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Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]
23 Jun 2012, 10:15

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Expert's post

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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]
03 Jul 2012, 07:47

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

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]
03 Jul 2012, 08:01

Expert's post

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 orm=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]
15 Aug 2012, 08:39

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]
27 Jan 2013, 20:42

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. _________________

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]
27 Jan 2013, 23:56

Expert's post

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.

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]
30 Jan 2013, 23:13

2

This post received KUDOS

Expert's post

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]
21 Nov 2013, 20:12

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? _________________

Please consider giving 'kudos' if you like my post and want to thank

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]
27 Dec 2013, 08:52

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?

Re: If x is an integer greater than 1, is x equal to the 12th [#permalink]
12 May 2014, 23:56

Expert's post

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?