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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.

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.

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.
_________________

Please contact me for super inexpensive quality private tutoring

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

(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.

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