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If n is a nonnegative integer such that 12^n is a divisor
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If n is a nonnegative integer such that 12^n is a divisor of 3,176,793, what is the value of n^1212^n? A. 11 B. 1 C. 0 D. 1 E. 11
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Originally posted by Orange08 on 18 Sep 2010, 12:48.
Last edited by Bunuel on 29 Mar 2013, 02:41, edited 2 times in total.
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Re: Divisor of 3,176,793
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18 Sep 2010, 20:11
Orange08 wrote: If n is a nonnegative integer such that 12n is a divisor of 3,176,793, what is the value of n^12 – 12^n ?
a. 11 b. 1 c. 0 d. 1 e. 11 If the answer is B then I think it should be \(12^n\) instead of \(12n\) So the question would be: If n is a nonnegative integer such that 12^n is a divisor of 3,176,793, what is the value of n^1212^n?3,176,793 is an odd number. The only way it to be a multiple of \(12^n\) (even number in integer power) is when \(n=0\), in this case \(12^n=12^0=1\) and 1 is a factor of every integer. Then \(n^{12}12^n=0^{12}12^0=1\). Answer: B. Hope it helps.
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Re: Divisor of 3,176,793
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18 Sep 2010, 13:00
3176793 is odd 12n is even How can 12n be a divisor ? The only answer I can think is n=0 which means 1 But I don't think you can count 0 as a "divisor"
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Re: Divisor of 3,176,793
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18 Sep 2010, 13:06
Precisely, for this reason, I have posted this question here. I am unclear is 0 should be considered as divisor.



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Re: Divisor of 3,176,793
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Updated on: 18 Sep 2010, 16:50
Orange08 wrote: Precisely, for this reason, I have posted this question here. I am unclear is 0 should be considered as divisor. What's the source of the question ? I am sure the only possible answer is 1, just not sure about the validity of the question
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Originally posted by shrouded1 on 18 Sep 2010, 13:10.
Last edited by shrouded1 on 18 Sep 2010, 16:50, edited 1 time in total.



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Re: Divisor of 3,176,793
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25 Feb 2011, 09:56
Nice question!
Bunuel's approach is very good.
Thanks!



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Re: Divisor of 3,176,793
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25 Feb 2011, 22:22
Thanks Bunnel's for this in depth explanation!!



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12^n will always be an even number because it will be a multiple of 12. however 3,176,793 is odd and there is no case when a positive number of n would be a factor of 3,176,793. Only number that would match is when n is zero.
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Re: If n is a nonnegative integer such that 12n is a divisor of
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28 Mar 2013, 22:12
nave81 wrote: If n is a nonnegative integer such that \(12^n\) is a divisor of 3,176,793, what is the value of n^12  12^n?
A. 11 B.  1 C. 0 D. 1 E. 11 n is any integer \(>=0\). Also, \(12^n\) is a divisor of the given number. \(12^0\) = 1 is a divisor of the given number. Replacing n = 0 in the given expression, we have 0^12  12^0 = 1. Note that for any other value of n, there will be a factor of 2 in \(12^n\). But the given number is odd and thus, has no factor of 2. Therefore, any other power of 12, can not be a divisor of the given number. B.
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Re: If n is a nonnegative integer such that 12n is a divisor of
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29 Mar 2013, 00:24
nave81 wrote: If n is a nonnegative integer such that \(12^n\) is a divisor of 3,176,793, what is the value of n^12  12^n?
A. 11 B.  1 C. 0 D. 1 E. 11 The only way that \(12^n\) can be a divisor of 3 is if \(n=0, 12^0=1\). So \(n=0\) 0^(12)  12^0=01=1 B
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Re: Divisor of 3,176,793
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05 Jul 2013, 09:08
3,176,793 is an odd number. The only way it to be a multiple of \(12^n\) (even number in integer power) is when \(n=0\), in this case \(12^n=12^0=1\) and 1 is a factor of every integer.
Can you elaborate on this.. The sum of the digits add up to 9 the only example I thought of 12^2 = 144 does sum of the digits have any relation to this question or it isn't related?
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Re: Divisor of 3,176,793
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05 Jul 2013, 09:17
fozzzy wrote: 3,176,793 is an odd number. The only way it to be a multiple of \(12^n\) (even number in integer power) is when \(n=0\), in this case \(12^n=12^0=1\) and 1 is a factor of every integer.
Can you elaborate on this.. The sum of the digits add up to 9 the only example I thought of 12^2 = 144
does sum of the digits have any relation to this question or it isn't related? No, the sum of the digits is not relevant for this question. 3,176,793 is an odd number. An odd number cannot be a multiple of any even number, and 12^n is even for any positive integer n. Therefore n cannot be positive which means that n can only be 0. Hope it's clear. Similar question to practice: newtoughandtrickyexponentsandrootsquestions12595640.html#p1029223
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Re: If n is a nonnegative integer such that 12^n is a divisor
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09 Mar 2016, 11:36
Bunuel wrote: fozzzy wrote: 3,176,793 is an odd number. The only way it to be a multiple of \(12^n\) (even number in integer power) is when \(n=0\), in this case \(12^n=12^0=1\) and 1 is a factor of every integer.
Can you elaborate on this.. The sum of the digits add up to 9 the only example I thought of 12^2 = 144
does sum of the digits have any relation to this question or it isn't related? No, the sum of the digits is not relevant for this question. 3,176,793 is an odd number. An odd number cannot be a multiple of any even number, and 12^n is even for any positive integer n. Therefore n cannot be positive which means that n can only be 0. Hope it's clear. Similar question to practice: newtoughandtrickyexponentsandrootsquestions12595640.html#p1029223Hi Bunuel, I did not notice that the number given is odd and do the thinking in mind. Rather I read the Q and understood that 12^n should be a divisor on the huge number. 1 is a divisor of the number. and 12^0=1 and hence n=0 satisfies the Q. So I realized that n&^1212^n = 1. if I follow this approach, Will I face a pit fall in any other question similar to this one?
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Re: If n is a nonnegative integer such that 12^n is a divisor
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20 Aug 2017, 22:46
0^anything=0 anything^0=1 Therefore the only value for n=0. Answer : 01=1(B)
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Re: If n is a nonnegative integer such that 12^n is a divisor
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17 Jan 2018, 16:53
Orange08 wrote: If n is a nonnegative integer such that 12^n is a divisor of 3,176,793, what is the value of n^1212^n?
A. 11 B. 1 C. 0 D. 1 E. 11 First notice the big hint right from the start: n is a nonnegative integerYour first reaction should be " Why not just tell us that n is positive?" The reason is that the testmaker wants to include zero as a possible value for n (and zero is neither positive nor negative). Since the testmaker went to the trouble to keep zero as a possible value for n, let's check to see whether n = 0 works. Well, 12^ 0 = 1, and 1 is a divisor of 3,176,793. So n must equal 0. Now that we know the value of n, we can evaluate n^12  12^n n^12  12^n = 0^12  12^ 0 = 0  1 = 1 Answer: B Cheers, Brent
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Re: If n is a nonnegative integer such that 12^n is a divisor
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08 Mar 2018, 12:10
Hi All, This question is built around a number of interesting Number Property rules. Here's how you can use those rules to avoid doing a lot of 'math' on this question. 12^N implies that we're probably dealing with an EVEN number (unless N = 0, in which 12^0 = 1). But we're told that 12^N is a divisor of 3,176,793, which is a big ODD number. EVEN numbers DO NOT divide evenly into ODD numbers, so N CANNOT be a positive number. Since we're told that N is A NONNEGATIVE INTEGER, the only other possibility is when N = 0. Knowing this, the rest of the math is fairly straightforward: (0^12)  (12^0) = 0  1 = 1 Final Answer: GMAT assassins aren't born, they're made, Rich
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