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What is the value of x? (1) X3 is a 2-digit positive odd integer. (2) X4 is a 2-digit positive odd integer.

I don't know whether the answer is correct. I got a different one.

What is the value of x?

Note that we are not told that x is an integer

(1) x^3 is a 2-digit positive odd integer --> now, if \(x\) is an integer then \(x=3\) as \(x^3=27\) is the only odd 2-digit positive cube of an integer (1^3=1 and 5^3=125) but if \(x\) is not an integer then it can be cube root of any 2-digit positive odd integer, for example if \(x=\sqrt[3]{11}\) then \(x^3=11\). Not sufficient.

(2) x^4 is a 2-digit positive odd integer --> basically the same here: if \(x\) is an integer then \(x=3\) or \(x=-3\) as \(x^4=81\) is the only odd 2-digit positive integer which is in fourth power of an integer (1^4=1 and 5^4=625) (so even if \(x\) is an integer this statement is still insufficient as it gives two values for \(x\): 3 and -3). \(x\) also can be non-integer as above: it can be fourth root from any 2-digit positive odd integer, for example if \(x=\sqrt[4]{11}\) then \(x^4=11\). Not sufficient.

(1)+(2) \(x\) cannot be an irrational number (so that both x^3 and x^4 to be integers), so \(x\) must be 3. Sufficient.

Always watch out for ZIP trap (assuming Zero, Integer, Positive) -> (Make sure to check for 0, factions and negatives) Especially for inequalities, algebraic, number/fraction problems.
_________________

(1)+(2) \(x\) cannot be an irrational number (so that both x^3 and x^4 to be integers), so \(x\) must be 3. Sufficient.

Answer: C.

Hi Bunuel: Just like others, I also have a hard time visualizing that there does not exist an irrational number whose 3rd and 4th power both result in an odd digit integer. I mean integer is a smaller set compared to irrational numbers and we still have 3 (an integer) whose 3rd and 4th power both result in an odd 2-digit integer. On the other hand in terms of irrational numbers we have tremendous possibilities even between two integers we have infinite irrational numbers and we cannot have such a number. It some how feels odd to me. I have no doubt what you are saying is right but I have hard time imagining it. Maybe my understanding of irrational numbers and their powers is still primordial.

Say x IS an irrational number and x*x*x=x^3=integer. In this case x*x*x*x=x^3*x=integer*irrational=irrational.

If x is an irrational number and x*x*x*x=x^4=integer, then x^3=x^4/x=integer/irrational=irrational.

So, as you can see if x is an irrational number, then both x^3 and x^4 cannot be rational.

What is the value of x? (1) X3 is a 2-digit positive odd integer. (2) X4 is a 2-digit positive odd integer.

I don't know whether the answer is correct. I got a different one.

What is the value of x?

Note that we are not told that x is an integer

(1) x^3 is a 2-digit positive odd integer --> now, if \(x\) is an integer then \(x=3\) as \(x^3=27\) is the only odd 2-digit positive cube of an integer (1^3=1 and 5^3=125) but if \(x\) is not an integer then it can be cube root of any 2-digit positive odd integer, for example if \(x=\sqrt[3]{11}\) then \(x^3=11\). Not sufficient.

(2) x^4 is a 2-digit positive odd integer --> basically the same here: if \(x\) is an integer then \(x=3\) or \(x=-3\) as \(x^4=81\) is the only odd 2-digit positive integer which is in fourth power of an integer (1^4=1 and 5^4=625) (so even if \(x\) is an integer this statement is still insufficient as it gives two values for \(x\): 3 and -3). \(x\) also can be non-integer as above: it can be fourth root from any 2-digit positive odd integer, for example if \(x=\sqrt[4]{11}\) then \(x^4=11\). Not sufficient.

(1)+(2) \(x\) can not be an irrational number (so that both x^3 and x^4 to be integers), so \(x\) must be 3. Sufficient.

Tricky one, I considered the integer constraint that didn't exist. Must take care with this.

Bunuel wrote:

shan123 wrote:

What is the value of x? (1) X3 is a 2-digit positive odd integer. (2) X4 is a 2-digit positive odd integer.

I don't know whether the answer is correct. I got a different one.

What is the value of x?

Note that we are not told that x is an integer

(1) x^3 is a 2-digit positive odd integer --> now, if \(x\) is an integer then \(x=3\) as \(x^3=27\) is the only odd 2-digit positive cube of an integer (1^3=1 and 5^3=125) but if \(x\) is not an integer then it can be cube root of any 2-digit positive odd integer, for example if \(x=\sqrt[3]{11}\) then \(x^3=11\). Not sufficient.

(2) x^4 is a 2-digit positive odd integer --> basically the same here: if \(x\) is an integer then \(x=3\) or \(x=-3\) as \(x^4=81\) is the only odd 2-digit positive integer which is in fourth power of an integer (1^4=1 and 5^4=625) (so even if \(x\) is an integer this statement is still insufficient as it gives two values for \(x\): 3 and -3). \(x\) also can be non-integer as above: it can be fourth root from any 2-digit positive odd integer, for example if \(x=\sqrt[4]{11}\) then \(x^4=11\). Not sufficient.

(1)+(2) \(x\) can not be an irrational number (so that both x^3 and x^4 to be integers), so \(x\) must be 3. Sufficient.

basically 1) is insuff because we have to consider integers and non integers (so irrational numbers). Same for 2)

Bothe statements are suff because we have only 3 that mettes the criteria so we have to consider only the 3 (the integer). So sufficient

But why we C is sufficient ?' why we can not consider the irrational numbers ??

Thanks. Now I hope is more clear what I mean. I'm sorry if I have explained myself badly

If x is an irrational number then x^3 and x^4 cannot both be integers as given in the statements, so x can only be 3.

Hi Bunnel,

Still did not get this part: If x is an irrational number then x^3 and x^4 cannot both be integers as given in the statements, so x can only be 3

Irrational no cannot be expressed as p/q, where p and q are integers.

I made my understand it like this: Their is only 1 number possible whose cube is 27 and only one number has fourth power equal to 81. Which is integer 3.

basically 1) is insuff because we have to consider integers and non integers (so irrational numbers). Same for 2)

Bothe statements are suff because we have only 3 that mettes the criteria so we have to consider only the 3 (the integer). So sufficient

But why we C is sufficient ?' why we can not consider the irrational numbers ??

Thanks. Now I hope is more clear what I mean. I'm sorry if I have explained myself badly

If x is an irrational number then x^3 and x^4 cannot both be integers as given in the statements, so x can only be 3.

Hi Bunnel,

Still did not get this part: If x is an irrational number then x^3 and x^4 cannot both be integers as given in the statements, so x can only be 3

Irrational no cannot be expressed as p/q, where p and q are integers.

I made my understand it like this: Their is only 1 number possible whose cube is 27 and only one number has fourth power equal to 81. Which is integer 3.

Please explain why have you mentioned it here.

I don't understand your question. Please elaborate.
_________________

(1)+(2) \(x\) cannot be an irrational number (so that both x^3 and x^4 to be integers), so \(x\) must be 3. Sufficient.

Answer: C.

Hi Bunuel: Just like others, I also have a hard time visualizing that there does not exist an irrational number whose 3rd and 4th power both result in an odd digit integer. I mean integer is a smaller set compared to irrational numbers and we still have 3 (an integer) whose 3rd and 4th power both result in an odd 2-digit integer. On the other hand in terms of irrational numbers we have tremendous possibilities even between two integers we have infinite irrational numbers and we cannot have such a number. It some how feels odd to me. I have no doubt what you are saying is right but I have hard time imagining it. Maybe my understanding of irrational numbers and their powers is still primordial.
_________________

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My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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