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M26-20

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M26-20  [#permalink]

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New post 16 Sep 2014, 01:25
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

49% (02:02) correct 51% (02:25) wrong based on 185 sessions

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Re M26-20  [#permalink]

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New post 16 Sep 2014, 01:25
1
3
Official Solution:

If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then \(y\) is NOT divisible by which of the following?

A. \(6^4\)
B. \(62^2\)
C. \(65^2\)
D. \(15^4\)
E. \(52^4\)


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=\)

\(=3^4*(3^3-1)^2*5^8*(5^3-1)^2=\)

\(=3^4*26^2*5^8*124^2=\)

\(=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).


Answer: E
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Re: M26-20  [#permalink]

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New post 21 Jul 2015, 05:37
1
Bunuel wrote:
Official Solution:

If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then \(y\) is NOT divisible by which of the following?

A. \(6^4\)
B. \(62^2\)
C. \(65^2\)
D. \(15^4\)
E. \(52^4\)


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=\)

\(=3^4*(3^3-1)^2*5^8*(5^3-1)^2=\)

\(=3^4*26^2*5^8*124^2=\)

\(=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).


Answer: E


Bunuel,

How did you get from: \((3^5-3^2)^2*(5^7-5^4)^2\)
to \(3^4*(3^3-1)^2*5^8*(5^3-1)^2\) ?

BR,
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Re: M26-20  [#permalink]

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New post 21 Jul 2015, 05:43
3
lucassandrade wrote:
Bunuel wrote:
Official Solution:

If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then \(y\) is NOT divisible by which of the following?

A. \(6^4\)
B. \(62^2\)
C. \(65^2\)
D. \(15^4\)
E. \(52^4\)


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=\)

\(=3^4*(3^3-1)^2*5^8*(5^3-1)^2=\)

\(=3^4*26^2*5^8*124^2=\)

\(=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).


Answer: E


Bunuel,

How did you get from: \((3^5-3^2)^2*(5^7-5^4)^2\)
to \(3^4*(3^3-1)^2*5^8*(5^3-1)^2\) ?

BR,


Consider this: \((ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2\), so \((3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2\).

The same for (5^7-5^4)^2

Hope it helps.
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Re: M26-20  [#permalink]

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New post 19 Mar 2016, 14:33
Bunuel wrote:
Official Solution:

If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then \(y\) is NOT divisible by which of the following?

A. \(6^4\)
B. \(62^2\)
C. \(65^2\)
D. \(15^4\)
E. \(52^4\)


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=\)

\(=3^4*(3^3-1)^2*5^8*(5^3-1)^2=\)

\(=3^4*26^2*5^8*124^2=\)

\(=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).


Answer: E



Bunuel, could you please elaborate the following steps?

\(=3^4*26^2*5^8*124^2=\)

\(=2^6*3^4*5^8*13^2*31^2\).

Thank you!
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M26-20  [#permalink]

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New post 19 Mar 2016, 19:28
paddy41 wrote:
Bunuel wrote:
Official Solution:

If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then \(y\) is NOT divisible by which of the following?

A. \(6^4\)
B. \(62^2\)
C. \(65^2\)
D. \(15^4\)
E. \(52^4\)


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=\)

\(=3^4*(3^3-1)^2*5^8*(5^3-1)^2=\)

\(=3^4*26^2*5^8*124^2=\)

\(=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).


Answer: E



Bunuel, could you please elaborate the following steps?

\(=3^4*26^2*5^8*124^2=\)

\(=2^6*3^4*5^8*13^2*31^2\).

Thank you!



hi,
let me do it for you from \(=3^4*26^2*5^8*124^2=\)
\(=3^4*(2*13)^2*5^8*(4*31)^2\)
=> \(=3^4*2^2*13^2*5^8*4^2*31^2\)
=> \(=3^4*2^{(2+4)}*13^2*5^8*31^2\)
=> \(2^6*3^4*5^8*13^2*31^2\)
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Re: M26-20  [#permalink]

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New post 14 Aug 2018, 11:21
In order to solve the problem take the common terms out and keep the square as it is:

y=(3^5−3^2)^2(5^7−5^4)^−2

this will become [{3^2(3^3-1))^2].[{5^4(5^3-1)}^2] and then you will reach the solution shared by Bunuel

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M26-20  [#permalink]

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New post 20 Aug 2018, 21:54
Bunuel wrote:
If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then \(y\) is NOT divisible by which of the following?

A. \(6^4\)
B. \(62^2\)
C. \(65^2\)
D. \(15^4\)
E. \(52^4\)


Wow!! What a question!
Took me 4 minutes to solve and check all the options.

Hi Bunuel ,

In actual GMAT, is it advisable to commit this much time to a tough question like this? I am asking this because of the trade off between the time and the higher marks associated to it (given that this is a tough one to crack).
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Re: M26-20  [#permalink]

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New post 20 Aug 2018, 22:37
1
anuj04 wrote:
Bunuel wrote:
If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then \(y\) is NOT divisible by which of the following?

A. \(6^4\)
B. \(62^2\)
C. \(65^2\)
D. \(15^4\)
E. \(52^4\)


Wow!! What a question!
Took me 4 minutes to solve and check all the options.

Hi Bunuel ,

In actual GMAT, is it advisable to commit this much time to a tough question like this? I am asking this because of the trade off between the time and the higher marks associated to it (given that this is a tough one to crack).


You don't want to miss hard questions on the test because they cost more than easier ones. So, you can spend more than an average time on them (more than 2 mins) because ideally you should be able to save time on easier question by solving them in less time. 4 minutes though is still seems too much.
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Re: M26-20  [#permalink]

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New post 20 Aug 2018, 23:54
Bunuel wrote:
anuj04 wrote:
Bunuel wrote:
If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then \(y\) is NOT divisible by which of the following?

A. \(6^4\)
B. \(62^2\)
C. \(65^2\)
D. \(15^4\)
E. \(52^4\)


Wow!! What a question!
Took me 4 minutes to solve and check all the options.

Hi Bunuel ,

In actual GMAT, is it advisable to commit this much time to a tough question like this? I am asking this because of the trade off between the time and the higher marks associated to it (given that this is a tough one to crack).


You don't want to miss hard questions on the test because they cost more than easier ones. So, you can spend more than an average time on them (more than 2 mins) because ideally you should be able to save time on easier question by solving them in less time. 4 minutes though is still seems too much.


Thanks Bunuel!!

I got your point. If tougher one can be solved within 3 minutes then, it's ok. Otherwise, take an informed guess and leave. :)
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Re: M26-20   [#permalink] 20 Aug 2018, 23:54
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