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Algebra

by Aman verma » Thu Aug 11, 2011 5:02 am
Q: The greatest possible divisor of [3^(2n+3)] - 24n -27 for every n belonging to natural number, which necessarily divides is :

a)24

b)64

c)96

d)97

e)none of these

Now the problem has something to do with the formula a^3 + b^3 + c^3 ,maybe where a+b+c = 0, but I am still unable to figure out the solution.
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by Frankenstein » Thu Aug 11, 2011 5:26 am
Hi,
I don't know about the method you stated.
My solution:
[3^(2n+3)] - 24n -27 = 27.9^n - 24n - 27 = 27(9^n - 1) -24n
9^n - 1 = (8+1)^n - 1 = (8^n + n.8^(n-1) + ... + nC2.8^2 + n.8 + 1) - 1 = 64*k +8n
So, 27(9^n - 1) -24n = 27(64k + 8n) - 24n = 27*64k + 24*9n - 24n = 27*64k + 24*8n
= 24*8(9k+n)
24*8 = 192
Hence, E

P.S. I guess these questions are tougher than those required for GMAT. Please make sure you are working with correct sources.
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by gmatboost » Thu Aug 11, 2011 7:38 am
I would like to second Frankenstein's recommendation that you consider working with other sources, since this is definitely not a GMAT question.
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by Anurag@Gurome » Thu Aug 11, 2011 7:46 am
Aman verma wrote:Q: The greatest possible divisor of [3^(2n+3)] - 24n -27 for every n belonging to natural number, which necessarily divides is :
Well, there is a pretty much faster and easier way to solve this.
The question asks to determine the greatest possible divisor of [3^(2n+3)] - 24n - 27 for every n belonging to natural number. So we can pick the lowest value possible for n, which is 1 and analyze.

For n = 1, [3^(2n+3)] - 24n - 27 = 3^5 - 24 - 27 = 243 - 51 = 192

Thus, 192 is the greatest possible divisor.

The correct answer is E.
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by Aman verma » Thu Aug 11, 2011 9:14 am
Hi!! all, thanks for the response. Now I can figure out my mistake. I was amazed by the Frankenstein's approach particularly the Binomial Expansion used, I was expecting a simpler approach. Nevertheless, this is one of the several ways to approach a problem . One should have all the tools in his arsenal for better preparation. Also I was inspired by Anurag sir's approach. So simple and effective.

Now regarding Gmatboost's concerns, I agree that this might not replicate exactly a GMAT question, but nevertheless one should prepare for any eventuality. The GMAT may throw an awkward question when one takes the actual test. I would like to mention my own experience- when I took the test one question from CONIC SECTION involving Focus, Latus Rectum did appeared and I was completely outwitted. Although Conic Section is not in the syllabus for GMAT, it did appeared on the actual test. It might have been an experimental question but nobody can tell beforehand which ones are experimental. So I would say one should prepare for any sort of eventuality. Besides an extra effort never hurts anyone !!
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by gmatboost » Thu Aug 11, 2011 10:35 am
I feel pretty strongly that it's not worth spending time worrying about topics that are not in the Official Guide's math review chapter (e.g. binomial distribution, conic sections). Of course there is nothing wrong with knowing them, but I maintain that you will not see a question on which you MUST know them.

Also, I would caution that plugging in n = 1 to get 192 does NOT in itself prove that 192 "necessarily divides" 3^(2n+3) - 24n - 27
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by Frankenstein » Thu Aug 11, 2011 10:43 am
gmatboost wrote: Also, I would caution that plugging in n = 1 to get 192 does NOT in itself prove that 192 "necessarily divides" 3^(2n+3) - 24n - 27
Exactly! This is not justified and cannot be generalized although it works in this particular example.
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by Anurag@Gurome » Thu Aug 11, 2011 11:19 am
Hi Greg and Frankenstein!

Number theory as a topic is very much logical and manipulative one. Fluke or luck has a very small role to play here. In my experience I've seen these problems (particularly those which involves integers or natural numbers) do follow a pattern which is true for all the integers in the range if they are true for the lowest permitted ones. And almost always there exist a mathematical logic behind them.

In this case also my argument is purely based on mathematical logic which does not involve any concept beyond GMAT (like binomial theorem etc). My only 'mistake' was that I didn't show all the logical details to support my claim. Anyway here it goes...

We can rewrite the expression as follows,
  • T(n) = 3^(2n+3) - 24n - 27
    = 9*[3^{2(n - 1) + 3} - 24(n - 1) - 27] + 192n
    = 9*T(n - 1) + 192n
So for n = 1, the expression, i.e. T(1) = 192
Therefore,
  • for n = 2, T(2) = 9*T(1) + 192*2 = 9*192 + 192*2 = 11*192
    for n = 3, T(3) = 9*T(2) + 192*3 = 9*11*192 + 192*3 = 102*192
    ... so on and so forth
So in this case plugging in n = 1 to get 192 does in itself prove that 192 "necessarily divides" 3^(2n+3) - 24n - 27 for all possible values of n.
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by Frankenstein » Thu Aug 11, 2011 11:49 am
Hi Anurag,
Thanks for posting. There is no disputing your knowledge/experience(in fact any experts). But, without proof, which has been posted now, at least some of the readers won't be convinced as we can always find exceptions or counter examples for similar cases. All of us definitely want to know the underlying concept and make sure it works *always*. Thanks again!
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