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# M26-20

Author Message
Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
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19 Feb 2013, 22:52
Hi
Sorry , I don't know how to format it correctly here, so pasting the image.

My doubt is .. If we take 5 raised to 4 as common in the denominator,
we will be left with

[5 raised to 4 ( 5 raised to 3 - 1 ) ] raised to -2.

Now this can be deduced to

5 raised to 2 * [5 raised to 3 - 1 ] raised to -2 ..

So 25 remains in the denominator .

However , if we do not take 5 raised to 4 as common in the denominator, the entire denominator can be taken above in the numerator by changing the sign of the exponent to positive 2.
In this case no 25 remains in the denominator.

I don't understand where I am going wrong . Kindly help.
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21 Feb 2013, 10:06
There is no denominator to worry about, move the denominator to the numerator and change the sign to positive.

(3^5 - 3^2)^2 * (5^7 - 5^4)^2

Then factor

(3^2(3^3 - 1))^2 * (5^4(5^3 - 1))^2

Simplify

(9*26)^2 * (625*124)^2

E is the answer because there are only 2 factors of 13 in the above expression and E has 4.
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22 Feb 2013, 00:54
OE is below:
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 $$y$$.

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22 Feb 2013, 01:16
If we observe the expression, it can be deduced to:

y= 〖3^2 (26)(5^4 (124))〗^2

Now let us eliminate options:

A: 6^4 can be eliminated as we have four 3s and four 2s
B: 62^2 can be eliminated as we have 124^2
C: 65^2 can be eliminated as we have 26^2 and 5^4
D: 15^4 can easily be eliminated
E: 52^4 cannot be eliminated as we do not have enough factors of 2

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Pushpinder Gill

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30 Mar 2013, 22:08
Bunuel wrote:
OE is below:
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 $$y$$.

Hi Brunel, How do you get 3^4(3^3-1)^2 from (3^5-3^2)^2?
Math Expert
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Posts: 36601
Followers: 7097

Kudos [?]: 93484 [0], given: 10563

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31 Mar 2013, 07:25
mp2469 wrote:
Bunuel wrote:
OE is below:
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 $$y$$.

Hi Brunel, How do you get 3^4(3^3-1)^2 from (3^5-3^2)^2?

Factor out 3^2 from (3^5-3^2)^2: (3^2(3^3-1))^2=3^4(3^3-1)^2.

Hope it's clear.
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Re: M26-20   [#permalink] 31 Mar 2013, 07:25
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# M26-20

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