How many subsets of a set contain a difference between elements that are divisible by $4$?How many integers in the range [1,999] are divisible by exactly 1 of 7 and 11?how many subsets with 4-elements?How many subsets of $S$ are there that contain $x$ but do not contain $y$?How many elements in $X$ are divisible by $3$?Lines in the plane from Concrete Mathematics. How many bounded regions are there?How many subsets of an $n$-element set contain $k$ or more elements?How many subsets contain no consecutive elements?Putting Bayes Theorem to workHow many subsets are there that contain a specific set?How many subsets contain $x_1$? How many subsets contain $x_2$ and $x_3$, but not $x_5$?

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How many subsets of a set contain a difference between elements that are divisible by $4$?


How many integers in the range [1,999] are divisible by exactly 1 of 7 and 11?how many subsets with 4-elements?How many subsets of $S$ are there that contain $x$ but do not contain $y$?How many elements in $X$ are divisible by $3$?Lines in the plane from Concrete Mathematics. How many bounded regions are there?How many subsets of an $n$-element set contain $k$ or more elements?How many subsets contain no consecutive elements?Putting Bayes Theorem to workHow many subsets are there that contain a specific set?How many subsets contain $x_1$? How many subsets contain $x_2$ and $x_3$, but not $x_5$?













2












$begingroup$


I'm practicing for an exam and I'm stuck on one question. I hope I have translated this question properly, here it goes:




Question: How many subsets of the set $1,...,25$ have the property such that
no difference between two of its elements is divisible by $4$?




The solutions suggestions don't really suggest a solution on this, it just states the answer to be $8cdot7^3=2744.$ I don't even know to what chapter this question relates to in my discrete mathematics book.



If I understand this correctly, I should find subsets like these: $1,7,16$ since $4nmid (7-1), 4nmid (16-1)$ and $4nmid (16-7)?$



Well obviously it would take a long time to write them all up on an exam so I sense there is some nice trick to this?










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    I'm practicing for an exam and I'm stuck on one question. I hope I have translated this question properly, here it goes:




    Question: How many subsets of the set $1,...,25$ have the property such that
    no difference between two of its elements is divisible by $4$?




    The solutions suggestions don't really suggest a solution on this, it just states the answer to be $8cdot7^3=2744.$ I don't even know to what chapter this question relates to in my discrete mathematics book.



    If I understand this correctly, I should find subsets like these: $1,7,16$ since $4nmid (7-1), 4nmid (16-1)$ and $4nmid (16-7)?$



    Well obviously it would take a long time to write them all up on an exam so I sense there is some nice trick to this?










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      I'm practicing for an exam and I'm stuck on one question. I hope I have translated this question properly, here it goes:




      Question: How many subsets of the set $1,...,25$ have the property such that
      no difference between two of its elements is divisible by $4$?




      The solutions suggestions don't really suggest a solution on this, it just states the answer to be $8cdot7^3=2744.$ I don't even know to what chapter this question relates to in my discrete mathematics book.



      If I understand this correctly, I should find subsets like these: $1,7,16$ since $4nmid (7-1), 4nmid (16-1)$ and $4nmid (16-7)?$



      Well obviously it would take a long time to write them all up on an exam so I sense there is some nice trick to this?










      share|cite|improve this question











      $endgroup$




      I'm practicing for an exam and I'm stuck on one question. I hope I have translated this question properly, here it goes:




      Question: How many subsets of the set $1,...,25$ have the property such that
      no difference between two of its elements is divisible by $4$?




      The solutions suggestions don't really suggest a solution on this, it just states the answer to be $8cdot7^3=2744.$ I don't even know to what chapter this question relates to in my discrete mathematics book.



      If I understand this correctly, I should find subsets like these: $1,7,16$ since $4nmid (7-1), 4nmid (16-1)$ and $4nmid (16-7)?$



      Well obviously it would take a long time to write them all up on an exam so I sense there is some nice trick to this?







      combinatorics elementary-number-theory discrete-mathematics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 18 at 12:52









      awkward

      6,71511025




      6,71511025










      asked Mar 17 at 22:43









      ParsevalParseval

      3,0771719




      3,0771719




















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          There are $7$ integers in the set $S = 1,dots,25$ which are equal to $1$ modulo $4$, and there are $6$ integers in $S$ which are equal to $0$ modulo $4$, and similarly $6$ for $2$ modulo $4$ and $6$ for $3$ modulo $4$. A subset of $S$ has the desired property if and only if it contains at most one number from each of these four groups (any difference of them is then not equal to $0$ modulo $4$, i.e., not divisible by $4$; conversely, if two integers are included which are the same modulo $4$, their difference is divisible by $4$). So you have $8$ choices for the first group: $7$ to include one of the $7$ integers which are equal to $1$ modulo $4$ plus one choice to leave them out entirely; you have $7$ choices for the second group: $6$ to include one of the $6$ which are $0$ modulo $4$ plus one choice to leave them out entirely; and so on. That gives $8cdot 7cdot 7 cdot 7$ possibilities.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            I get that $S_1=4,9,13,17,21$ are the elements that are equal to $1$ in mod $4$. So $|S_1|=5$. Which numbers did I miss?
            $endgroup$
            – Parseval
            Mar 18 at 13:50






          • 1




            $begingroup$
            They should be 1, 5, 9, 13, 17, 21, 25
            $endgroup$
            – csprun
            Mar 18 at 13:53


















          1












          $begingroup$

          If no $2$ elements of a subset can have a difference divisible by $4$, those elements must belong to different remainder classes modulo $4$. The set $1,2,ldots,25$ contains $7$ elements $equiv 1 pmod 4$ , namly $1,5,9,ldots,25$ and $6$ elements $equiv 2,3,4 pmod 4$ each.



          That means if we wanted to find all $4$-element subsets witht the required property, it would be $7cdot6^3$, because we need one element from each remainder class, and for $equiv 1 pmod 4$ there are 7 choices, for $equiv 2 pmod 4$ there are 6 choices a.s.o.



          But since we are looking for arbitrary size subsets (well, arbitrary means size $le 4$, as otherwise 2 elements would be from the same remainder class), we use a trick. We add the numbers $26,27,28$ and $29$ to the set. If we have a subset of size smaller than $4$, we add the missing remainder classes from those 'new' numbers.



          That is a one-to-one correspondence between all subsets of $1,2,ldots,25$ with the required property and all 4-element subsets of $1,2,ldots,29$ with the required property. Since we added 1 element to each remainder class, that number is $8cdot7^3$






          share|cite|improve this answer









          $endgroup$












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            2 Answers
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            2 Answers
            2






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            active

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            active

            oldest

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            5












            $begingroup$

            There are $7$ integers in the set $S = 1,dots,25$ which are equal to $1$ modulo $4$, and there are $6$ integers in $S$ which are equal to $0$ modulo $4$, and similarly $6$ for $2$ modulo $4$ and $6$ for $3$ modulo $4$. A subset of $S$ has the desired property if and only if it contains at most one number from each of these four groups (any difference of them is then not equal to $0$ modulo $4$, i.e., not divisible by $4$; conversely, if two integers are included which are the same modulo $4$, their difference is divisible by $4$). So you have $8$ choices for the first group: $7$ to include one of the $7$ integers which are equal to $1$ modulo $4$ plus one choice to leave them out entirely; you have $7$ choices for the second group: $6$ to include one of the $6$ which are $0$ modulo $4$ plus one choice to leave them out entirely; and so on. That gives $8cdot 7cdot 7 cdot 7$ possibilities.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              I get that $S_1=4,9,13,17,21$ are the elements that are equal to $1$ in mod $4$. So $|S_1|=5$. Which numbers did I miss?
              $endgroup$
              – Parseval
              Mar 18 at 13:50






            • 1




              $begingroup$
              They should be 1, 5, 9, 13, 17, 21, 25
              $endgroup$
              – csprun
              Mar 18 at 13:53















            5












            $begingroup$

            There are $7$ integers in the set $S = 1,dots,25$ which are equal to $1$ modulo $4$, and there are $6$ integers in $S$ which are equal to $0$ modulo $4$, and similarly $6$ for $2$ modulo $4$ and $6$ for $3$ modulo $4$. A subset of $S$ has the desired property if and only if it contains at most one number from each of these four groups (any difference of them is then not equal to $0$ modulo $4$, i.e., not divisible by $4$; conversely, if two integers are included which are the same modulo $4$, their difference is divisible by $4$). So you have $8$ choices for the first group: $7$ to include one of the $7$ integers which are equal to $1$ modulo $4$ plus one choice to leave them out entirely; you have $7$ choices for the second group: $6$ to include one of the $6$ which are $0$ modulo $4$ plus one choice to leave them out entirely; and so on. That gives $8cdot 7cdot 7 cdot 7$ possibilities.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              I get that $S_1=4,9,13,17,21$ are the elements that are equal to $1$ in mod $4$. So $|S_1|=5$. Which numbers did I miss?
              $endgroup$
              – Parseval
              Mar 18 at 13:50






            • 1




              $begingroup$
              They should be 1, 5, 9, 13, 17, 21, 25
              $endgroup$
              – csprun
              Mar 18 at 13:53













            5












            5








            5





            $begingroup$

            There are $7$ integers in the set $S = 1,dots,25$ which are equal to $1$ modulo $4$, and there are $6$ integers in $S$ which are equal to $0$ modulo $4$, and similarly $6$ for $2$ modulo $4$ and $6$ for $3$ modulo $4$. A subset of $S$ has the desired property if and only if it contains at most one number from each of these four groups (any difference of them is then not equal to $0$ modulo $4$, i.e., not divisible by $4$; conversely, if two integers are included which are the same modulo $4$, their difference is divisible by $4$). So you have $8$ choices for the first group: $7$ to include one of the $7$ integers which are equal to $1$ modulo $4$ plus one choice to leave them out entirely; you have $7$ choices for the second group: $6$ to include one of the $6$ which are $0$ modulo $4$ plus one choice to leave them out entirely; and so on. That gives $8cdot 7cdot 7 cdot 7$ possibilities.






            share|cite|improve this answer











            $endgroup$



            There are $7$ integers in the set $S = 1,dots,25$ which are equal to $1$ modulo $4$, and there are $6$ integers in $S$ which are equal to $0$ modulo $4$, and similarly $6$ for $2$ modulo $4$ and $6$ for $3$ modulo $4$. A subset of $S$ has the desired property if and only if it contains at most one number from each of these four groups (any difference of them is then not equal to $0$ modulo $4$, i.e., not divisible by $4$; conversely, if two integers are included which are the same modulo $4$, their difference is divisible by $4$). So you have $8$ choices for the first group: $7$ to include one of the $7$ integers which are equal to $1$ modulo $4$ plus one choice to leave them out entirely; you have $7$ choices for the second group: $6$ to include one of the $6$ which are $0$ modulo $4$ plus one choice to leave them out entirely; and so on. That gives $8cdot 7cdot 7 cdot 7$ possibilities.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 17 at 22:58

























            answered Mar 17 at 22:52









            cspruncsprun

            2,670210




            2,670210











            • $begingroup$
              I get that $S_1=4,9,13,17,21$ are the elements that are equal to $1$ in mod $4$. So $|S_1|=5$. Which numbers did I miss?
              $endgroup$
              – Parseval
              Mar 18 at 13:50






            • 1




              $begingroup$
              They should be 1, 5, 9, 13, 17, 21, 25
              $endgroup$
              – csprun
              Mar 18 at 13:53
















            • $begingroup$
              I get that $S_1=4,9,13,17,21$ are the elements that are equal to $1$ in mod $4$. So $|S_1|=5$. Which numbers did I miss?
              $endgroup$
              – Parseval
              Mar 18 at 13:50






            • 1




              $begingroup$
              They should be 1, 5, 9, 13, 17, 21, 25
              $endgroup$
              – csprun
              Mar 18 at 13:53















            $begingroup$
            I get that $S_1=4,9,13,17,21$ are the elements that are equal to $1$ in mod $4$. So $|S_1|=5$. Which numbers did I miss?
            $endgroup$
            – Parseval
            Mar 18 at 13:50




            $begingroup$
            I get that $S_1=4,9,13,17,21$ are the elements that are equal to $1$ in mod $4$. So $|S_1|=5$. Which numbers did I miss?
            $endgroup$
            – Parseval
            Mar 18 at 13:50




            1




            1




            $begingroup$
            They should be 1, 5, 9, 13, 17, 21, 25
            $endgroup$
            – csprun
            Mar 18 at 13:53




            $begingroup$
            They should be 1, 5, 9, 13, 17, 21, 25
            $endgroup$
            – csprun
            Mar 18 at 13:53











            1












            $begingroup$

            If no $2$ elements of a subset can have a difference divisible by $4$, those elements must belong to different remainder classes modulo $4$. The set $1,2,ldots,25$ contains $7$ elements $equiv 1 pmod 4$ , namly $1,5,9,ldots,25$ and $6$ elements $equiv 2,3,4 pmod 4$ each.



            That means if we wanted to find all $4$-element subsets witht the required property, it would be $7cdot6^3$, because we need one element from each remainder class, and for $equiv 1 pmod 4$ there are 7 choices, for $equiv 2 pmod 4$ there are 6 choices a.s.o.



            But since we are looking for arbitrary size subsets (well, arbitrary means size $le 4$, as otherwise 2 elements would be from the same remainder class), we use a trick. We add the numbers $26,27,28$ and $29$ to the set. If we have a subset of size smaller than $4$, we add the missing remainder classes from those 'new' numbers.



            That is a one-to-one correspondence between all subsets of $1,2,ldots,25$ with the required property and all 4-element subsets of $1,2,ldots,29$ with the required property. Since we added 1 element to each remainder class, that number is $8cdot7^3$






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              If no $2$ elements of a subset can have a difference divisible by $4$, those elements must belong to different remainder classes modulo $4$. The set $1,2,ldots,25$ contains $7$ elements $equiv 1 pmod 4$ , namly $1,5,9,ldots,25$ and $6$ elements $equiv 2,3,4 pmod 4$ each.



              That means if we wanted to find all $4$-element subsets witht the required property, it would be $7cdot6^3$, because we need one element from each remainder class, and for $equiv 1 pmod 4$ there are 7 choices, for $equiv 2 pmod 4$ there are 6 choices a.s.o.



              But since we are looking for arbitrary size subsets (well, arbitrary means size $le 4$, as otherwise 2 elements would be from the same remainder class), we use a trick. We add the numbers $26,27,28$ and $29$ to the set. If we have a subset of size smaller than $4$, we add the missing remainder classes from those 'new' numbers.



              That is a one-to-one correspondence between all subsets of $1,2,ldots,25$ with the required property and all 4-element subsets of $1,2,ldots,29$ with the required property. Since we added 1 element to each remainder class, that number is $8cdot7^3$






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                If no $2$ elements of a subset can have a difference divisible by $4$, those elements must belong to different remainder classes modulo $4$. The set $1,2,ldots,25$ contains $7$ elements $equiv 1 pmod 4$ , namly $1,5,9,ldots,25$ and $6$ elements $equiv 2,3,4 pmod 4$ each.



                That means if we wanted to find all $4$-element subsets witht the required property, it would be $7cdot6^3$, because we need one element from each remainder class, and for $equiv 1 pmod 4$ there are 7 choices, for $equiv 2 pmod 4$ there are 6 choices a.s.o.



                But since we are looking for arbitrary size subsets (well, arbitrary means size $le 4$, as otherwise 2 elements would be from the same remainder class), we use a trick. We add the numbers $26,27,28$ and $29$ to the set. If we have a subset of size smaller than $4$, we add the missing remainder classes from those 'new' numbers.



                That is a one-to-one correspondence between all subsets of $1,2,ldots,25$ with the required property and all 4-element subsets of $1,2,ldots,29$ with the required property. Since we added 1 element to each remainder class, that number is $8cdot7^3$






                share|cite|improve this answer









                $endgroup$



                If no $2$ elements of a subset can have a difference divisible by $4$, those elements must belong to different remainder classes modulo $4$. The set $1,2,ldots,25$ contains $7$ elements $equiv 1 pmod 4$ , namly $1,5,9,ldots,25$ and $6$ elements $equiv 2,3,4 pmod 4$ each.



                That means if we wanted to find all $4$-element subsets witht the required property, it would be $7cdot6^3$, because we need one element from each remainder class, and for $equiv 1 pmod 4$ there are 7 choices, for $equiv 2 pmod 4$ there are 6 choices a.s.o.



                But since we are looking for arbitrary size subsets (well, arbitrary means size $le 4$, as otherwise 2 elements would be from the same remainder class), we use a trick. We add the numbers $26,27,28$ and $29$ to the set. If we have a subset of size smaller than $4$, we add the missing remainder classes from those 'new' numbers.



                That is a one-to-one correspondence between all subsets of $1,2,ldots,25$ with the required property and all 4-element subsets of $1,2,ldots,29$ with the required property. Since we added 1 element to each remainder class, that number is $8cdot7^3$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 17 at 23:01









                IngixIngix

                5,087159




                5,087159



























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