders) will have the same value ...".
            $endgroup$
            – rtybase
            Dec 13 '18 at 14:01








          • 1




            $begingroup$
            @rtybase That makes a lot of sense. Thank you very much.
            $endgroup$
            – JulianGi
            Dec 13 '18 at 14:07










          1




          1




          $begingroup$
          This is very extensive and well written. Thanks!
          $endgroup$
          – Empty2k12
          Dec 13 '18 at 12:59




          $begingroup$
          This is very extensive and well written. Thanks!
          $endgroup$
          – Empty2k12
          Dec 13 '18 at 12:59












          $begingroup$
          Thanks for your answer. Now I understand how this works. However I have trouble to understand how you reach the conclusion, that "If none is 0, then at least 2 will have the same value". If this would be given the rest of your calculations fall into place.
          $endgroup$
          – JulianGi
          Dec 13 '18 at 13:46






          $begingroup$
          Thanks for your answer. Now I understand how this works. However I have trouble to understand how you reach the conclusion, that "If none is 0, then at least 2 will have the same value". If this would be given the rest of your calculations fall into place.
          $endgroup$
          – JulianGi
          Dec 13 '18 at 13:46






          1




          1




          $begingroup$
          @JulianGi recall divisibility, remainder is $0leq r <n$. We excluded $0$ so $1leq r <n$ or we have $n-1$ possible values from $1$ to $n-1$. The sequence constructed $a_1, a_1+a_2,...,a_1+a_2+...+a_n$ contains $n$ elements, taking the remainder of $pmod{n}$ yields $n$ remainders, taking values from $1$ to $n-1$. There is no one to one mapping between $n$ remainders taking values from ${1,2,...n-1}$ and ${1,2,...n-1}$, so "at least 2 (remainders) will have the same value ...".
          $endgroup$
          – rtybase
          Dec 13 '18 at 14:01






          $begingroup$
          @JulianGi recall divisibility, remainder is $0leq r <n$. We excluded $0$ so $1leq r <n$ or we have $n-1$ possible values from $1$ to $n-1$. The sequence constructed $a_1, a_1+a_2,...,a_1+a_2+...+a_n$ contains $n$ elements, taking the remainder of $pmod{n}$ yields $n$ remainders, taking values from $1$ to $n-1$. There is no one to one mapping between $n$ remainders taking values from ${1,2,...n-1}$ and ${1,2,...n-1}$, so "at least 2 (remainders) will have the same value ...".
          $endgroup$
          – rtybase
          Dec 13 '18 at 14:01






          1




          1




          $begingroup$
          @rtybase That makes a lot of sense. Thank you very much.
          $endgroup$
          – JulianGi
          Dec 13 '18 at 14:07






          $begingroup$
          @rtybase That makes a lot of sense. Thank you very much.
          $endgroup$
          – JulianGi
          Dec 13 '18 at 14:07




















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