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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 --> \(x=m^3\) for some positive integer \(m\). If \(m\) itself is 4th power of some integer (for example if \(m=2^4\)), then the answer will be YES (since in this case \(x=(2^4)^3=2^{12}\)), but if it's not (for example if \(m=2\)), then the answer will be NO. Not sufficient.

(i) Notice that from this statement we have that \(x^4=m^{12}\).

(2) x is equal to the 4th Power of an integer --> \(x=n^4\) for some positive integer \(n\). If \(n\) itself is 3rd power of some integer (for example if \(n=2^3\)), then the answer will be YES (since in this case \(x=(2^3)^4=2^{12}\)), but if it's not (for example if \(n=2\)), then the answer will be NO. Not sufficient.

(ii) Notice that from this statement we have that \(x^3=n^{12}\).

(1)+(2) Divide (i) by (ii): \(x=(\frac{m}{n})^{12}=integer\). Now, \(\frac{m}{n}\) can be neither an irrational number (since it's the ratio of two integers) nor some reduced fraction (since no reduced fraction, like 1/2 or 3/2, when raised to some positive integer power can give an integer), therefore \(\frac{m}{n}\) must be an integer, hence \(x=(\frac{m}{n})^{12}=integer^{12}\). Sufficient.

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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31 Jan 2013, 00:13

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Given that x>1 and an integer.

From F.S 1, we have \(x=t^3\),t is a positive integer. Now for t=16, we will have a sufficient condition but not for say t=8. Thus not sufficient.

From F.S 2, we have \(x=z^4\). z is a positive integer. Now just as above, for z=8, we will have a sufficient condition but not for say z=16. Thus not sufficient.

Combining both of them, we have;

\(x=t^3; x=z^4\). Hence, \(t^3 = z^4\). Now this can be written as \(t = z^{\frac{4}{3}}\) \(\to t = z^{\frac{3+1}{3}} \to t = z*z^{\frac{1}{3}}\) Now, as both t and z are integers, we must have \(z^{\frac{1}{3}}\) as an integer.Thus, t = kz , where \(k = z^{\frac{1}{3}}\) Cubing on both sides, we have \(z = k^3.\)

Replace this value of z,\(x = z^4 or x = (k^3)^4 = k^{12}\).

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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21 Nov 2013, 21:12

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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? _________________

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Re: Data Sufficiency problem - exponents [#permalink]

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07 Aug 2014, 22:11

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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.

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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03 Jul 2012, 08:47

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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

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 \(x=m^3=n^4=a^{12}\).

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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15 Aug 2012, 09:39

i agree that indivisualy we cannot answer this question ...but how abt this approach .if we combine both statement then we can be sure that x =(int ) ^12 becoz under this condition only can both the conditions be met .so we can now be sure that this int can be expressed as some int raised to the power of 12 .expert plz evaluate this !!

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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27 Jan 2013, 21:42

Bunuel wrote:

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 --> \(x=m^3\) for some positive integer \(m\). If \(m\) itself is 4th power of some integer (for example if \(m=2^4\)), then the answer will be YES (since in this case \(x=(2^4)^3=2^{12}\)), but if it's not (for example if \(m=2\)), then the answer will be NO. Not sufficient.

(i) Notice that from this statement we have that \(x^4=m^{12}\).

(2) x is equal to the 4th Power of an integer --> \(x=n^4\) for some positive integer \(n\). If \(n\) itself is 3rd power of some integer (for example if \(n=2^3\)), then the answer will be YES (since in this case \(x=(2^3)^4=2^{12}\)), but if it's not (for example if \(n=2\)), then the answer will be NO. Not sufficient.

(ii) Notice that from this statement we have that \(x^3=n^{12}\).

(1)+(2) Divide (i) by (ii): \(x=(\frac{m}{n})^{12}=integer\). Now, \(\frac{m}{n}\) can be neither an irrational number (since it's the ratio of two integers) nor some reduced fraction (since no reduced fraction, like 1/2 or 3/2, when raised to some positive integer power can give an integer), therefore \(\frac{m}{n}\) must be an integer, hence \(x=(\frac{m}{n})^{12}=integer^{12}\). Sufficient.

Is this true in all cases that it must be an integer ( is there a theorem or something along those lines) , could you please provide an example. _________________

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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 --> \(x=m^3\) for some positive integer \(m\). If \(m\) itself is 4th power of some integer (for example if \(m=2^4\)), then the answer will be YES (since in this case \(x=(2^4)^3=2^{12}\)), but if it's not (for example if \(m=2\)), then the answer will be NO. Not sufficient.

(i) Notice that from this statement we have that \(x^4=m^{12}\).

(2) x is equal to the 4th Power of an integer --> \(x=n^4\) for some positive integer \(n\). If \(n\) itself is 3rd power of some integer (for example if \(n=2^3\)), then the answer will be YES (since in this case \(x=(2^3)^4=2^{12}\)), but if it's not (for example if \(n=2\)), then the answer will be NO. Not sufficient.

(ii) Notice that from this statement we have that \(x^3=n^{12}\).

(1)+(2) Divide (i) by (ii): \(x=(\frac{m}{n})^{12}=integer\). Now, \(\frac{m}{n}\) can be neither an irrational number (since it's the ratio of two integers) nor some reduced fraction (since no reduced fraction, like 1/2 or 3/2, when raised to some positive integer power can give an integer), therefore \(\frac{m}{n}\) must be an integer, hence \(x=(\frac{m}{n})^{12}=integer^{12}\). Sufficient.

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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27 Dec 2013, 09:52

MensaNumber wrote:

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?

Well that's what I'm asking myself but think about it for a sec

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

Hence, for both we need all the prime factors to have multiples of 12 at least

So IMHO I think this should be correct under this scenario

Bunuel, would you give your blessing on this statement?

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?

Well that's what I'm asking myself but think about it for a sec

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

Hence, for both we need all the prime factors to have multiples of 12 at least

So IMHO I think this should be correct under this scenario

Bunuel, would you give your blessing on this statement?

Re: Data Sufficiency problem - exponents [#permalink]

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07 Aug 2014, 21:26

taransambi wrote:

Source: Question Pack 1

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.

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. _________________

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Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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04 Jun 2015, 09:16

Bunuel wrote:

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 --> \(x=m^3\) for some positive integer \(m\). If \(m\) itself is 4th power of some integer (for example if \(m=2^4\)), then the answer will be YES (since in this case \(x=(2^4)^3=2^{12}\)), but if it's not (for example if \(m=2\)), then the answer will be NO. Not sufficient.

(i) Notice that from this statement we have that \(x^4=m^{12}\).

(2) x is equal to the 4th Power of an integer --> \(x=n^4\) for some positive integer \(n\). If \(n\) itself is 3rd power of some integer (for example if \(n=2^3\)), then the answer will be YES (since in this case \(x=(2^3)^4=2^{12}\)), but if it's not (for example if \(n=2\)), then the answer will be NO. Not sufficient.

(ii) Notice that from this statement we have that \(x^3=n^{12}\).

(1)+(2) Divide (i) by (ii): \(x=(\frac{m}{n})^{12}=integer\). Now, \(\frac{m}{n}\) can be neither an irrational number (since it's the ratio of two integers) nor some reduced fraction (since no reduced fraction, like 1/2 or 3/2, when raised to some positive integer power can give an integer), therefore \(\frac{m}{n}\) must be an integer, hence \(x=(\frac{m}{n})^{12}=integer^{12}\). Sufficient.

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 --> \(x=m^3\) for some positive integer \(m\). If \(m\) itself is 4th power of some integer (for example if \(m=2^4\)), then the answer will be YES (since in this case \(x=(2^4)^3=2^{12}\)), but if it's not (for example if \(m=2\)), then the answer will be NO. Not sufficient.

(i) Notice that from this statement we have that \(x^4=m^{12}\).

(2) x is equal to the 4th Power of an integer --> \(x=n^4\) for some positive integer \(n\). If \(n\) itself is 3rd power of some integer (for example if \(n=2^3\)), then the answer will be YES (since in this case \(x=(2^3)^4=2^{12}\)), but if it's not (for example if \(n=2\)), then the answer will be NO. Not sufficient.

(ii) Notice that from this statement we have that \(x^3=n^{12}\).

(1)+(2) Divide (i) by (ii): \(x=(\frac{m}{n})^{12}=integer\). Now, \(\frac{m}{n}\)can be neither an irrational number (since it's the ratio of two integers) nor some reduced fraction (since no reduced fraction, like 1/2 or 3/2, when raised to some positive integer power can give an integer), therefore \(\frac{m}{n}\) must be an integer, hence \(x=(\frac{m}{n})^{12}=integer^{12}\). Sufficient.

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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04 Jun 2015, 09:26

Bunuel wrote:

honchos wrote:

Bunuel wrote:

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 --> \(x=m^3\) for some positive integer \(m\). If \(m\) itself is 4th power of some integer (for example if \(m=2^4\)), then the answer will be YES (since in this case \(x=(2^4)^3=2^{12}\)), but if it's not (for example if \(m=2\)), then the answer will be NO. Not sufficient.

(i) Notice that from this statement we have that \(x^4=m^{12}\).

(2) x is equal to the 4th Power of an integer --> \(x=n^4\) for some positive integer \(n\). If \(n\) itself is 3rd power of some integer (for example if \(n=2^3\)), then the answer will be YES (since in this case \(x=(2^3)^4=2^{12}\)), but if it's not (for example if \(n=2\)), then the answer will be NO. Not sufficient.

(ii) Notice that from this statement we have that \(x^3=n^{12}\).

(1)+(2) Divide (i) by (ii): \(x=(\frac{m}{n})^{12}=integer\). Now, \(\frac{m}{n}\)can be neither an irrational number (since it's the ratio of two integers) nor some reduced fraction (since no reduced fraction, like 1/2 or 3/2, when raised to some positive integer power can give an integer), therefore \(\frac{m}{n}\) must be an integer, hence \(x=(\frac{m}{n})^{12}=integer^{12}\). Sufficient.

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. _________________

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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17 Mar 2016, 12:18

Hi everyone,

A question. And I already had this doubt several times on different questions.

To test both statements together, in this case, you say "now divide them". How do you know that you have to divide? Why not multiply? or subtract? or add? It seems that from quesion stem there is nothing that indicates that a division could be a solution to the problem.

I remember that I have this doubt on inequalities DS very often. "sum the inequalities" or subtract, etc.

How to decide on the best course of action to correctly test both statements together?

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