Check if $k$ is divisible by $2^9$ or $2^{10}$












1












$begingroup$


$k=frac{512!}{256!*128!*...*2!*1! } $



I need to check if the expression k is divisible by $2^9$ or $2^{10}$. This is a multiple choice question and the options and the question goes like this:



Statement I : $2^9 divides$ $k$



Statement II : $2^{10} divides$ $k$



(a) Statement I is true, Statement II is true



(b) Statement I is false, Statement II is true



(c) Statement I is true, Statement II is false



(d) Statement I is false, Statement II is false



I almost have no lead in this problem. I just figured out that the numbers are in power of 2 and $2^8!$ can get cancelled from numerator and denominator. :(






divisibility factorial







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$endgroup$








  • 2




    $begingroup$
    You need Legendre's formula for the power of a prime dividing a factorial Clearly II implies I. From a test taking perspective, the fact that the question is asked means the answer is c. They would have chosen different exponents if not.
    $endgroup$
    – Ross Millikan
    Jan 9 at 4:23


















1












$begingroup$


$k=frac{512!}{256!*128!*...*2!*1! } $



I need to check if the expression k is divisible by $2^9$ or $2^{10}$. This is a multiple choice question and the options and the question goes like this:



Statement I : $2^9 divides$ $k$



Statement II : $2^{10} divides$ $k$



(a) Statement I is true, Statement II is true



(b) Statement I is false, Statement II is true



(c) Statement I is true, Statement II is false



(d) Statement I is false, Statement II is false



I almost have no lead in this problem. I just figured out that the numbers are in power of 2 and $2^8!$ can get cancelled from numerator and denominator. :(






divisibility factorial







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You need Legendre's formula for the power of a prime dividing a factorial Clearly II implies I. From a test taking perspective, the fact that the question is asked means the answer is c. They would have chosen different exponents if not.
    $endgroup$
    – Ross Millikan
    Jan 9 at 4:23
















1












1








1





$begingroup$


$k=frac{512!}{256!*128!*...*2!*1! } $



I need to check if the expression k is divisible by $2^9$ or $2^{10}$. This is a multiple choice question and the options and the question goes like this:



Statement I : $2^9 divides$ $k$



Statement II : $2^{10} divides$ $k$



(a) Statement I is true, Statement II is true



(b) Statement I is false, Statement II is true



(c) Statement I is true, Statement II is false



(d) Statement I is false, Statement II is false



I almost have no lead in this problem. I just figured out that the numbers are in power of 2 and $2^8!$ can get cancelled from numerator and denominator. :(






divisibility factorial







share|cite|improve this question









$endgroup$




$k=frac{512!}{256!*128!*...*2!*1! } $



I need to check if the expression k is divisible by $2^9$ or $2^{10}$. This is a multiple choice question and the options and the question goes like this:



Statement I : $2^9 divides$ $k$



Statement II : $2^{10} divides$ $k$



(a) Statement I is true, Statement II is true



(b) Statement I is false, Statement II is true



(c) Statement I is true, Statement II is false



(d) Statement I is false, Statement II is false



I almost have no lead in this problem. I just figured out that the numbers are in power of 2 and $2^8!$ can get cancelled from numerator and denominator. :(






divisibility factorial




divisibility factorial






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 4:12









tomriddle99tomriddle99

1157




1157








  • 2




    $begingroup$
    You need Legendre's formula for the power of a prime dividing a factorial Clearly II implies I. From a test taking perspective, the fact that the question is asked means the answer is c. They would have chosen different exponents if not.
    $endgroup$
    – Ross Millikan
    Jan 9 at 4:23
















  • 2




    $begingroup$
    You need Legendre's formula for the power of a prime dividing a factorial Clearly II implies I. From a test taking perspective, the fact that the question is asked means the answer is c. They would have chosen different exponents if not.
    $endgroup$
    – Ross Millikan
    Jan 9 at 4:23










2




2




$begingroup$
You need Legendre's formula for the power of a prime dividing a factorial Clearly II implies I. From a test taking perspective, the fact that the question is asked means the answer is c. They would have chosen different exponents if not.
$endgroup$
– Ross Millikan
Jan 9 at 4:23






$begingroup$
You need Legendre's formula for the power of a prime dividing a factorial Clearly II implies I. From a test taking perspective, the fact that the question is asked means the answer is c. They would have chosen different exponents if not.
$endgroup$
– Ross Millikan
Jan 9 at 4:23












2 Answers
2






active

oldest

votes


















1












$begingroup$

Well, to start with, any number that's divisible by $2^{10}$ is divisible by $2^n$ - so it's not (b).



To count how many times $2$ is a factor in $n!$, we ask a series of questions. That factorial is a product of $n$ terms; of those terms, how many are divisible by $2$? How many are divisible by $2^2$? How many are divisible by $2^3$? And so on, until we reach a power of $2$ that's larger than $n$. As an example, here's the power of $2$ that goes into $2019!$:

There are $1009$ even numbers $le 2019$, so that's $1009$ terms divisible by $2$. $504$ are divisible by $4$, $252$ are divisible by $8$, $126$ by $16$, $63$ by $32$, $31$ by $64$, $15$ by $128$, $7$ by $256$, $3$ by $512$, and finally one term is divisible by $1024$. Add those up: $1009+504+252+126+63+31+15+7+3+1=2011$. $2019!$ is divisible by $2^{2011}$ but not by $2^{2012}$.



Can you see how it will go for the factorials you're working with? They're all simpler than the example I gave. Once we have that, take the exponents we get for each term in the denominator and add them to get the exponent in the product. Subtract that from the exponent in the numerator to get the total here.



An additional note: the quantity $frac{512!}{256!cdot 128!cdot 64!cdots 2!cdot 1!}$ is an integer because it's a multinomial coefficient - or at least it is if we add another $1!$ term to that product in the denominator.



Oh, and a side note - there is a clever formula for the exponent of $2$ in a factorial. We don't need to know it to do this; it makes things a little easier, but we can just calculate them anyway.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$


    • $1! = 2^0$

    • $2! = 2^1$

    • $4! = 2^3 × 3$

    • $8! = 2^7 × 315$

    • $16! = 2^{15} × 638512875$


    Do you see the pattern here? Can you use it to find exactly how many powers of 2 are in the numerator and denominator?






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      If that pattern is maintained, the denominator has the total power of: $1+3+7+15+31+63+127+255 = 502$ and the numerator has a total power of: $255+256=511$ in the base of $2$. So, $k$ will be divisible by $2^9$ but not by $2^{10}$. Therefore, correct choice is: (c) .
      $endgroup$
      – tomriddle99
      Jan 9 at 4:54











    Your Answer





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    2 Answers
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    2 Answers
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    active

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    1












    $begingroup$

    Well, to start with, any number that's divisible by $2^{10}$ is divisible by $2^n$ - so it's not (b).



    To count how many times $2$ is a factor in $n!$, we ask a series of questions. That factorial is a product of $n$ terms; of those terms, how many are divisible by $2$? How many are divisible by $2^2$? How many are divisible by $2^3$? And so on, until we reach a power of $2$ that's larger than $n$. As an example, here's the power of $2$ that goes into $2019!$:

    There are $1009$ even numbers $le 2019$, so that's $1009$ terms divisible by $2$. $504$ are divisible by $4$, $252$ are divisible by $8$, $126$ by $16$, $63$ by $32$, $31$ by $64$, $15$ by $128$, $7$ by $256$, $3$ by $512$, and finally one term is divisible by $1024$. Add those up: $1009+504+252+126+63+31+15+7+3+1=2011$. $2019!$ is divisible by $2^{2011}$ but not by $2^{2012}$.



    Can you see how it will go for the factorials you're working with? They're all simpler than the example I gave. Once we have that, take the exponents we get for each term in the denominator and add them to get the exponent in the product. Subtract that from the exponent in the numerator to get the total here.



    An additional note: the quantity $frac{512!}{256!cdot 128!cdot 64!cdots 2!cdot 1!}$ is an integer because it's a multinomial coefficient - or at least it is if we add another $1!$ term to that product in the denominator.



    Oh, and a side note - there is a clever formula for the exponent of $2$ in a factorial. We don't need to know it to do this; it makes things a little easier, but we can just calculate them anyway.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Well, to start with, any number that's divisible by $2^{10}$ is divisible by $2^n$ - so it's not (b).



      To count how many times $2$ is a factor in $n!$, we ask a series of questions. That factorial is a product of $n$ terms; of those terms, how many are divisible by $2$? How many are divisible by $2^2$? How many are divisible by $2^3$? And so on, until we reach a power of $2$ that's larger than $n$. As an example, here's the power of $2$ that goes into $2019!$:

      There are $1009$ even numbers $le 2019$, so that's $1009$ terms divisible by $2$. $504$ are divisible by $4$, $252$ are divisible by $8$, $126$ by $16$, $63$ by $32$, $31$ by $64$, $15$ by $128$, $7$ by $256$, $3$ by $512$, and finally one term is divisible by $1024$. Add those up: $1009+504+252+126+63+31+15+7+3+1=2011$. $2019!$ is divisible by $2^{2011}$ but not by $2^{2012}$.



      Can you see how it will go for the factorials you're working with? They're all simpler than the example I gave. Once we have that, take the exponents we get for each term in the denominator and add them to get the exponent in the product. Subtract that from the exponent in the numerator to get the total here.



      An additional note: the quantity $frac{512!}{256!cdot 128!cdot 64!cdots 2!cdot 1!}$ is an integer because it's a multinomial coefficient - or at least it is if we add another $1!$ term to that product in the denominator.



      Oh, and a side note - there is a clever formula for the exponent of $2$ in a factorial. We don't need to know it to do this; it makes things a little easier, but we can just calculate them anyway.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Well, to start with, any number that's divisible by $2^{10}$ is divisible by $2^n$ - so it's not (b).



        To count how many times $2$ is a factor in $n!$, we ask a series of questions. That factorial is a product of $n$ terms; of those terms, how many are divisible by $2$? How many are divisible by $2^2$? How many are divisible by $2^3$? And so on, until we reach a power of $2$ that's larger than $n$. As an example, here's the power of $2$ that goes into $2019!$:

        There are $1009$ even numbers $le 2019$, so that's $1009$ terms divisible by $2$. $504$ are divisible by $4$, $252$ are divisible by $8$, $126$ by $16$, $63$ by $32$, $31$ by $64$, $15$ by $128$, $7$ by $256$, $3$ by $512$, and finally one term is divisible by $1024$. Add those up: $1009+504+252+126+63+31+15+7+3+1=2011$. $2019!$ is divisible by $2^{2011}$ but not by $2^{2012}$.



        Can you see how it will go for the factorials you're working with? They're all simpler than the example I gave. Once we have that, take the exponents we get for each term in the denominator and add them to get the exponent in the product. Subtract that from the exponent in the numerator to get the total here.



        An additional note: the quantity $frac{512!}{256!cdot 128!cdot 64!cdots 2!cdot 1!}$ is an integer because it's a multinomial coefficient - or at least it is if we add another $1!$ term to that product in the denominator.



        Oh, and a side note - there is a clever formula for the exponent of $2$ in a factorial. We don't need to know it to do this; it makes things a little easier, but we can just calculate them anyway.






        share|cite|improve this answer











        $endgroup$



        Well, to start with, any number that's divisible by $2^{10}$ is divisible by $2^n$ - so it's not (b).



        To count how many times $2$ is a factor in $n!$, we ask a series of questions. That factorial is a product of $n$ terms; of those terms, how many are divisible by $2$? How many are divisible by $2^2$? How many are divisible by $2^3$? And so on, until we reach a power of $2$ that's larger than $n$. As an example, here's the power of $2$ that goes into $2019!$:

        There are $1009$ even numbers $le 2019$, so that's $1009$ terms divisible by $2$. $504$ are divisible by $4$, $252$ are divisible by $8$, $126$ by $16$, $63$ by $32$, $31$ by $64$, $15$ by $128$, $7$ by $256$, $3$ by $512$, and finally one term is divisible by $1024$. Add those up: $1009+504+252+126+63+31+15+7+3+1=2011$. $2019!$ is divisible by $2^{2011}$ but not by $2^{2012}$.



        Can you see how it will go for the factorials you're working with? They're all simpler than the example I gave. Once we have that, take the exponents we get for each term in the denominator and add them to get the exponent in the product. Subtract that from the exponent in the numerator to get the total here.



        An additional note: the quantity $frac{512!}{256!cdot 128!cdot 64!cdots 2!cdot 1!}$ is an integer because it's a multinomial coefficient - or at least it is if we add another $1!$ term to that product in the denominator.



        Oh, and a side note - there is a clever formula for the exponent of $2$ in a factorial. We don't need to know it to do this; it makes things a little easier, but we can just calculate them anyway.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 9 at 4:53

























        answered Jan 9 at 4:33









        jmerryjmerry

        3,875514




        3,875514























            1












            $begingroup$


            • $1! = 2^0$

            • $2! = 2^1$

            • $4! = 2^3 × 3$

            • $8! = 2^7 × 315$

            • $16! = 2^{15} × 638512875$


            Do you see the pattern here? Can you use it to find exactly how many powers of 2 are in the numerator and denominator?






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              If that pattern is maintained, the denominator has the total power of: $1+3+7+15+31+63+127+255 = 502$ and the numerator has a total power of: $255+256=511$ in the base of $2$. So, $k$ will be divisible by $2^9$ but not by $2^{10}$. Therefore, correct choice is: (c) .
              $endgroup$
              – tomriddle99
              Jan 9 at 4:54
















            1












            $begingroup$


            • $1! = 2^0$

            • $2! = 2^1$

            • $4! = 2^3 × 3$

            • $8! = 2^7 × 315$

            • $16! = 2^{15} × 638512875$


            Do you see the pattern here? Can you use it to find exactly how many powers of 2 are in the numerator and denominator?






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              If that pattern is maintained, the denominator has the total power of: $1+3+7+15+31+63+127+255 = 502$ and the numerator has a total power of: $255+256=511$ in the base of $2$. So, $k$ will be divisible by $2^9$ but not by $2^{10}$. Therefore, correct choice is: (c) .
              $endgroup$
              – tomriddle99
              Jan 9 at 4:54














            1












            1








            1





            $begingroup$


            • $1! = 2^0$

            • $2! = 2^1$

            • $4! = 2^3 × 3$

            • $8! = 2^7 × 315$

            • $16! = 2^{15} × 638512875$


            Do you see the pattern here? Can you use it to find exactly how many powers of 2 are in the numerator and denominator?






            share|cite|improve this answer









            $endgroup$




            • $1! = 2^0$

            • $2! = 2^1$

            • $4! = 2^3 × 3$

            • $8! = 2^7 × 315$

            • $16! = 2^{15} × 638512875$


            Do you see the pattern here? Can you use it to find exactly how many powers of 2 are in the numerator and denominator?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 4:37









            DanDan

            4,50511517




            4,50511517








            • 1




              $begingroup$
              If that pattern is maintained, the denominator has the total power of: $1+3+7+15+31+63+127+255 = 502$ and the numerator has a total power of: $255+256=511$ in the base of $2$. So, $k$ will be divisible by $2^9$ but not by $2^{10}$. Therefore, correct choice is: (c) .
              $endgroup$
              – tomriddle99
              Jan 9 at 4:54














            • 1




              $begingroup$
              If that pattern is maintained, the denominator has the total power of: $1+3+7+15+31+63+127+255 = 502$ and the numerator has a total power of: $255+256=511$ in the base of $2$. So, $k$ will be divisible by $2^9$ but not by $2^{10}$. Therefore, correct choice is: (c) .
              $endgroup$
              – tomriddle99
              Jan 9 at 4:54








            1




            1




            $begingroup$
            If that pattern is maintained, the denominator has the total power of: $1+3+7+15+31+63+127+255 = 502$ and the numerator has a total power of: $255+256=511$ in the base of $2$. So, $k$ will be divisible by $2^9$ but not by $2^{10}$. Therefore, correct choice is: (c) .
            $endgroup$
            – tomriddle99
            Jan 9 at 4:54




            $begingroup$
            If that pattern is maintained, the denominator has the total power of: $1+3+7+15+31+63+127+255 = 502$ and the numerator has a total power of: $255+256=511$ in the base of $2$. So, $k$ will be divisible by $2^9$ but not by $2^{10}$. Therefore, correct choice is: (c) .
            $endgroup$
            – tomriddle99
            Jan 9 at 4:54


















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