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

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

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23 Jun 2012, 10:44
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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

(2) x is equal to the 4th Power of an integer.
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Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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23 Jun 2012, 11:15
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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.

Hope it's clear.

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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,$$x = z^4 or x = (k^3)^4 = k^{12}$$.

C.
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Re: Data Sufficiency problem - exponents [#permalink]

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07 Aug 2014, 21:26
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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

Hope it helps.
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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.

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

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03 Jul 2012, 09:01
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shekharverma wrote:
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.
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Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]

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08 Nov 2017, 09:44
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DanielAustin wrote:
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

How is my thinking flawed here?

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

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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 !!

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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.

Hope it's clear.

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

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28 Jan 2013, 00:56
fozzzy 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.

Hope it's clear.

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.

What you mean by "all cases"? Anyway, if m and n are integers and $$x=(\frac{m}{n})^{12}=integer$$, then m/n=integer.
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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?

Cheers!
J

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

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13 May 2014, 00:56
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jlgdr wrote:
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?

Cheers!
J

Yes, that's correct.
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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.

Hope it's clear.

What if m/n = √2 type that is quite possible.
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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, 09:21
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.

Hope it's clear.

What if m/n = √2 type that is quite possible.

Have you read the highlighted part?
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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, 09:26
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.

Hope it's clear.

What if m/n = √2 type that is quite possible.

Have you read the highlighted part?

Ratio of two Integers is never an Irrational number, right?
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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, 09:29
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honchos wrote:

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

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23 Sep 2016, 09:53
C is correct. Here's why:

(1) x = n^3 --> Try plugging in arbitrary value for n.

x = 2^3 --> x = 8 --> now ask yourself what integer raised to the 12th power could give you 8. Answer = none

INSUFFICIENT

(2) x = n^4 --> repeat the same process as in (1)

x = 2^4 --> x= 16 --> There is no integer that could give us this value using main equation

INSUFFICIENT

(1) + (2) Together - SUFFICIENT

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

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08 Dec 2016, 15:23
Bunuel wrote:
[b] Now, $$\frac{m}{n}$$ can be neither an irrational number (since it's the ratio of two integers)

I am unable to understand this part can you please explain or perhaps provide me a link where I can understand my knowledge gap. Thanks!
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Re: If x is an integer greater than 1, is x equal to the 12th   [#permalink] 08 Dec 2016, 15:23

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