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What's the secret to approaching this problem?

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What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 10:12
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A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

25% (03:37) correct 75% (00:47) wrong based on 19 sessions
If (6^2)(44)(5^x)(20) / (8^2)(9) = 1375, what is the value of x?

A -1
B 0
C 1
D 2
E 3

My approach was the "blunt force" method, by slowly multiplying it all out then trying to solve; the numbers quickly become untenable. And aside from slowly plugging-in each answer choice until I get lucky, what's the REAL way to solve this?

EDIT: (62) should be (6^2), fixed.
EDIT: (82) was changed to (8^2). fixed. (Problem is now accurate)

Last edited by lifeisshort on 20 Aug 2010, 16:34, edited 4 times in total.
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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 10:18
Is answer 2?

If just looked at 5 then we have 5^(x+1) = 5^3

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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 10:28
mainhoon wrote:
Is answer 2?

If just looked at 5 then we have 5^(x+1) = 5^3

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How can we assume five-to-the-x-power, (5^x), is actually 5^x plus "some other number"? If that was the intention of the problem, then the problem would show (5^(x+1)) not (5^x)...right?

And in anycase, how did you decide on the answer choice? How do you know what the answer is....

Last edited by lifeisshort on 20 Aug 2010, 10:31, edited 1 time in total.
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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 10:30
I got the extra 5 from 20. I wa isolating all the 5s in one place.. So is it 2?

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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 10:57
I don't know the answer, unfortunately. Anyone able to offer a solution/explanation for this problem?
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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 11:02
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Ok so I took a calculator out and here is the bottom-line
5^x = 18.59, it's not a clean figure, I don't know why
But closest has to be 2..

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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 11:19
Ooops, I must apologize because I made a blunder in my copy/paste of this equation.
(62) should be (6^2), six-to-the-power-of-two.

So, does this change your answer...if not, can you elaborate more on how exactly you arrived at your solution? THanks.
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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 12:29
Are you sure that the denominator is correct? 1,375 factors to 11 * 5^3. The denominator factors to 41*2*3*3. There is no other 41 in the equation to cancel out. Further, 41 is a prime number. It just seems to me that this is the sort of problem where most other numbers will cancel out and you would be left with 5^x = 5^3.
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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 13:03
I would say x is still 2 as your correction did not introduce any more 5s... Doublecheck with calc but 6^2 still doesn't work

lifeisshort wrote:
Ooops, I must apologize because I made a blunder in my copy/paste of this equation.
(62) should be (6^2), six-to-the-power-of-two.

So, does this change your answer...if not, can you elaborate more on how exactly you arrived at your solution? THanks.


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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 16:33
runnergirl683 wrote:
Are you sure that the denominator is correct? 1,375 factors to 11 * 5^3. The denominator factors to 41*2*3*3. There is no other 41 in the equation to cancel out. Further, 41 is a prime number. It just seems to me that this is the sort of problem where most other numbers will cancel out and you would be left with 5^x = 5^3.


Gosh, yes I see the denominator was incorrect; it should be (8^2), I made the change.
The problem is now correct....and with a quick calculator check, the answer should be 2 (answer D).

But is there anyway to do this OTHER than doing ALL of that LONG-hand multiplication and LONG-hand division? It takes too much time and increases the chance I'll make an error.

What is the shortcut?
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Re: What's the secret to approaching this problem? [#permalink] New post 20 Aug 2010, 16:53
Well, as I have been saying in my earlier posts above the shortest method indeed is:
5^(x+1) = 5^3
x=>2

Just compare the powers of 5 on both sides. I can't imagine anything shorter than this.
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Re: What's the secret to approaching this problem? [#permalink] New post 23 Aug 2010, 11:38
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The attached should help.
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Re: What's the secret to approaching this problem? [#permalink] New post 24 Aug 2010, 09:40
seekmba wrote:
The attached should help.


Yup, that's how I did it. I think what we must remember in this case is that most of the time, GMAT problems that appear to require complex calculations can often be simplified to very manageable proportions.
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Re: What's the secret to approaching this problem? [#permalink] New post 27 Aug 2010, 06:34
You can do it two ways Either by solving the whole equation or by considering 1375 and it's factor with the factors on the other side
for full explanation Seekmba has given right but long explanation
another the shorter version
The whole equation on the left side when solved should be equal to the 1375
the factors of 1375 are 5*5*5*11
that means all this value will be present on the left side of the equation
and rest all will cancel each other out
Moreover we are not concerned about any powers of 2,3 and 11
we are only concerned with the power of 5
so we will search for the 5 as factor for the equation on the left
it comes out to be 5^x and 5 as a factor of 20
Hence 5^x*5 = 5^3
that is all what is required therefore the answer for x is 2
i.e. D
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Re: What's the secret to approaching this problem? [#permalink] New post 04 Nov 2010, 11:58
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easy one if you prime factorise every term with 2 and 3 , just discard all possible and you are left with only one prime "5" to be operated
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Re: What's the secret to approaching this problem?   [#permalink] 04 Nov 2010, 11:58
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