Prove that $forall n in Bbb N, ~40^n n! mid (5n)!$












3












$begingroup$


I'm having trouble trying to solve this problem:




Prove that $forall n in Bbb N,~ 40^n n! mid (5n)!$




I must be overlooking something simple because I can't go through it with induction.



Base case $P(1)$ works.



I want to see that it holds for $P(n+1)$.



So I'm looking at $40^{n+1} (n+1)! | (5n+5)!$ with the hypothesis that what I want to prove is right.



But operating I end up with: $dfrac{(5n+5)!}{40^{n+1} (n+1)!}$



If I expand the factorial in the numerator I see my inductive hypothesis but I have an 8 in the denominator and some other factors above that I don't know how to deal with: $$dfrac{(5n+4)(5n+3)(5n+2)(5n+1)5n!}{40^{n} (n)! cdot 8}$$



Perhaps this is not the way to do it. I've also tried starting with my hypothesis and then multiplying by $dfrac{40}{40}$ and $dfrac{(n+1)}{(n+1)}$ but it's almost the same thing.



I'll be thankful with any suggestion!






elementary-number-theory proof-verification proof-writing divisibility factorial







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
    $endgroup$
    – Thomas Andrews
    Apr 25 '17 at 23:42






  • 1




    $begingroup$
    You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
    $endgroup$
    – Marcelo Fornet
    Apr 25 '17 at 23:44






  • 1




    $begingroup$
    The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
    $endgroup$
    – Thomas Andrews
    Apr 25 '17 at 23:46
















3












$begingroup$


I'm having trouble trying to solve this problem:




Prove that $forall n in Bbb N,~ 40^n n! mid (5n)!$




I must be overlooking something simple because I can't go through it with induction.



Base case $P(1)$ works.



I want to see that it holds for $P(n+1)$.



So I'm looking at $40^{n+1} (n+1)! | (5n+5)!$ with the hypothesis that what I want to prove is right.



But operating I end up with: $dfrac{(5n+5)!}{40^{n+1} (n+1)!}$



If I expand the factorial in the numerator I see my inductive hypothesis but I have an 8 in the denominator and some other factors above that I don't know how to deal with: $$dfrac{(5n+4)(5n+3)(5n+2)(5n+1)5n!}{40^{n} (n)! cdot 8}$$



Perhaps this is not the way to do it. I've also tried starting with my hypothesis and then multiplying by $dfrac{40}{40}$ and $dfrac{(n+1)}{(n+1)}$ but it's almost the same thing.



I'll be thankful with any suggestion!






elementary-number-theory proof-verification proof-writing divisibility factorial







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
    $endgroup$
    – Thomas Andrews
    Apr 25 '17 at 23:42






  • 1




    $begingroup$
    You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
    $endgroup$
    – Marcelo Fornet
    Apr 25 '17 at 23:44






  • 1




    $begingroup$
    The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
    $endgroup$
    – Thomas Andrews
    Apr 25 '17 at 23:46














3












3








3





$begingroup$


I'm having trouble trying to solve this problem:




Prove that $forall n in Bbb N,~ 40^n n! mid (5n)!$




I must be overlooking something simple because I can't go through it with induction.



Base case $P(1)$ works.



I want to see that it holds for $P(n+1)$.



So I'm looking at $40^{n+1} (n+1)! | (5n+5)!$ with the hypothesis that what I want to prove is right.



But operating I end up with: $dfrac{(5n+5)!}{40^{n+1} (n+1)!}$



If I expand the factorial in the numerator I see my inductive hypothesis but I have an 8 in the denominator and some other factors above that I don't know how to deal with: $$dfrac{(5n+4)(5n+3)(5n+2)(5n+1)5n!}{40^{n} (n)! cdot 8}$$



Perhaps this is not the way to do it. I've also tried starting with my hypothesis and then multiplying by $dfrac{40}{40}$ and $dfrac{(n+1)}{(n+1)}$ but it's almost the same thing.



I'll be thankful with any suggestion!






elementary-number-theory proof-verification proof-writing divisibility factorial







share|cite|improve this question











$endgroup$




I'm having trouble trying to solve this problem:




Prove that $forall n in Bbb N,~ 40^n n! mid (5n)!$




I must be overlooking something simple because I can't go through it with induction.



Base case $P(1)$ works.



I want to see that it holds for $P(n+1)$.



So I'm looking at $40^{n+1} (n+1)! | (5n+5)!$ with the hypothesis that what I want to prove is right.



But operating I end up with: $dfrac{(5n+5)!}{40^{n+1} (n+1)!}$



If I expand the factorial in the numerator I see my inductive hypothesis but I have an 8 in the denominator and some other factors above that I don't know how to deal with: $$dfrac{(5n+4)(5n+3)(5n+2)(5n+1)5n!}{40^{n} (n)! cdot 8}$$



Perhaps this is not the way to do it. I've also tried starting with my hypothesis and then multiplying by $dfrac{40}{40}$ and $dfrac{(n+1)}{(n+1)}$ but it's almost the same thing.



I'll be thankful with any suggestion!






elementary-number-theory proof-verification proof-writing divisibility factorial




elementary-number-theory proof-verification proof-writing divisibility factorial






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 25 '17 at 23:54









Jaideep Khare

17.8k32669




17.8k32669










asked Apr 25 '17 at 23:38









Joaquin RomeraJoaquin Romera

883520




883520








  • 2




    $begingroup$
    So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
    $endgroup$
    – Thomas Andrews
    Apr 25 '17 at 23:42






  • 1




    $begingroup$
    You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
    $endgroup$
    – Marcelo Fornet
    Apr 25 '17 at 23:44






  • 1




    $begingroup$
    The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
    $endgroup$
    – Thomas Andrews
    Apr 25 '17 at 23:46














  • 2




    $begingroup$
    So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
    $endgroup$
    – Thomas Andrews
    Apr 25 '17 at 23:42






  • 1




    $begingroup$
    You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
    $endgroup$
    – Marcelo Fornet
    Apr 25 '17 at 23:44






  • 1




    $begingroup$
    The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
    $endgroup$
    – Thomas Andrews
    Apr 25 '17 at 23:46








2




2




$begingroup$
So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:42




$begingroup$
So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:42




1




1




$begingroup$
You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
$endgroup$
– Marcelo Fornet
Apr 25 '17 at 23:44




$begingroup$
You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
$endgroup$
– Marcelo Fornet
Apr 25 '17 at 23:44




1




1




$begingroup$
The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:46




$begingroup$
The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:46










2 Answers
2






active

oldest

votes


















3












$begingroup$

You are actually done:
$$
frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
$$



The first is an integer, by the inductive hypothesis.



To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).



Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.



Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$



    Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.



    Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      3












      $begingroup$

      You are actually done:
      $$
      frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
      $$



      The first is an integer, by the inductive hypothesis.



      To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).



      Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.



      Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        You are actually done:
        $$
        frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
        $$



        The first is an integer, by the inductive hypothesis.



        To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).



        Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.



        Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          You are actually done:
          $$
          frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
          $$



          The first is an integer, by the inductive hypothesis.



          To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).



          Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.



          Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.






          share|cite|improve this answer











          $endgroup$



          You are actually done:
          $$
          frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
          $$



          The first is an integer, by the inductive hypothesis.



          To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).



          Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.



          Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 3 at 17:51

























          answered Apr 25 '17 at 23:45









          астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

          39.1k33477




          39.1k33477























              2












              $begingroup$

              You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$



              Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.



              Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$



                Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.



                Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$



                  Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.



                  Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$






                  share|cite|improve this answer









                  $endgroup$



                  You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$



                  Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.



                  Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 25 '17 at 23:45









                  Jaideep KhareJaideep Khare

                  17.8k32669




                  17.8k32669






























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