What is the value of x?

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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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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Thanks for the answer and detailed explanation.

Carelessly, I overlooked the possibility that x could be negative. Thanks Bunuel!
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Tricky one, I considered the integer constraint that didn't exist. Must take care with this.

I always forget about radical roots. Thanks for the explanation Bunnel.

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?

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.

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.

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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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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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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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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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.

Wow! Makes complete sense. This explanation is superb. Thanks!

Innocuous looking deadly snake this one !!!
bowled me ....
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since x can be any number than an integer, and x can have negative value, combine 2 statement will yield the answer.

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