If x is an integer greater than 1, is x equal to the 12th

By Statement I: x= m^3 not sufficient since m= 16 satisfies but m= 8 does not satisfy
By Statement II: x= n^4 not sufficient since n= 8 satisfies but n = 16 does not satisfy
By I & II: X= m^3 and x = n^4 => m^3= n^4 which is only true for 1 or 0, in both cases original condition is satisfied

Notice that we are told that \(x\) is an integer greater than 1, so \(m=n=0\) or \(m=n=1\) are not possible since in this case \(x\) becomes 0 or 1.

Though if we proceed the way you propose, then from \(x=m^3\) and \(x=n^4\) we can conclude that those two conditions also hold true when \(m=a^{4}\) and \(n=a^3\) (for some positive integer \(a\)), so when \(x=m^3=n^4=a^{12}\).

Hope it helps.

So the important take away here is: if X = nth power of an integer and x= mth power of an integer simultaneously, x= (LCM of m and n)th power of an integer?

Statement 1: x= a^3. For example, x = 2^3 = 8 --> cannot equal to 12th power of an integer--> INSUFFICIENT
Statement 2: x= a^4. For example, x = 2^4 = 16--> cannot equal to 12th power of an integer--> INSUFFICIENT
Combine 2 statements:
x= a^3 --> x^4= a^12
x= b^4 --> x^3=b^12
-> x^4/x^3 = x = a^12/b^12 = (a/b)^12
x is an integer, so (a/b)^12 is an integer, so (a/b) has to be an integer also, called c
so x= c^12 --> SUFFICIENT

C is the answer.
Hope it helps.

If X is an integer greater than 1, is X equal to 12th power of an integer?

1. X is equal to 3rd power of an integer.
2. X is equal to 4th power for an integer.


Here is how i solved it. From statements 1 and 2, we know that X=a^3 as well as b^4. Therefore, a^3=b^4.

This is only possible when either 1) a=b=1 OR 2) a=b=0.

The questions says that X>1, so none of the above cases are true.

So, for a^3 to be equal to b^4, a needs to have a 4th power of b in it AND b needs to have a 3rd power of a in it. In either case, X will have a 12th power of an integer in it. Hence, C.

Ratio of two Integers is never an Irrational number, right?

Yes. In mathematics, an irrational number is any real number that cannot be expressed as a ratio of integers. Irrational numbers cannot be represented as terminating or repeating decimals.

Hi, I need some help.

I follow the above posts on (i) and (ii) as not sufficient. However, I am struggling with a case where, if true, leads to the answer E, not C:

If \(x=m^3=n^4\), where \(m=2^8\) and \(n=2^6\), then

\(x=(2^8)^3=(2^6)^4=2^{24}\) which does NOT equal \(2^{12}\); NOT SUFFICIENT

But where where \(m=2^4\) and \(n=2^3\), then

\(x=(2^4)^3=(2^3)^4=2^{12}\), SUFFICIENT

THUS answer E.

How is my thinking flawed here?

If \(x=2^{24}\) it's still is the 12th power of an integer: \(x=2^{24}=4^{12}\)

We are given that x is an integer greater than 1 and must determine whether x is equal to the 12th power of an integer.

Statement One Alone:

x is equal to the 3rd power of an integer.

Using the information in statement one, we cannot determine whether x is equal to the 12th power of an integer. For example, if x = 8 = 2^3, then it’s not equal to the 12th power of an integer. However, if x = (2^4)^3 = 2^12, then it is equal to the 12th power of an integer. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

x is equal to the 4th power of an integer

Using the information in statement two, we cannot determine whether x is equal to the 12th power of an integer. For example, if x = 16 = 2^4, then it’s not equal to the 12th power of an integer. However, if x = (2^3)^4 = 2^12, then it is equal to the 12th power of an integer. Statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Using the information from statements one and two, we know that x is equal to the 3rd power of an integer and that x is also equal to the 4th power of some other integer. Let’s represent x as a^3 where a is an integer > 1. Since a^3 is also a 4th power, the 4th root of a^3 is an integer. The only way this could happen is if a is also the 4th power of an integer; in other words, a by itself is a 4th power, say a = b^4 where b is an integer > 1.

Thus, x = a^3 = (b^4)^3 = b^12. Therefore, x is equal to the 12th power of an integer.

Answer: C

Test an EASY CASE.
Test POWERS OF 2.

Statement 1:
x = 2³, 2⁶, 2⁹, 2¹²...
If x = 2³, then x is NOT equal to the 12th power of an integer.
If x = 2¹², then x IS equal to the 12th power of an integer.
INSUFFICIENT.

Statement 2:
x = 2⁴, 2⁸, 2¹²...
If x = 2⁴, then x is NOT equal to the 12th power of an integer.
If x = 2¹², then x IS equal to the 12th power of an integer.
INSUFFICIENT.

Statements combined:
The smallest value common to both the red list and the blue list is 2¹², which is the 12th power of an integer.
If we extend the two lists, we get:
x = 2¹⁵, 2¹⁸, 2²¹, 2²⁴...
x = 2¹⁶, 2²⁰, 2²⁴...
The next value common to both lists is 2²⁴ = 4¹², which is the 12th power of an integer.
Implication:
To satisfy both statements, x must be the 12th power of an integer.
SUFFICIENT.



Alternate approach:

Statement 1: x = a³, where a is an integer[/b]
If a=2, then x = 2³, which is not the 12th power of an integer.
If a=2⁴, then x = (2⁴)³ = 2¹², which is the 12th power of an integer.
INSUFFICIENT.

Statement 2: x = b⁴, where b is an integer
If b=2, then x = 2⁴, which is not the 12th power of an integer.
If b=2³, then x = (2³)⁴ = 2¹², which is the 12th power of an integer.
INSUFFICIENT.

Statements 1 and 2 combined:
Since x = a³ and x = b⁴, we get:
a³ = b⁴
a³ = (b³)b
b = (a/b)³.
Since b is an integer, (a/b)³ is an integer.
Since a/b = integer/integer -- the definition of a rational number -- it is not possible that a/b is equal to an irrational value such as ³√2.
Thus, in order for (a/b)³ to be an integer, a/b must be an integer, implying that b is the CUBE OF AN INTEGER.
Thus, x = b⁴ = (integer³)⁴ = integer¹².
SUFFICIENT.

Bunuel
can you help me understand this:
"For perfect cube we need all prime factors to have a multiple of 3
For perfect fourth powers we need all the same prime factors to have a multiple of 4"
Thanks

Consider the following examples: 2^3, 6^6 = (6^2)^3 = 2^6*3^6, 10^12 = (10^4)^3 = 2^12*5^12, ... All those numbers are perfect squares and the powers of the primes of all those numbers are multiples of 3.

I have an example here:
(1) x=m^5 -> x^7=m^35
(2) x=n^7 -> x^5=n^35

(2)/(1) -> x^2=(n/m)^35
x is not yet proven to be equal to the 35th power of an integer

Could you explain more about this take away?

If x is an integer greater than 1, is x equal to the 12th power of an integer ?

(1) x is equal to the 3rd power of an integer

(2) x is equal to the 4th power of an integer