Is x an odd number? (1) 1 < x < 8 (2) 7x^2 has four positive factors

Answer is E ,

A - Itself not sufficient . We don't know that X is integer or not .
B -

since we X^2 has to be a prime for 4 total factors and X^0.5 will not be an integer . - No

Or
if X = 7 , then 7^3 will have 4 factors total . Ans - Yes .

Not sufficient ..
Taking both A and B , also not sufficient .
So answer is E.

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Is x an odd number?

(1) 1<x<8
(2) 7*X^2 has four positive factors

evaluating option A:
X=2 no
X=3 yes
insufficient

evaluation option B
to have 4 factors the power must be 3
to make the power 3; X=7

therefore 7^3 = 4 factors; X=7

option B is sufficient!

In st 2 what if x=√3

Yes, you’re right.

in order to make the no of factors to equal 4 , we can do either of the two things.

1> make the no a cube....... let x=7 so 7x^2 = 7^3 ...... therefore 7x^2 will have 2*2 factors = 4
2> make it an irrational square---- ex x=sqrt(2) ---> x^2 = 2 .... therefore 7x^2 will have 2*2 factors = 4


E

Can you please guide how to think of the Exception like you have thought of 3^(1/2) a the value of x here.... how did you choose amonst so many values....Please guide

Hi, avirup2018
Firstly, nothing tells us whether x is integer or not in the question stem. Well, we need to check non-integer values of x too.
Secondly, statement2 says there are 4 factors of \(7x^{2}\).
—> there are two cases of finding factors (formula)
Case1: (1+1)(1+1) = 4
\(7x^{2}\) => \(7^{1}\)— Ok. \(x^{2}\) should be equal to any prime number except 7.
—> \(x^{2}= 5\) —> \(x= 5^{1/2}\)
—> \(7x^{2} = 7^{1}*5^{1}\) —> (1+1)(1+1) = 4 ( x is not an integer)

Case2: 3+1= 4
\(7x^{2}\) = should be equal to \(7^{3}\)
—> x= 7 (x is an integer)

Hi, would you the consider root 3 to be even?

√3 is irrational number and irrational numbers are not expressed in odd or even.

Posted from my mobile device

Thanks so much for this!

Is x an odd number?

Stat1: 1<x<8, So x can be 2, 3, 4,5, 6,7. Not sufficient, as x can be even or odd.

Stat2: 7*x^2 has four positive factors, then x= 7 or sqrt (2), sqrt (3)....Not sufficient, as x can be odd or irrational no.

Combining both, x= 7 or sqrt (2), sqrt (3), again Not sufficient!

Ans. E.

sqrt(3) cannot be expressed as even or odd. And the question asks to find whether x is odd. So the case of non-integer values must be eliminated right?

Is X Odd?

the key is noticing the big hint the question leaves us: there are NO Constraints on the Value of X.

X can be any real value on the number line: Fractional/Decimal, Irrational Number, Positive/Negative, etc.


s1: 1 < X < 8

X can = Even, Odd, or not even an Integer at all.

S1 NOT Suff.


s2: 7 (x)^2 -----> have 4 positive Factors


For a Given Integer Value to have 4 positive factors, the number must take either 1 of 2 following forms:


N = (p) (q) -------> where p and q are unique Prime numbers

or

N = (p)^3--------> where p is a Prime Number


CASE 1: X can = 7, in which case we get a YES to the question --> X is Odd

7 * (7)^2 = (7)^3 ------> Total Number of Positive Factors = (3 + 1) = 4


CASE 2: as stated above, X does not have to be an Integer. X can be an Irrational Number, such as the Square Root of a Prime Number

if X = sqrt(2)

7 * [ sqrt(2) ]^2 = 7 * 2 ------> Total Number of Positive Factors = (1 + 1) * (1 + 1) = 2 * 2 = 4

Since X is equal to the Square Root of 2 and NOT even an Integer, X can not be ODD

S2 NOT Sufficient


(1)+(2) Together

Case 1 and Case 2 satisfy statement 1 as well.

the Square Root of 2 and 7 each lie between 1 and 8 on the Number Line

-E- Neither together nor alone

(1) \(1 < x < 8\)
x could be anything: even, odd, irrational or a fraction
Insufficient

(2) \(7x^2\) has four positive factors
Number of factors =\((a+1)(b+1)….\), where the number in its prime factorisation =\(x^a*y^b….\)
four factors
a) 4=1*4=(0+1)(3+1)
So the number could be a perfect cube, that is \(7x^2=y^3\), where y is an integer.
Clearly x=7 is one case, but there will be infinite more solutions as x need not be integer.
\(x=\frac{\sqrt{8}}{\sqrt{7}}\)
\(7x^2=7*\frac{8}{7}=8=2^3\)
b) 4=2*2=(1+1)(1+1)
So x could be square root of any prime number other than 7
\(x=\sqrt{5}……..7x^2=7*5=35\)…factors :- 1, 5, 7, 35
Insufficient


Combined
x could be 7 or x could be square root of 2 or of 3 or of 5.
Insufficient


E

(1) \(1 < x < 8\)
x=3 or x=4 odd or even possibility exists
Clearly insufficient

(2) \(7x^2\) has four positive factors
x=7^2 yes x=odd
when x= 7^1/2 x is not odd

Even when 1 and 2 is combined we cannot get a common ground

THerefore IMO E

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