What is the value of positive integer x?

(1) clearly insuff

(2) x can be 45 or 90

combining we get x =15 or 3*5*2 =30 ( which is <45)

insuff

Ans E

Yes of course
you are correct
post edited!!!!!

thanks

How are we getting x=30 by combining both statements?

Given x>0 & Integer

Find Value of 'x'

Statement 1 x<45
=> So x=1,2,3,4....44
=> Many Values of 'x'. Therefore INSUFFICIENT

Statement 2 45 is a factor of \(4x^2\)
=> So \(4x^2\) is a multiple of 45
=> OR \(4x^2=45*k\) ----- 'k' some constant
=> OR \(x^2=\frac{(45*k)}{4}\)

=> For 'x' to be Positive Integer \(\frac{(45*k)}{4}\) should be a PERFECT SQUARE & Integer.

=> Now Prime Factorising 45 we have \(45=3^2*5\)

=> Therefore 'k' should be \((4*5)\) or \((4*5*2^2)\) or \((4*5*2^2*3^3)\) or \((4*5^3*2^2)\) and so on

=> Thus \(x^2\)=\(\frac{(45*4*5)}{4}\) or \(\frac{(45*4*5*2^2)}{4}\) or \(\frac{(45*4*5*2^2*3^2)}{4}\) or \(\frac{(45*4*5^3*2^2)}{4}\) and so on

=> OR x=15,30,90,150..... so on
=> Many Values of 'x'. Therefore INSUFFICIENT

BOTH Stat 1 & 2
=> x=15,30,90,150..... so on ---- from Stat 2
=> x=15,30 --------------- constraint of Stat 1
=> AGAIN 'x' has 2 values. Therefore INSUFFICIENT

Therefore 'E'

Thanks
Dinesh

I think it's E
x could be 15 or 30, both <45

Is it?

What if x=30? It suffices both conditions A and B. Also it is the only value other than 15 which fits both criteria.

Hi TheMastermind ,

For 45 to be factor of 4x^2 , I need atleast two 3s and one 5 in 4x^2.

Now, this could be possible x = 15. Its square will give me what I want.

Similarly, x could be 30 , again its square will give me what I want. Since we have two possible values of x, answer is E.

Does that make sense?
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Tried by plugging in numbers only.

1) St x < 45--> clearly NS
2) 4x ^2/ 45 = no remainder. just by hit and trial (factors of 9,5,3) --> 4 *22500/45 (no remainder) and 4*900/45 (no remainder).
x = 30 or 15. NS

1) 2) Both giving multiple values of x. So insufficient

ok people.

So, A is insufficient. B is insufficient.

However, combining the two statements, you get 4x^2 = 45I
The question already states that x is a positive integer. Therefore, x^2 also HAS to be a positive integer.
THerefore, x^2 = 45I/4 = an integer

So, x has to be an even integer.

So, out of 15 and 30, only 30 suffices.

Hence C.

YAHOOOO

Good one. X=30 or 15 because 45 = 3*3*5. Even if X has one 5 & 3, it can be divided by 45 (5*3*3) if it is squared. X^2 = 5*5*3*3. Thus, The least possible value of X = 5*3=15. And it is less than 45 according to I. Therefore answer = 15. Wait, another possible value is 30 which also fits the equation. Hence, E.

I'm super lost on this one - I'm sorry for missing what appears to be a simple step.

How do we go from finding the factors of 45 to knowing that x must be 15? Is there a rule that says the product of two prime factors equals another factor of a given number?

Solution :-

Check first statement :-
X<45

It can take any positive value less than 45 right .------- Insufficient


Check Second statement :-

45 is a factor of 4x^2
which means 3^2 * 5 is a factor of 4 * x *x

So,
x should contain atleast 3 * 5 right to be able to get divided by 45.

So, x is of the form : 3*5k=15k
It can take any value :- 15,30,45,60 ......-------Insufficient


Combining 1st and 2nd statements:-

X can take values 15, 30 -----Insufficient to arrive at a unique answer

Hence E

(1) It's apparent this is not sufficient
(2) 3*3*5 divides into 2*2*x*x
x = 3*5, or any multiple of that. Not sufficient.

(1) and (2) together
x = 3*5 which is less than 45
x = 3*5*2 which is also less than 45

Answer E

On a tangent almost failed number theory Die professor jk jk. 49% correct.

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