Let $n$ be the product of all positive factors of $10^4$. Find $log_{10} n$












2












$begingroup$


I can't see any relationship between the products of the positive factors of powers of 10.
The answer given here is 50, by the way, but I simply don't know where to start.






logarithms







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: Almost all factors can be paired up $(a,b)$ such that $ab = 10^4$. So this really boils down to: which one(s) can't? And how many pairs of factors can we make?
    $endgroup$
    – platty
    Dec 12 '18 at 0:04










  • $begingroup$
    $10^4=2^4cdot 5^4$, now all combinations of $2^kcdot 5^j$ for $j,kleq 4$ are your options here. If you use brute force (basically some technical combinatorics) you can now get to the result you want using $log$'s properties (and use some symmetries to make it easier to calculate $log_{10}$ of the things you get). Might be a more elegant way but this should also work.
    $endgroup$
    – NL1992
    Dec 12 '18 at 0:08
















2












$begingroup$


I can't see any relationship between the products of the positive factors of powers of 10.
The answer given here is 50, by the way, but I simply don't know where to start.






logarithms







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: Almost all factors can be paired up $(a,b)$ such that $ab = 10^4$. So this really boils down to: which one(s) can't? And how many pairs of factors can we make?
    $endgroup$
    – platty
    Dec 12 '18 at 0:04










  • $begingroup$
    $10^4=2^4cdot 5^4$, now all combinations of $2^kcdot 5^j$ for $j,kleq 4$ are your options here. If you use brute force (basically some technical combinatorics) you can now get to the result you want using $log$'s properties (and use some symmetries to make it easier to calculate $log_{10}$ of the things you get). Might be a more elegant way but this should also work.
    $endgroup$
    – NL1992
    Dec 12 '18 at 0:08














2












2








2





$begingroup$


I can't see any relationship between the products of the positive factors of powers of 10.
The answer given here is 50, by the way, but I simply don't know where to start.






logarithms







share|cite|improve this question









$endgroup$




I can't see any relationship between the products of the positive factors of powers of 10.
The answer given here is 50, by the way, but I simply don't know where to start.






logarithms




logarithms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 0:00









Nameless KingNameless King

306




306








  • 1




    $begingroup$
    Hint: Almost all factors can be paired up $(a,b)$ such that $ab = 10^4$. So this really boils down to: which one(s) can't? And how many pairs of factors can we make?
    $endgroup$
    – platty
    Dec 12 '18 at 0:04










  • $begingroup$
    $10^4=2^4cdot 5^4$, now all combinations of $2^kcdot 5^j$ for $j,kleq 4$ are your options here. If you use brute force (basically some technical combinatorics) you can now get to the result you want using $log$'s properties (and use some symmetries to make it easier to calculate $log_{10}$ of the things you get). Might be a more elegant way but this should also work.
    $endgroup$
    – NL1992
    Dec 12 '18 at 0:08














  • 1




    $begingroup$
    Hint: Almost all factors can be paired up $(a,b)$ such that $ab = 10^4$. So this really boils down to: which one(s) can't? And how many pairs of factors can we make?
    $endgroup$
    – platty
    Dec 12 '18 at 0:04










  • $begingroup$
    $10^4=2^4cdot 5^4$, now all combinations of $2^kcdot 5^j$ for $j,kleq 4$ are your options here. If you use brute force (basically some technical combinatorics) you can now get to the result you want using $log$'s properties (and use some symmetries to make it easier to calculate $log_{10}$ of the things you get). Might be a more elegant way but this should also work.
    $endgroup$
    – NL1992
    Dec 12 '18 at 0:08








1




1




$begingroup$
Hint: Almost all factors can be paired up $(a,b)$ such that $ab = 10^4$. So this really boils down to: which one(s) can't? And how many pairs of factors can we make?
$endgroup$
– platty
Dec 12 '18 at 0:04




$begingroup$
Hint: Almost all factors can be paired up $(a,b)$ such that $ab = 10^4$. So this really boils down to: which one(s) can't? And how many pairs of factors can we make?
$endgroup$
– platty
Dec 12 '18 at 0:04












$begingroup$
$10^4=2^4cdot 5^4$, now all combinations of $2^kcdot 5^j$ for $j,kleq 4$ are your options here. If you use brute force (basically some technical combinatorics) you can now get to the result you want using $log$'s properties (and use some symmetries to make it easier to calculate $log_{10}$ of the things you get). Might be a more elegant way but this should also work.
$endgroup$
– NL1992
Dec 12 '18 at 0:08




$begingroup$
$10^4=2^4cdot 5^4$, now all combinations of $2^kcdot 5^j$ for $j,kleq 4$ are your options here. If you use brute force (basically some technical combinatorics) you can now get to the result you want using $log$'s properties (and use some symmetries to make it easier to calculate $log_{10}$ of the things you get). Might be a more elegant way but this should also work.
$endgroup$
– NL1992
Dec 12 '18 at 0:08










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint:



We can see, either through investigation or just trying a few examples if you want, that, if a number $n$ has $d(n)$ divisors (sometimes denoted $sigma_0(n)$ instead), then we can give the product of $n$'s divisors by



$$prod_{d|n} d = n^{d(n) / 2}$$



This is discussed in more length at Product of Divisors of some $n$ proof



Personally I've seen this occasionally denoted $pi(n)$ and used such a notation myself when I was looking into the product of a number's divisors. Either way, not important. But with that in mind, you are seeking



$$log_{10} left( pi(10^4) right) ;;; text{or equivalently} ;;; log_{10} left( prod_{d|10^4} d right)$$



With the previous fact in mind, this problem is essentially reduced to simply finding the number of divisors of $10^4$ after using a few basic properties of logarithms and exponents.



You should be able to easily continue from here.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    As @NL1992 pointed out, the key is write $10^4 = 2^4 times 5^4$.



    Consider the factors which have no contribution from the "2". These are $5^0, 5^1, 5^2, 5^3$, and $5^4$. Their product is $5^{10}$.



    Now, consider the factors which exactly one "2". These are $2 times 5^0$, $2 times 5^1$, $2 times 5^2$, $2 times 5^3$, and $2 times 5^4$. Their product is $2^5 times 5^{10}$.



    Generalizing, you can see that if you multiply all factors which have a $2^k$ term, you get $2^{5k} times 5^{10}$.



    So the final answer is $n = prod_{k=0}^4 2^{5k}5^{10} = 2^{50} 5^{50}$, implying $log_{10}(n) = 50$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Hint:



      We can see, either through investigation or just trying a few examples if you want, that, if a number $n$ has $d(n)$ divisors (sometimes denoted $sigma_0(n)$ instead), then we can give the product of $n$'s divisors by



      $$prod_{d|n} d = n^{d(n) / 2}$$



      This is discussed in more length at Product of Divisors of some $n$ proof



      Personally I've seen this occasionally denoted $pi(n)$ and used such a notation myself when I was looking into the product of a number's divisors. Either way, not important. But with that in mind, you are seeking



      $$log_{10} left( pi(10^4) right) ;;; text{or equivalently} ;;; log_{10} left( prod_{d|10^4} d right)$$



      With the previous fact in mind, this problem is essentially reduced to simply finding the number of divisors of $10^4$ after using a few basic properties of logarithms and exponents.



      You should be able to easily continue from here.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Hint:



        We can see, either through investigation or just trying a few examples if you want, that, if a number $n$ has $d(n)$ divisors (sometimes denoted $sigma_0(n)$ instead), then we can give the product of $n$'s divisors by



        $$prod_{d|n} d = n^{d(n) / 2}$$



        This is discussed in more length at Product of Divisors of some $n$ proof



        Personally I've seen this occasionally denoted $pi(n)$ and used such a notation myself when I was looking into the product of a number's divisors. Either way, not important. But with that in mind, you are seeking



        $$log_{10} left( pi(10^4) right) ;;; text{or equivalently} ;;; log_{10} left( prod_{d|10^4} d right)$$



        With the previous fact in mind, this problem is essentially reduced to simply finding the number of divisors of $10^4$ after using a few basic properties of logarithms and exponents.



        You should be able to easily continue from here.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint:



          We can see, either through investigation or just trying a few examples if you want, that, if a number $n$ has $d(n)$ divisors (sometimes denoted $sigma_0(n)$ instead), then we can give the product of $n$'s divisors by



          $$prod_{d|n} d = n^{d(n) / 2}$$



          This is discussed in more length at Product of Divisors of some $n$ proof



          Personally I've seen this occasionally denoted $pi(n)$ and used such a notation myself when I was looking into the product of a number's divisors. Either way, not important. But with that in mind, you are seeking



          $$log_{10} left( pi(10^4) right) ;;; text{or equivalently} ;;; log_{10} left( prod_{d|10^4} d right)$$



          With the previous fact in mind, this problem is essentially reduced to simply finding the number of divisors of $10^4$ after using a few basic properties of logarithms and exponents.



          You should be able to easily continue from here.






          share|cite|improve this answer











          $endgroup$



          Hint:



          We can see, either through investigation or just trying a few examples if you want, that, if a number $n$ has $d(n)$ divisors (sometimes denoted $sigma_0(n)$ instead), then we can give the product of $n$'s divisors by



          $$prod_{d|n} d = n^{d(n) / 2}$$



          This is discussed in more length at Product of Divisors of some $n$ proof



          Personally I've seen this occasionally denoted $pi(n)$ and used such a notation myself when I was looking into the product of a number's divisors. Either way, not important. But with that in mind, you are seeking



          $$log_{10} left( pi(10^4) right) ;;; text{or equivalently} ;;; log_{10} left( prod_{d|10^4} d right)$$



          With the previous fact in mind, this problem is essentially reduced to simply finding the number of divisors of $10^4$ after using a few basic properties of logarithms and exponents.



          You should be able to easily continue from here.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 0:24

























          answered Dec 12 '18 at 0:18









          Eevee TrainerEevee Trainer

          5,7571936




          5,7571936























              2












              $begingroup$

              As @NL1992 pointed out, the key is write $10^4 = 2^4 times 5^4$.



              Consider the factors which have no contribution from the "2". These are $5^0, 5^1, 5^2, 5^3$, and $5^4$. Their product is $5^{10}$.



              Now, consider the factors which exactly one "2". These are $2 times 5^0$, $2 times 5^1$, $2 times 5^2$, $2 times 5^3$, and $2 times 5^4$. Their product is $2^5 times 5^{10}$.



              Generalizing, you can see that if you multiply all factors which have a $2^k$ term, you get $2^{5k} times 5^{10}$.



              So the final answer is $n = prod_{k=0}^4 2^{5k}5^{10} = 2^{50} 5^{50}$, implying $log_{10}(n) = 50$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                As @NL1992 pointed out, the key is write $10^4 = 2^4 times 5^4$.



                Consider the factors which have no contribution from the "2". These are $5^0, 5^1, 5^2, 5^3$, and $5^4$. Their product is $5^{10}$.



                Now, consider the factors which exactly one "2". These are $2 times 5^0$, $2 times 5^1$, $2 times 5^2$, $2 times 5^3$, and $2 times 5^4$. Their product is $2^5 times 5^{10}$.



                Generalizing, you can see that if you multiply all factors which have a $2^k$ term, you get $2^{5k} times 5^{10}$.



                So the final answer is $n = prod_{k=0}^4 2^{5k}5^{10} = 2^{50} 5^{50}$, implying $log_{10}(n) = 50$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  As @NL1992 pointed out, the key is write $10^4 = 2^4 times 5^4$.



                  Consider the factors which have no contribution from the "2". These are $5^0, 5^1, 5^2, 5^3$, and $5^4$. Their product is $5^{10}$.



                  Now, consider the factors which exactly one "2". These are $2 times 5^0$, $2 times 5^1$, $2 times 5^2$, $2 times 5^3$, and $2 times 5^4$. Their product is $2^5 times 5^{10}$.



                  Generalizing, you can see that if you multiply all factors which have a $2^k$ term, you get $2^{5k} times 5^{10}$.



                  So the final answer is $n = prod_{k=0}^4 2^{5k}5^{10} = 2^{50} 5^{50}$, implying $log_{10}(n) = 50$.






                  share|cite|improve this answer









                  $endgroup$



                  As @NL1992 pointed out, the key is write $10^4 = 2^4 times 5^4$.



                  Consider the factors which have no contribution from the "2". These are $5^0, 5^1, 5^2, 5^3$, and $5^4$. Their product is $5^{10}$.



                  Now, consider the factors which exactly one "2". These are $2 times 5^0$, $2 times 5^1$, $2 times 5^2$, $2 times 5^3$, and $2 times 5^4$. Their product is $2^5 times 5^{10}$.



                  Generalizing, you can see that if you multiply all factors which have a $2^k$ term, you get $2^{5k} times 5^{10}$.



                  So the final answer is $n = prod_{k=0}^4 2^{5k}5^{10} = 2^{50} 5^{50}$, implying $log_{10}(n) = 50$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 0:24









                  Aditya DuaAditya Dua

                  1,11418




                  1,11418






























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