Minimum value of unknown exponents for a perfect cube











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The following question showed up in an aptitude test I took.




If $N = 2197^P × 144^2 × 2^R × 3^S$ is the perfect cube of a natural
number, where $P$, $R$ and $S$ are distinct positive integers, then find the
minimum value of $(P + R + S)$.




I solved it as follows:
My solution



However, the correct answer is 6 as per the answer key released. Could anyone please tell where am I going wrong?



Here is the solution provided by the testing agency:
Provided solution










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  • The answer you provide appears to be $4$ as well, no?
    – lulu
    Nov 15 at 18:53










  • @lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
    – Aamir Khan
    Nov 15 at 18:54












  • Oh, the problem is that $P,Q,R$ are required to be distinct.
    – lulu
    Nov 15 at 18:56










  • Maybe. Let me just also attach the image of their solution.
    – Aamir Khan
    Nov 15 at 18:57






  • 1




    Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
    – lulu
    Nov 15 at 19:01















up vote
0
down vote

favorite












The following question showed up in an aptitude test I took.




If $N = 2197^P × 144^2 × 2^R × 3^S$ is the perfect cube of a natural
number, where $P$, $R$ and $S$ are distinct positive integers, then find the
minimum value of $(P + R + S)$.




I solved it as follows:
My solution



However, the correct answer is 6 as per the answer key released. Could anyone please tell where am I going wrong?



Here is the solution provided by the testing agency:
Provided solution










share|cite|improve this question
























  • The answer you provide appears to be $4$ as well, no?
    – lulu
    Nov 15 at 18:53










  • @lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
    – Aamir Khan
    Nov 15 at 18:54












  • Oh, the problem is that $P,Q,R$ are required to be distinct.
    – lulu
    Nov 15 at 18:56










  • Maybe. Let me just also attach the image of their solution.
    – Aamir Khan
    Nov 15 at 18:57






  • 1




    Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
    – lulu
    Nov 15 at 19:01













up vote
0
down vote

favorite









up vote
0
down vote

favorite











The following question showed up in an aptitude test I took.




If $N = 2197^P × 144^2 × 2^R × 3^S$ is the perfect cube of a natural
number, where $P$, $R$ and $S$ are distinct positive integers, then find the
minimum value of $(P + R + S)$.




I solved it as follows:
My solution



However, the correct answer is 6 as per the answer key released. Could anyone please tell where am I going wrong?



Here is the solution provided by the testing agency:
Provided solution










share|cite|improve this question















The following question showed up in an aptitude test I took.




If $N = 2197^P × 144^2 × 2^R × 3^S$ is the perfect cube of a natural
number, where $P$, $R$ and $S$ are distinct positive integers, then find the
minimum value of $(P + R + S)$.




I solved it as follows:
My solution



However, the correct answer is 6 as per the answer key released. Could anyone please tell where am I going wrong?



Here is the solution provided by the testing agency:
Provided solution







number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 15 at 19:15

























asked Nov 15 at 18:51









Aamir Khan

33




33












  • The answer you provide appears to be $4$ as well, no?
    – lulu
    Nov 15 at 18:53










  • @lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
    – Aamir Khan
    Nov 15 at 18:54












  • Oh, the problem is that $P,Q,R$ are required to be distinct.
    – lulu
    Nov 15 at 18:56










  • Maybe. Let me just also attach the image of their solution.
    – Aamir Khan
    Nov 15 at 18:57






  • 1




    Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
    – lulu
    Nov 15 at 19:01


















  • The answer you provide appears to be $4$ as well, no?
    – lulu
    Nov 15 at 18:53










  • @lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
    – Aamir Khan
    Nov 15 at 18:54












  • Oh, the problem is that $P,Q,R$ are required to be distinct.
    – lulu
    Nov 15 at 18:56










  • Maybe. Let me just also attach the image of their solution.
    – Aamir Khan
    Nov 15 at 18:57






  • 1




    Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
    – lulu
    Nov 15 at 19:01
















The answer you provide appears to be $4$ as well, no?
– lulu
Nov 15 at 18:53




The answer you provide appears to be $4$ as well, no?
– lulu
Nov 15 at 18:53












@lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
– Aamir Khan
Nov 15 at 18:54






@lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
– Aamir Khan
Nov 15 at 18:54














Oh, the problem is that $P,Q,R$ are required to be distinct.
– lulu
Nov 15 at 18:56




Oh, the problem is that $P,Q,R$ are required to be distinct.
– lulu
Nov 15 at 18:56












Maybe. Let me just also attach the image of their solution.
– Aamir Khan
Nov 15 at 18:57




Maybe. Let me just also attach the image of their solution.
– Aamir Khan
Nov 15 at 18:57




1




1




Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
– lulu
Nov 15 at 19:01




Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
– lulu
Nov 15 at 19:01










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$$2197 = 13^3$$
$$144 = 2^4 3^2$$



exponent of $13$ is $3P$



exponent of $2$ is $4 + R$



exponent of $3$ is $2+S$



So $P$ is $1,2,3,4,5...$



$R$ is $2,5,8,...$



$S$ is $1,4,7,...$



If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$



With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$






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    $$2197 = 13^3$$
    $$144 = 2^4 3^2$$



    exponent of $13$ is $3P$



    exponent of $2$ is $4 + R$



    exponent of $3$ is $2+S$



    So $P$ is $1,2,3,4,5...$



    $R$ is $2,5,8,...$



    $S$ is $1,4,7,...$



    If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$



    With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$






    share|cite|improve this answer

























      up vote
      0
      down vote













      $$2197 = 13^3$$
      $$144 = 2^4 3^2$$



      exponent of $13$ is $3P$



      exponent of $2$ is $4 + R$



      exponent of $3$ is $2+S$



      So $P$ is $1,2,3,4,5...$



      $R$ is $2,5,8,...$



      $S$ is $1,4,7,...$



      If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$



      With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        $$2197 = 13^3$$
        $$144 = 2^4 3^2$$



        exponent of $13$ is $3P$



        exponent of $2$ is $4 + R$



        exponent of $3$ is $2+S$



        So $P$ is $1,2,3,4,5...$



        $R$ is $2,5,8,...$



        $S$ is $1,4,7,...$



        If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$



        With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$






        share|cite|improve this answer












        $$2197 = 13^3$$
        $$144 = 2^4 3^2$$



        exponent of $13$ is $3P$



        exponent of $2$ is $4 + R$



        exponent of $3$ is $2+S$



        So $P$ is $1,2,3,4,5...$



        $R$ is $2,5,8,...$



        $S$ is $1,4,7,...$



        If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$



        With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 15 at 19:44









        Will Jagy

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