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I can't solve this problem:

Suppose $n$ and $p$ are integers greater than $1$, $5n$ is the square of a number, and $75np$ is the cube of a number. What is the smallest value for $n+p$?

(Answer given is $14$)

I don't even understand if $5n$ is the square of the same number which has a cube of $75np$. Any suggestions? How would I solve this problem?

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  • $\begingroup$ No, the problem does not tell you that $5n$ must be the square of the same number that $75np$ is a cube of. $\endgroup$ – Zev Chonoles Jul 24 '12 at 7:48
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$5n$ is a square implies that $5n = 5^2 \times a^2$. This gives us that $n = 5a^2$, where $a \in \mathbb{Z}$.

Similarly, $75np$ is a cube implies that $75np = 3^3 5^3 b^3 \implies np = 3^2 5 b^3$, where $b \in \mathbb{Z}$.

Can you now conclude what $a$ and $b$ should be for $n+p$ to be a minimum?

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  • $\begingroup$ Thank you for your answer.I guess I can take it from here.. since $n=5a^2$ and $p=\frac{9b^3}{a^2}$ put a and b =1 we get 14. However I still dont get how you got $75np = 3^35^3b^3$. It says 75np is the cube of a number. How can that number be $(3.5.b)^3$.? $\endgroup$ – Rajeshwar Jul 24 '12 at 9:33
  • $\begingroup$ @Rajeshwar: What you need to do is "prime factorization", $75np$ consists of one $3$ and two $5$s so to be the cube of a number it needs two $3$s and one $5$, when you already have $n=5k$ then you only need $p$ to be $9k'$ ($k \in \mathbb N$ since $n$ and $p$ are integers greater than $1$). $\endgroup$ – Gigili Jul 24 '12 at 9:54
  • $\begingroup$ @Gigili I understand the prime factorization part but I still dont get the part so to be the cube of a number it needs two 3s and one 5 ? $\endgroup$ – Rajeshwar Jul 24 '12 at 13:58
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    $\begingroup$ @Rajeshwar: As long as every cube is in the form of $a^3$, the prime factors should at least be raised to the power of 3. $75 = 3^1 \cdot 5^2$, so ... $\endgroup$ – Gigili Jul 24 '12 at 14:12
  • $\begingroup$ That answers it. Thanks. $\endgroup$ – Rajeshwar Jul 24 '12 at 14:15
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  • $5n$ is the square of a number, so $n=5k$ ($k \in \mathbb N$).
  • $75np=25n \cdot 3p=5^3k \cdot 3p $ is the cube of a number when $n=5k$, so $p=9k'$ ($k' \in \mathbb N$).

The smallest value of $n+p$ is when $k =k'=1$.

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  • $\begingroup$ @DavidWallace: You're free to do so, not that I need your upvote or something like that. $\endgroup$ – Gigili Jul 24 '12 at 11:08
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You're given that $5n$ is a square number. If it's square, each of its prime factors must be to an even power. This means that $n$ must be divisible by $5$.

Now you're given that $75np$ is a cube. If it's a cube, each of its prime factors must be to a power that's a multiple of $3$. Can you finish the argument from here?

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