n and p are integers greater than 1 5n is the square of a number 75np

Question

n and p are integers greater than 1;
5n is the square of a number;
75np is the cube of a number.
The smallest value for n + p is

A. 14 
B. 18 
C. 20 
D. 30 
E. 50

DensetsuNo · Accepted Answer

n > 1 
 p > 1 
 5*n = x^2 
 75np = y^3 
 n+p = ?

Well, a square number is usually the result of himself times himself so in the case of n the least value possible (being 5 a prime) is 5*5
 n=5 
as for p
 75*5*p= y^3 
we know  375  is equal =  5^3*3^1  therefore to obtain a cube we need  3^2   or  9  ->   p=9

p+n=14

A)

quantumliner · Answer

n and p are integers greater than 1; 5n is the square of a number; 75np is the cube of a number. The smallest value for n + p is

Given :  5n is the square of a number 
This means that n=5 a^2 , where a>=1

Given :  75np is the cube of a number 
This means that 3*25*5 a^2 p is the cube of a number. This implies that p= 3^2 a

Now n+p = 5 a^2  + 9a = a

The smallest value of n+p, will be when a=1. Therefore, n+p = 5+9 = 14

Answer is  A

ScottTargetTestPrep · Answer

The smallest value for n is 5.

Recall that a perfect cube has prime factors that must each be raised to a multiple of 3. We will use this fact to solve for p.

So 75np = 375p, and since 375p = 5^3 x 3p is a cube, we see that the smallest value for p is 9 (notice that 375p = 5^3 x 3^3). Therefore, the smallest value for n + p is 5 + 9 = 14.

Answer: A

fireagablast · Answer

5n = x^2
n = 5 because in order for x^2 to be an integer square, , the root must also be an integer

75np = x^3
75 = 3*5*5
75n = 3*5*5*5

in order for x^3 to be cubed, we need an even distribution of factors.
75n has three 5's and one 3, so p needs to add an additional two 3's

75np = 3*3*3*5*5*5

therefore:
n=5
p=3*3=9

n+p = 14

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n and p are integers greater than 1 5n is the square of a number 75np

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n and p are integers greater than 1 5n is the square of a number 75np

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