Is x an odd number? (1) 1

Question

Is x an odd number?

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

Are You Up For the Challenge: 700 Level Questions

lacktutor · Answer

Is x an odd number??

(Statement1) 1< x < 8
x could be odd or even number in the range from 1 to 8.

Insufficient

(Statement2):
\(7x^{2}\) has four positive number factors

If the value of x is equal to 7, statement2 will have 4 positive factors
—> x=7 —> \(7*7^{2}= 7^{3}\)
(3+1)= 4 ( 1, 7,49, 343)
—> x is odd (YES)

If \(x= 3^{1/2}\), then 7*3 has also 4 positive factors (—> 1,3,7,21)
x is not odd ( NO)

Insufficient

Taken together 1&2,
—> x could be \(3^{1/2}\) or 7 in the range from 1 to 8.

Insufficient

The answer is E

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Archit3110 · Answer

#1
 1 < x < 8 
x can be odd or even
insufficient
#2
  7x^2  has four positive factors
x can be prime integer ; 7*2^2 ;yes
or x can be a fraction ; 7*(1/7)^2 ; 4 factors
from 1 &2
nothing in common
IMO E

DasAshish1 · Answer

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.

Bunuel · Answer

Just a note: this is a 700+ question from our new collection  Are You Up For the Challenge: 700 Level Questions . Aren't you missing something there?

sampriya · Answer

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!

Peddi · Answer

In st 2 what if x=√3

lacktutor · Answer

Yes, you’re right.

ShankSouljaBoi · Answer

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

avirup2018 · Answer

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

lacktutor · Answer

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)

veron90 · Answer

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

yashikaaggarwal · Answer

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

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veron90 · Answer

Thanks so much for this!

TarunKumar1234 · Answer

Is x an odd number? Stat1: 1

Rohithkumar17998 · Answer

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?

Fdambro294 · Answer

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

chetan2u · Answer

(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

Crytiocanalyst · Answer

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

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