Arithmetic

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Arithmetic

by N:Dure » Sun Dec 12, 2010 11:05 am

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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
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by shovan85 » Sun Dec 12, 2010 11:30 am
N:Dure wrote: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
5n is a square => n = 5 (lowest value)

75np is a cube
75 = 3 * 5^2 then np = 3^2*5 = 45 (lowest value)
then taking n = 5 p = 9

Thus n + p = 14

IMO A
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by Rahul@gurome » Sun Dec 12, 2010 11:50 am
Minimum value of 5n such that it is a square of an integer = 25. Thus minimum possible value of n = 5

75np = (3)*(5²)*(n)*(p)
After placing the minimum possible value of n, 75np = (3)*(5³)*(p)

Thus minimum possible value of 75np such that it is cube of an integer is (3³)*(5³). Thu minimum possible value of p is 3² = 9

Minimum possible value of (n + p) = (5 + 9) = 14

The correct answer is A.
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by N:Dure » Sun Dec 12, 2010 12:28 pm
shovan85 wrote:
N:Dure wrote: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
5n is a square => n = 5 (lowest value)

75np is a cube
75 = 3 * 5^2 then np = 3^2*5 = 45 (lowest value)
then taking n = 5 p = 9

Thus n + p = 14

IMO A
Thanks Shovan!
Minimum value of 5n such that it is a square of an integer = 25. Thus minimum possible value of n = 5

75np = (3)*(5²)*(n)*(p)
After placing the minimum possible value of n, 75np = (3)*(5³)*(p)

Thus minimum possible value of 75np such that it is cube of an integer is (3³)*(5³). Thu minimum possible value of p is 3² = 9

Minimum possible value of (n + p) = (5 + 9) = 14

The correct answer is A.
Thanks Rahul, but I don't get this part: "Thus minimum possible value of 75np such that it is cube of an integer is (3³)*(5³). Thu minimum possible value of p is 3² = 9" ?

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by shovan85 » Sun Dec 19, 2010 9:09 pm
5n is a square => n = 5 (lowest value)

75np is a cube
75 = 3 * 5^2 then np = 3^2*5 = 45 (lowest value)
then taking n = 5 p = 9


Thus n + p = 14
n = 5

We know 75np is a cube of an integer.

Now we have the LOWEST value of n=5, put this value in 75np.

Thus, 75np becomes 75*5*p.

Hence, 375p is a cube.

Factorize 375 to know what can be the minimum integer required to be mulitplied to 375 to make 375p a perfect cube.

375 = 125 * 3 = (5^3) * 3

From factorization we can see that 5 already have a power of 3, So it is already in the form of a cube. We need 3 to be in a perfect cube. To get 3 in the perfect cube from 3^2 = 9 is needed to be multiplied.

Thus p = 9
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Re: Arithmetic

by Scott@TargetTestPrep » Sun Mar 15, 2020 7:25 am
N:Dure wrote:
Sun Dec 12, 2010 11:05 am
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
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

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