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hey guys this is the first question i got from the prep program and i am having trouble figuring it out: \((\frac{1}{4})^{18} * (\frac{1}{5})^{m} = \frac{1}{2*10^{35}}\) find M A) 17 B) 18 C) 34 D) 35 E) 36 any help will greatly be appreciated
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Last edited by steliossb on 27 May 2009, 18:23, edited 3 times in total.



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Re: find M [#permalink]
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27 May 2009, 05:00
What are the different options available?



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Re: find M [#permalink]
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27 May 2009, 05:51
sorry for not including them:
A) 17 B) 18 C) 34 D) 35 E) 36



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Re: find M [#permalink]
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27 May 2009, 15:22
I would like to know how to solve this one.
(1/2)(1/5)=(1/10)
since (1/5)^18, I don't see how we can get (1/10)^35. no power of 2 ends in 0, which must be the ending digit for (1/10)^n



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Re: find M [#permalink]
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27 May 2009, 16:19
I guess the first step would be to replace the fractions with their reciprocals, which you can do if you do the entire equation, so (1/4)^M * (1/5)^18 = 10^35 * 2 4^M * 5^18 = 10^35 * 2 you then turn both sides into factors, 10 into 5s and 2s, and you get (parenthesis means amount of that number in the multiple, sth like 5*5*5*5*5*5 = 5(6)) 4 (M) * 5 (18) = 5 (35) * 2 (36) divide by the eighteen fives 4(M) = 5(17) * 2(36) we can say that 4(18) = 2(36), so we know that M>18, eliminate A and B by taking these out of the equation we say the solution is N+18 4(N) = 5(17) if you subtract 18 from C,D,and E they are 16, 17. and 18 it actually doesn't equal out to any of them..imagine that. I would answer E tho, because it is def. larger than 17. yeah IDK maybe someone sees my mistake... is this a reliable question? bc I have seen errors in Kaplan and other test providers. I actually entered all of the possible answers into the original equation in a calculator and couldn't get them to equal so I am pretty sure the question is flawed. the answer is somewhere around 37.74 roughly



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Re: find M [#permalink]
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27 May 2009, 17:03
this question was from the official prep software of the GMAT website.
The answer is supposedly 35, no idea how to derive that though.



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Re: find M [#permalink]
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27 May 2009, 18:00
could you make sure you wrote it down right? I have never seen a GMATPREP question flawed, but if you plug 35 into that formula with a calculator it does not work. i would check your software again. Or maybe a parenthesis or sth...



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Re: find M [#permalink]
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27 May 2009, 18:04
that is the accepted method of solving those types of problems tho I believe, it is for me anyway.



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Re: find M [#permalink]
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27 May 2009, 18:08
4^19.74 is more or less equal to 5^17, so if the answer were there, that method would find it
plug in 37.74 into that equation, and you will see what I mean, 35 is nowhere close.



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Re: find M [#permalink]
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27 May 2009, 18:17
i swear i checked that question 3 times to make sure i had it correct. It was only when i was about to post the picture that i saw the mistake:
(1/4)^18 (1/5)^M
which using your method can be pretty straightforward.
I am really sorry about this, i appreciate all your help.



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Re: find M [#permalink]
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27 May 2009, 18:51
(1/4)^18 * (1/5)^m = 1/(2*10^35)
equals
2*10^35 = 4^18*5^m
=>2*2^35*5^35 = 2^36*5^m (splitting 10^35 on LHS to 2^35 *5^35 and splitting 4^18 into 2^36)
which now becomes simple by equating quotients of 5
and answer would be m = 35



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Re: find M [#permalink]
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27 May 2009, 18:53
(1/4)^18 *(1/5)^m =(1/2*10^35) 1/(2*5^35 * 2^35)=1/(2^35+1*5^35)=1/(2^36 * 5^35)= 1/(4^18 * 5^35) ==>m=35
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Re: find M [#permalink]
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27 May 2009, 21:06
Yes correct M=35.
(1/4^18)*(1/5^M) = (1/2*10^35) => 4^18 * 5^M = 2 * 2^35 * 5^35 => 2^36 * 5^M = 2^36 * 5^35
So derived M = 35



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Re: find M [#permalink]
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28 May 2009, 08:05
steliossb wrote: hey guys
this is the first question i got from the prep program and i am having trouble figuring it out:
\((\frac{1}{4})^{18} * (\frac{1}{5})^{m} = \frac{1}{2*10^{35}}\)
find M
A) 17 B) 18 C) 34 D) 35 E) 36
any help will greatly be appreciated There are some questions that are discussed frequently and it is one of them. I believe it tests a basic concept of power. Do you guys also think that it is really difficult or is just a coincident ? Administrator, can we short the questions according to the response or frequency of posting?
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Re: find M [#permalink]
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28 May 2009, 14:24
Yeah this question is def pretty straightforward. I guess it tests your understanding of reciprocals, powers, etc. If it was early in the test it isn't supposed to be one of the more difficult ones.



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Re: find M [#permalink]
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25 Oct 2010, 12:23
Question: isn't it 'dangerous' to just convert to their reciprocals without considering what's in the numerator... ?What's the reasoning to convert to reciprocals (apart from that it is easier to work with after conversion) ?



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Re: find M [#permalink]
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25 Oct 2010, 13:35
n2739178 wrote: Question: isn't it 'dangerous' to just convert to their reciprocals without considering what's in the numerator... ?What's the reasoning to convert to reciprocals (apart from that it is easier to work with after conversion) ? n2739178: The reasoning behind converting to reciprocals is the oft used 'cross multiplication' If I have \((\frac{1}{2})^3 = (\frac{1}{2})^m\) and I need to find the value of m, I can simply treat this as \(\frac{1}{2^3} = \frac{1}{2^m}\) and cross multiply to get \(1.2^m = 1.2^3\). So the left side denominator goes to right with the right side numerator and right side denominator goes to left with the left side numerator. Here, since the numerators are 1, it doesn't matter. Though, if you have all the 2s of the left side and the right side in the denominator, you can simply compare them there and do not need to cross multiply. Here I can simply say m =3. The important thing is that all powers of a certain prime should be consolidated in one place on the left side and same for the right side. e.g. \(3^5.(\frac{1}{2.3})^3 = (\frac{1}{2})^m.3^n\) \(\frac{3^{2}}{2^3} = (\frac{1}{2^m}).3^n\) Here m = 3 and n = 2
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Re: find M [#permalink]
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25 Oct 2010, 14:10
Awesome thanks for that!!



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Re: find M [#permalink]
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25 Oct 2010, 14:19
actually, is there a reason not to apply the exponent rule where for example (\frac{1}{2})exponent m = 1 exponent m / m2 exponent m ? (sorry I can't get the superscript to work...)
I'm getting confused because the rule states you have to do that... But I noticed you didn't give 1 the power of m for example...



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Re: find M [#permalink]
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25 Oct 2010, 17:49
n2739178 wrote: actually, is there a reason not to apply the exponent rule where for example (\frac{1}{2})exponent m = 1 exponent m / m2 exponent m ? (sorry I can't get the superscript to work...)
I'm getting confused because the rule states you have to do that... But I noticed you didn't give 1 the power of m for example... I am sorry, I didn't get your question. Once you type using fraction or powers, highlight the entire Math part and click 'm' given above. It lets the editor know that Math symbols are used here. Anyway, if your confusion is that why did I not give 1 the power of m, it is only because no matter what m is, 1 will remain 1. 1 to any power is still 1. Rather than 1, if the numerator was 2, I would have definitely written it as \(2^m\). The exponent has to be applied to both numerator and denominator in such a case. If your question is something else, let me know.
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