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What is the value of x?

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What is the value of x?

(1) x^3 is a 2-digit positive odd integer.
(2) x^4 is a 2-digit positive odd integer.
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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) \(x\) cannot be an irrational number (so that both x^3 and x^4 to be integers), so \(x\) must be 3. Sufficient.

Answer: C.
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Re: Value of X [#permalink]

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New post 01 Jan 2011, 03:48
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.

Answer: C.


Thanks for the answer and detailed explanation.
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New post 17 Feb 2011, 22:29
Carelessly, I overlooked the possibility that x could be negative. Thanks Bunuel!
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New post 18 Feb 2011, 09:25
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.

Answer: C.
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Re: Value of X [#permalink]

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New post 18 Feb 2011, 12:13
I always forget about radical roots. Thanks for the explanation Bunnel.
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New post 05 Jan 2013, 21:47
carcass wrote:
What is the value of\(x\) ?

(1) \(X^3\) is a 2-digit positive odd integer.

(2)\(X^4\) is a 2-digit positive odd integer.


Hi carcass,

Stat 1 :

Only 2 digit positive integers for S1 are :
\(x\)-------- 3 ------ 4
\(x^3\) ----27-----64

Here odd integer is x=3 and x^3 = 27
SUFFICIENT

Stat 2 :

Only 2 digit positive integers for S2 are :
\(x\)----------+/-2-------------+/-3
\(x^3\)----------16----------81

Here odd integer is x=+/-3 and x^3 = 81
INSUFFICIENT (two values for x)

IMO A.

But how come C?
did i missed out anything?
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New post 05 Jan 2013, 21:55
Shanmugam, the problem doesnt explicitly state that x is an integer. It can be fraction.

e.g. Choice (A), x can be fraction -> \(x^3 = 35\) i.e. x = \(\sqrt[3]{35}\)

Similarly Choice (B) alone is not sufficient.

Hence (C) is the answer.
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Glad that helped.

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.
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New post 06 Jan 2013, 04:47
Sorry Bunuel I do not "visualize" why in C \(x^3\) and \(x^4\)cannot be rational numbers aka integers

because an irrational can't be at the same time an 2 digits odd number ?' and of course only 3 meets both conditions ?'

Can you explain me please ?'

Thanks
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New post 07 Jan 2013, 03:12
carcass wrote:
Sorry Bunuel I do not "visualize" why in C \(x^3\) and \(x^4\)cannot be rational numbers aka integers

because an irrational can't be at the same time an 2 digits odd number ?' and of course only 3 meets both conditions ?'

Can you explain me please ?'

Thanks


Not sure I understand what you mean.

Anyway, rational numbers and integers are not the same. Also, irrational numbers are not integers, thus they can be neither odd nor even.

For more check here: math-number-theory-88376.html
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New post 07 Jan 2013, 04:29
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
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carcass wrote:
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.
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Re: What is the value of x? [#permalink]

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New post 16 Jun 2013, 02:59
Bunuel wrote:
carcass wrote:
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.
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New post 16 Jun 2013, 03:15
cumulonimbus wrote:
Bunuel wrote:
carcass wrote:
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.
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New post 23 May 2014, 06:25
Bunuel wrote:
shan123 wrote:

(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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MensaNumber wrote:
Bunuel wrote:
shan123 wrote:

(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.

Does this make sense?
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Bunuel wrote:
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.

Does this make sense?


Wow! Makes complete sense. This explanation is superb. Thanks!
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Re: What is the value of x? [#permalink]

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New post 26 Jun 2016, 04:14
Hello from the GMAT Club BumpBot!

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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Re: What is the value of x?   [#permalink] 26 Jun 2016, 04:14
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