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# M26-20

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Math Expert
Joined: 02 Sep 2009
Posts: 56366

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16 Sep 2014, 01:25
00:00

Difficulty:

95% (hard)

Question Stats:

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

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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. $$52^4$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 56366

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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 $$y$$.

_________________
Intern
Joined: 16 Jun 2015
Posts: 6

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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 $$y$$.

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,
Math Expert
Joined: 02 Sep 2009
Posts: 56366

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21 Jul 2015, 05:43
3
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 $$y$$.

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.
_________________
Intern
Joined: 11 Jan 2015
Posts: 33

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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 $$y$$.

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!
Math Expert
Joined: 02 Aug 2009
Posts: 7763

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19 Mar 2016, 19:28
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 $$y$$.

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 $$=3^4*26^2*5^8*124^2=$$
$$=3^4*(2*13)^2*5^8*(4*31)^2$$
=> $$=3^4*2^2*13^2*5^8*4^2*31^2$$
=> $$=3^4*2^{(2+4)}*13^2*5^8*31^2$$
=> $$2^6*3^4*5^8*13^2*31^2$$
_________________
Intern
Joined: 10 May 2018
Posts: 1

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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

Akanksha
Manager
Joined: 25 Jul 2017
Posts: 93

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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).
Math Expert
Joined: 02 Sep 2009
Posts: 56366

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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.
_________________
Manager
Joined: 25 Jul 2017
Posts: 93

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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.
Re: M26-20   [#permalink] 20 Aug 2018, 23:54
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