What is the smallest positive integer x, such that 1,152x is a perfect

prime factorization 0f 1152 = 2*2*2*2*2*2*2*3*3 = \(2^{7}\) *\(3^2\)
for a number to be perfect cube it should have powers of its prime factors in multiple of 3 (3,6 ,9 ...etc)
we need \(2^2\) *3 = 12 to make the given number a perfect cube so x =12
Correct Answer - D

1152 =2^7*3^2
for it to be a perfect cube we have to multiply it with 2^2 *3 = 12

1,152 is 2^7*3^2

Or 2^3 * 2^3 *2*3^2

We need 2^2 & 3, to make 1152 a perfect cube.
So 4*3=12 D.

It seems the question is asking if x is an integer placed in front of the number making it either a 5 digit perfect cube or a six digit perfect cube.
please post questions with clarity. The language here should have been : what should be the minimum value of integer x when multiplied by 1152 yields a perfect cube?
Thanks!

1,152x means 1,152*x.

1152 = (2^7)(3^2)

To get a perfect cube, we need to get the exponents (listed above) so that they are multiples of 3 --> closest multiple for 2^7 is 2^9 --> we need 2^2 to do this (KEEP)

Let's do the same for 3^2: closest multiple is 3^3 --> we need 3 to do this (KEEP)

X should have both 2^2 & 3 in order to transform 1152 into a perfect cube.

Thus, D is correct.

------ASIDE--------------------------------------
Let's examine a perfect CUBE.

1000 is a perfect cube since 1000 = 10^3
Now check out the prime factorization of 1000
1000 = (2)(2)(2)(5)(5)(5)
= [(2)(5)][(2)(5)][(2)(5)]
Notice that we can take the prime factorization of 1000 and divide the prime factors into THREE identical groups
-----NOW ONTO THE QUESTION------------------------

1,152 = (2)(2)(2)(2)(2)(2)(2)(3)(3)
So, 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(x)
Let's test the answer choices....

A. x = 4
We get: 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(4)
= (2)(2)(2)(2)(2)(2)(2)(3)(3)(2)(2)
In order for the above to be a perfect CUBE, we must be able to divide the prime factors into THREE identical groups.
Since we cannot do that here, we can ELIMINATE A

B. x = 6
We get: 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(6)
= (2)(2)(2)(2)(2)(2)(2)(3)(3)(2)(3)
Can we divide the above prime factors into THREE identical groups?
NO! ELIMINATE B

C. x = 8
We get: 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(8)
= (2)(2)(2)(2)(2)(2)(2)(3)(3)(2)(2)(2)
Can we divide the above prime factors into THREE identical groups?
NO! ELIMINATE C

D. x = 12
We get: 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(12)
= (2)(2)(2)(2)(2)(2)(2)(3)(3)(2)(2)(3)
Can we divide the above prime factors into THREE identical groups?
YES!
1152x = [(2)(2)(2)(3)][(2)(2)(2)(3)][(2)(2)(2)(3)]
So, if x = 12, then 1152x IS a perfect cube.

Answer: D

RELATED VIDEO

We need to find the value of x which will make 1152x a perfect cube

Let's start factorizing 1152x. We get
1152x = \(2^7 * 3^2 * x\)
So, for 1152x to be a perfect cube
x should be equal to \(2^2 * 3 \)
Making 1152x = \(2^7 * 3^2 * 2^2 * 3 \) = \(2^9 * 3^3\)

=> x = 4*3 = 12

So, Answer will be D
Hope it helps!

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