If x is a positive integer less than 100 such that x is

We have \(0<x<100\) and \(x=k\times 2^y\)

(1) \(x^2 > 3600 = 60^2 \implies x>60\)

If \(k=1 \implies x=2^y \implies 60 < 2^y < 100\)

Note that \(2^5=32\), \(2^6=64\), \(2^7=128\). Hence we have \(y=6\).

If \(k=2 \implies x=2^{y+1} \implies 60 < 2^{y+1} < 100 \implies y=5\).

Hence insufficient.

(2) \(\frac{x^2}{2^{y+2}}=\frac{k^2 \times 2^{2y}}{2^{y+2}}=k^2 \times 2^{y-2}\) is odd

Hence we must have \(y-2=0 \implies y=2\). Sufficient.

The answer is B
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Reserving this space to post the official solution.
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(1) since x >0
x>60
or 60<x<100
only x= 64 satisfies x is div. by 2^y
64= 2^6 thus y =6
88= 11*2^3 thus y=3

insuff..

(2) x can be 88 then y=2
or x=64 then y =10
insuff


combining as above ex. in (2)
Ans E

shud be E..
let me try,,

stat1 : says x>60 and stem says x is less than 10.,, numerous possibilities for y
consider x= 64 or x = 96 gives diff values of y

stat2: stem says x is less than 100

same examples above apply...

hence not suff

ans E
cheers

HI,
Your explanation is well understood, but I have a simple query, are we not supposed to understand from the question stem that x is a multiple of 2^y??
Thanx

Not sure where "k" comes from and why you went this route?

Because the question said that \(x\) is divisible by \(2^y\), then there must be an integer \(k\) that \(x=k * 2^y\)
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I am not sure how B is the answer.
consider two examples
when x=68,
\frac{x^2}{2^{y+2}},
\frac{68^2}{2^{y+2}},
\frac{17*17*4*4}{2^{y+2}},
in this case to make the above value odd, y will be 2.
And when x=80
\frac{x^2}{2^{y+2}}
\frac{80^2}{2^{y+2}}
\frac{5*5*2^4*2^4}{2^{y+2}}
in this case to make the above value odd, y will be 6.
there are multiple value for y.
Experts please help.

I am sorry but I get frustrated when I don't understand
Why first statement is not sufficient?
60<x<100
How would x be 32?
then y only could be 6

thank you so much

I did almost the same but not sure what is the loophole in my method.
For stat 2, (k^2* 2^2y)/(2^y*2^2) = odd
=> k^2* 2^2y / (2^y * 4)= odd
now we multiply both sides by 4. When 4 multiplied with odd number it gives even number.
so the equation becomes:
k^2* 2^2y/2^y = even
k^2 * 2^y = even
here y can take any value and the product remains even.

Can you please tell me where I am going wrong?

Statement 1: Clearly insufficient

Statement 2:
Let \(x = 2^m\)

so (x^2) / (2^(y+2)) = odd => 2 ^ 2m / ( 2 ^(y+2)) = odd

for above to be odd, 2 ^ 2m = ( 2 ^(y+2)) => 2m = y + 2 => m = (y/2) + 1
also, note that y <= m (since x is divisible by 2^y), substituting for m => y <= (y+2) + 1 => y <= 2,

So y can be 1 or 2
for m = (y/2) + 1;

if y = 1, m = 1.5, but this is not possible, as m is integer
if y = 2, m = 2, so we have found the value of y

Sufficient.

Answer (B)

if we take x = 96, we are getting y = 2 from statement 2.
if we take x = 80, we are getting y = 6 from statement 2 .
hence B is not correct...

Please correct me, if I am wrong.



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There is more than one solution to statement 2. Put y = 5 and x = 8 into it and you'll see it works.

Hi

x=8 and y=5 will Not work, because we are given that x should be divisible by 2^y. Here 2^y = 2^5 = 32, and 8 is Not divisible by 32.
So I think we cannot take this example value here.

Hi

If we take x=96, then x = 2^5 * 3, and x^2 = 2^10 * 3^2. Since x^2 should give an odd number when divided by 2^(y+2), it means 2^(y+2) = 2^10, this gives y+2 = 10 or y = 8. So we are NOT getting y=2 from here. Also, x should be divisible by 2^y as per the question stem, and 96 is NOT divisible by 2^8. So this example of x=96 and y=2 does not make sense.

Now, if we take x=80, then x = 2^4 * 5 and x^2 = 2^8 * 5^2. Since x^2 should give an odd number when divided by 2^(y+2), it means 2^(y+2) = 2^8, this gives y+2 = 8 or y = 6. So yes, this way we get y=6. BUT - x should be divisible by 2^y as per the question stem, and 80 is NOT divisible by 2^6. So this example of x=80 and y=6 also does not make sense.

As is typical with the high level E-GMAT questions, we have a larger question stem (great question 😁 )


X has to be a + integer < 100

Y has to be a + integer

And

X / (2)^y = Integer

What is the value of y = ?


S1: (X)^2 > 3,600

Since we are given that X is a positive integer, we do not have to worry about the (-)negative root when we take the square root both sides:

[X] > sqrt(3,600)

[X] > 60

since X > 0 ——-> [X] = X

X > 60

X can take the value of 64 = (2)^6

We just need to meet the constraint that:

64 / (2)^y

y could = 1, 2, 3, 4, etc. —— not a unique value for Y

S1 NOT sufficient


S2:

(X)^2 / (2)^(y + 2) = Odd Integer

Concept: since X must be a positive integer, anytime we square X it must have an Identical row of prime factors because (X)^2 = X * X —— thus the Powers of all the Prime Bases that make up (X)^2 will be EVEN Powers

Further, in order to have an Odd Integer Quotient result after diving (X)^2 by (2)^(y +2) ————>

the value of (X)^2 must include/cancel out all the Prime Factors of 2 in the DEN As well as having all of the Primes in the NUM cancel

Because of these two facts, Y can not take an Odd Value (inference 1)

For example, if Y took an Odd value:

Case 1: let Y = 1

(X)^2 / (2)^3 = Odd Integer

In order to have an Odd Integer quotient, two things must happen:

(1st) all the 2 prime factors in the DEN must cancel

And

(2nd) there must be no prime factors of 2 left in the NUM

If we let X = 2 * 2 * 3 for example, when we square X this will double the Prime Factors and result in:

(X)^2 = (2)^4 * (3)^2

When we divide by (2)^3, the DEN will cancel out and we will be left with the following as the quotient:

(2) * (3)^2 = 18 ——- which is NOT an Odd Integer

Therefore, y must take an Even Positive Integer Value:


Case 2: let Y = 2

(X)^2 / (2)^4 = ODD Int.

We can let X = (2)^2 * 3

When we square X it will double all the prime factors and we will have:

(X)^2 = (2)^4 * (3)^2

When we divide by (2)^4 ———> both the Primes of 2 in the NUM and DEN will cancel out and we will be left with a Quotient = 9 = Odd integer

Furthermore, we must satisfy the Given Constraint in which X = (2)^2 * 3 is divisible by (2)^y

In this case, since y = 2, we will meetthe constraint as:

(2)^2 * 3 is divisible by (2)^2


Thus: Y = 2 is a possible value


If we try to make Y any larger, we cannot have a Value of X that is both:

-less than 100

-satisfies statement 2 when squared

And

-divisible by (2)^y


Try case 3: let y = 4

(X)^2 / (2)^6 = Odd Integer

We need to make sure that the primes in X are just enough so that they cancel our (2)^6, otherwise we will have an Even Int. Quotient.

In other words, (X)^2 must contain no more than (2)^6 in its prime factorization.

Let X = (2)^3 * (3)

This will satisfy statement 2 once we square X ———> (X)^2 = (2)^6 (3)^2

And when we divide by (2)^6 ——-> quotient = 9 = Odd Integer. So we can satisfy statement 2.

BUT unfortunately the Constraint in the question stem will be violated if Y = 4:

X / (2)^y = Int.

(2)^3 * (3) / (2)^4 ————> not an integer

Again, we can not add any more (2) primes to X, because then we will not satisfy statement 2 ——— adding anymore (2) primes to X and they will not be canceled out by (2)^6 and we will be left with an Even Integer Quotient

As you move up and try higher and higher Even Values for Y, the same problem will keep happening. We can meet statement 2’a criteria, but we can not simultaneously satisfy the given constraint.


Therefore, under the constraint and statement 2, the only value that Y can take is:

Y = 2

B - s2 is sufficient alone

Hard question

Posted from my mobile device

I have a doubt...

Stmt2 - (x)^2/(2)^(y+2) = odd integer

(x)^2/(4)(2)^y = odd integer

(x)^2/(2)^y = 4(odd integer) = even integer

Can we deduce from this that the expression in the question will also be an integer?

Since x is a positive multiple of \(2^y\), we get:
\(x=2^yk\), where k is a positive integer.

Statement 2:
Case 1: \(y=1\), with the result that \(x=2k\)
Substituting \(x=2k\) and \(y=1\) into \(\frac{x^2}{2^{(y+2)}}\) = odd, we get:
\(\frac{2^2k^2}{2^3}\) = odd

\(\frac{k^2}{2}\) = odd
Not viable: half of a perfect square such as \(k^2\) cannot yield an odd integer.

Case 2: \(y=2\), with the result that \(x=2^2k\)
Substituting \(x=2^2k\) and \(y=2\) into \(\frac{x^2}{2^{(y+2)}}\) = odd, we get:
\(\frac{2^4k^2}{2^4}\) = odd

\(k^2\) = odd
Case 2 is viable for any odd integer k such that \(x=2^2k\) is less than 100.

Case 3: \(y=3\), with the result that \(x=2^3k\)
Substituting \(x=2^3k\) and \(y=3\) into \(\frac{x^2}{2^{(y+2)}}\) = odd, we get:
\(\frac{2^6k^2}{2^5}\) = odd

\(2k^2\) = odd
Not viable, since the left side is even.

Any greater value for y will yield the same result as Case 3:
\(\frac{x^2}{2^{(y+2)}}\) = even, violating the condition in Statement 2.

Implication:
Statement 2 is satisfied only by Case 2.
Thus, y=2.
SUFFICIENT.
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EgmatQuantExpert

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