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

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If x is a positive integer less than 100 such that x is  [#permalink]

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Q.

If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


Answer Choices



    A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
    B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
    C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
    D. EACH statement ALONE is sufficient to answer the question asked.
    E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Thanks,
Saquib
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Originally posted by EgmatQuantExpert on 26 May 2017, 05:50.
Last edited by EgmatQuantExpert on 07 Aug 2018, 04:34, edited 1 time in total.
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 26 May 2017, 08:18
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7
EgmatQuantExpert wrote:
If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


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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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 26 May 2017, 05:50
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post Updated on: 26 May 2017, 08:37
1
EgmatQuantExpert wrote:
Q.

If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


Answer Choices



    A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
    B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
    C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
    D. EACH statement ALONE is sufficient to answer the question asked.
    E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Thanks,
Saquib
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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

Originally posted by rohit8865 on 26 May 2017, 07:49.
Last edited by rohit8865 on 26 May 2017, 08:37, edited 2 times in total.
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 03 Jun 2017, 01:09
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
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 13 Jun 2017, 01:51
rohit8865 wrote:
EgmatQuantExpert wrote:
Q.

If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


Answer Choices



    A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
    B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
    C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
    D. EACH statement ALONE is sufficient to answer the question asked.
    E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Thanks,
Saquib
Quant Expert
e-GMAT

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts :)

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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


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
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 03 Jul 2017, 14:15
nguyendinhtuong wrote:
EgmatQuantExpert wrote:
If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


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


Not sure where "k" comes from and why you went this route?
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 03 Jul 2017, 17:53
1
Smokeybear00 wrote:
nguyendinhtuong wrote:
EgmatQuantExpert wrote:
If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


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


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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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 06 Aug 2017, 22:42
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.
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 08 Aug 2017, 02:38
broall wrote:
EgmatQuantExpert wrote:
If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


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


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
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 02 Nov 2017, 19:48
broall wrote:
EgmatQuantExpert wrote:
If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


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



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?
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 06 Nov 2017, 13:42
1
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)
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 16 Dec 2017, 07:53
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.


broall wrote:
EgmatQuantExpert wrote:
If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


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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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 20 Feb 2018, 11:47
There is more than one solution to statement 2. Put y = 5 and x = 8 into it and you'll see it works.
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 20 Feb 2018, 23:04
NigerianBankScammer wrote:
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.
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Re: If x is a positive integer less than 100 such that x is  [#permalink]

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New post 20 Feb 2018, 23:23
gmatbusters wrote:
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.


broall wrote:
EgmatQuantExpert wrote:
If x is a positive integer less than 100 such that x is divisible by \(2^y\), where y is a positive integer, what is the value of y?

    (1) \(x^2 > 3600\)

    (2) \(\frac{x^2}{2^{y+2}}\) is an odd integer


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



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.
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