Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
No, the problem does not tell you that $5n$ must be the square of the same number that $75np$ is a cube of. –  Zev Chonoles Jul 24 '12 at 7:48

3 Answers 3

up vote 3 down vote accepted

$5n$ 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?

share|improve this answer
    
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 $(3.5.b)^3$.? –  Rajeshwar Jul 24 '12 at 9:33
    
@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 $1$). –  Gigili Jul 24 '12 at 9:54
    
@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_ ? –  Rajeshwar Jul 24 '12 at 13:58
2  
@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 ... –  Gigili Jul 24 '12 at 14:12
    
That answers it. Thanks. –  Rajeshwar Jul 24 '12 at 14:15
  • $5n$ 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 $k =k'=1$.

share|improve this answer
    
@DavidWallace: You're free to do so, not that I need your upvote or something like that. –  Gigili Jul 24 '12 at 11:08

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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.