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 Post subject: GMAT Number Theory
PostPosted: Fri Jul 13, 2012 5:41 am 
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Let a be a positive integer. If n is divisible by both 2ª and 3²ª, then which of the following might NOT be a divisor of n?

A. 6
B. 3 × 2ª
C. 2 × 3²ª
D. 6ª
E. 6²ª

(E) n is divisible by 2ª means that the decomposition of n has at least a times the number “2” in it.
n is divisible by 3²ª means that the decomposition of n has at least 2a times the number “3” in it.

The smallest number that satisfies the both conditions is 2ª × 3²ª, because 2 and 3 are prime. We can see that this number is smaller than the choice (E),
6²ª = (2 × 3)²ª = 2²ª × 3²ª = 2ª × [2ª × 3²ª],
therefore it is possible that n is not divisible by 6²ª, choice (E).

You can ascertain that any of the numbers in choices (A) to (D) has less “2” and “3” in its decomposition than the number n. So it means, that n must be divisible by any of those.
---------

Why do we conclude from "that this number is smaller than choice (E)," that choice (E) is the right one?


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 Post subject: Re: math, t.1, qt.18: exponents, number theory
PostPosted: Fri Jul 13, 2012 5:42 am 
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We make a conclusion that choice (E) is the right one because we've shown that if n = 2ª × 3²ª (it fully satisfies the question statement) then it is not divisible by 6²ª simply because 6²ª is greater.

And furthermore I'd like to go over this question once again in details.
The question statement tells us that n is divisible by 2ª and 3²ª, where a is some positive integer. Therefore we can represent n as: n = 2ª × 3²ª × d, where d is some positive integer.

Note, that we consider n to be a positive integer for simplicity since divisibility of n or -n is the same.

We write the above mentioned formula for n because 2ª and 3²ª are relatively prime. If they where, for example, 10 and 15 (instead of 2ª and 3²ª) then n would equal 2 × 3 × 5 × d because 2 × 3 × 5 is the least common multiple of 10 and 15 while the least common multiple of 2ª and 3²ª is 2ª × 3²ª.

Let's return to our formula and consider all the answer choices:
n = 2ª × 3²ª × d

A. Is n divisible by 6?
Yes, because it is clearly divisible by 2 and 3.

B. Is n divisible by 3 × 2ª
Yes, because it is clearly divisible by 3 and 2ª is a multiplier in formula for n.

C. Is n divisible by 2 × 3²ª
Yes, because it is clearly divisible by 2 and 3²ª is a multiplier in formula for n.

D. Is n divisible by 6ª
Yes, because we can rewrite the formula for n:
n = 2ª × 3ª × 3ª × d and 6ª = (2 × 3)ª = 2ª × 3ª
So 2ª and 3ª are the multipliers in formula for n.

E. Is n divisible by 6²ª
It can be or it can be not. Let us divide n by 6²ª:
n / 6²ª = (2ª × 3ª × 3ª × d) / (2ª × 2ª × 3ª × 3ª) = d / 3ª
So n / 6²ª = d / 3ª

We clearly see that if d is divisible by 3ª then n is divisible by 6²ª. But if , let's say, d = 1 then n is not.

Therefore n MUST BE divisible by the choices (A) – (D) and NOT NECESSARILY by choice (E).


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 Post subject: Re: math, t.1, qt.18: exponents, number theory
PostPosted: Fri Jul 13, 2012 5:43 am 
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I'm not seeing this:
- 6 to the 2a is different than 6 to the a how?
- both 6's are multiples of each of the primes provided and how is C not the same? (2 × 3 to the 2a?)


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 Post subject: Re: math, t.1, qt.18: exponents, number theory
PostPosted: Fri Jul 13, 2012 5:43 am 
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Choice (C), 2 × 3²ª, is much smaller than Choice (E), 6²ª = (2 × 3)²ª = 2²ª × 3²ª.

Note that notations (2 × 3)²ª and 2 × 3²ª mean two different numbers.

Choices (D) and (E) also differ.
6²ª = (6²)ª = 6ª × 6ª or 36ª. It is clearly greater than choice (D), 6ª.


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 Post subject: Re: math, t.1, qt.18: exponents, number theory
PostPosted: Fri Jul 13, 2012 5:44 am 
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I did not understand the explanation to this question quite well. Why is it not B for example?


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 Post subject: Re: math, t.1, qt.18: exponents, number theory
PostPosted: Fri Jul 13, 2012 5:44 am 
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questioner wrote:
Why is it not B for example?
Because, if n is divisible by 2ª and n is also divisible by 3²ª, then n MUST be divisible by 3 × 2ª (B). While the question asks us to choose a value that might NOT be a divisor of n.

In other words, we don't know what the real value of n is. It can be any value that satisfies "n is divisible by 2ª and n is also divisible by 3²ª". Choices A, B, C, D are divisors of any such value of n. While there is at least one possible value of n (e.g. 2ª × 3²ª) for which E, 6²ª, is NOT a divisor.


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 Post subject: Re: math, t.1, qt.18: exponents, number theory
PostPosted: Fri Jul 13, 2012 5:46 am 
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Not sure I am following the logic here due to the fact that Answer Choice E seems to be larger than Answer Choice D? I could be completely missing the point here, and most likely the case, but if you have some additional explanation that would be really helpful. Would picking numbers at all make sense here? Also, is this a 700-800 level question or more in the range of 600-700? Please let me know when you have an opportunity, thank you very much.


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 Post subject: Re: math, t.1, qt.18: exponents, number theory
PostPosted: Fri Jul 13, 2012 6:23 am 
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questioner wrote:
Not sure I am following the logic here due to the fact that Answer Choice E seems to be larger than Answer Choice D? I could be completely missing the point here, and most likely the case, but if you have some additional explanation that would be really helpful. Would picking numbers at all make sense here? Also, is this a 700-800 level question or more in the range of 600-700?
This is more a 600-700 question, though it might look harder. But the concept is very simple:

1. The smallest possible number, which is divisible by both 2ª and 3²ª is 2ª × 3²ª .
2. Choice E, 6²ª , is greater than 2ª × 3²ª , so it is definitely NOT a divisor of 2ª × 3²ª.

We have shown that at least one possible value of n (2ª × 3²ª) is NOT divisible by E. So E might NOT be a divisor of n.

The question tests the following abilities:
- dealing with powers
- knowing what factorization is and how it is connected to divisibility.


ANOTHER APPROACH.
Take a look at the answer choices. Exactly one of them must be the correct answer.

Choice E is divisible by any other answer choice. So if E is a divisor of n, then all the other answer choices are divisors as well.

Thus E cannot be the divisor of all possible values of n, because in this case all the answer choices would be the divisors and we would have NO correct answer.

A. 6
B. 3 × 2ª
C. 2 × 3²ª
D. 6ª
E. 6²ª


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