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Bunuel
For which of the following positive integers is the square of that integer divided by the cube root of the same integer equal to nine times that integer?

(A) 4
(B) 8
(C) 16
(D) 27
(E) 125

Hi Bunuel,

I tried by taking each of the answer choices.

I started with 125 then went on to 27 and stopped.

Some other way to solve this?
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Bunuel
For which of the following positive integers is the square of that integer divided by the cube root of the same integer equal to nine times that integer?

(A) 4
(B) 8
(C) 16
(D) 27
(E) 125

Working with the options is the best strategy for solving such questions...

(A) Cube of 4 not possible ( For GMAT ) hence rejected.
(B) 8*8/2 = 32 ( Not equal to 9 times 8 = 72) , hence rejected.
(C) Cube of 16 not possible ( For GMAT ) hence rejected.
(D) 27*27/3 = 243 = 9*27 (Answer )
(E) 125*125/5 = 3125 ( Not equal to 9 times 125 = 1125) , hence rejected.

The answer must be (D)
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Bunuel
For which of the following positive integers is the square of that integer divided by the cube root of the same integer equal to nine times that integer?

(A) 4
(B) 8
(C) 16
(D) 27
(E) 125

Let the positive integer be x.

Given, square of integer 'x' divided by the cube root of the same integer(x) equal to nine times that integer(x).

Converting the problem from word to algebra:

\(\sqrt{x}/\sqrt[3]{x}=9x\)
Or, \(x^2/x^{\frac{1}{3}}=9x\)
Or, \(x^{2-\frac{1}{3}}=9x\)
Or, \(x^{\frac{5}{3}}=9x\)
Or, \((x^{\frac{5}{3}})^3=(9x)^3\)
Or, \(x^5=9^3*x^3\)

Since x is positive, we can cancel out common factor from both the sides, we have
\(x^2=9^3\)
Or,\(x^2=3^6=(3^3)^2\)
Or, \(x=3^3=27\)

Ans. (D)

Hi zanaik89,
I hope this helps.
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Bunuel
For which of the following positive integers is the square of that integer divided by the cube root of the same integer equal to nine times that integer?

(A) 4
(B) 8
(C) 16
(D) 27
(E) 125
Answer is 27
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Bunuel
For which of the following positive integers is the square of that integer divided by the cube root of the same integer equal to nine times that integer?

(A) 4
(B) 8
(C) 16
(D) 27
(E) 125

let x^3=integer
let x=cube root of integer
let x^6=square of integer
x^6/x=9x^3➡
x=3
x^3=27
D
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Solution



Given


To Find

    • The positive integers for which the square of that integer divided by the cube root of the same integer equal to nine times that integer.

Approach and Working Out

Let the integer be \(x^3\).
    • As we know it has a cube root.
The square of that integer is \(x^6\).
    • We can translate the information as \(x^6\)/x = 9\(x^3\)
    • \(x^5 \)= 9 \(x^3\)
    • x = 3
    • \(x^3\) = 27 (As the number was \(x^3\))

Correct Answer: Option D
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Bunuel
For which of the following positive integers is the square of that integer divided by the cube root of the same integer equal to nine times that integer?

(A) 4
(B) 8
(C) 16
(D) 27
(E) 125
Solution:

Since 9 is a power of 3 and 27 is also a power of 3, we will guess the integer is 27. Now, let’s check if that is indeed the case.

We see that 27^2 = 729 and ∛27 = 3 and 729 / 3 = 243. Since 243 is indeed 9 times 27, we see that 27 is the correct answer.

Answer: D
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