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If n is a non-negative 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 non-negative integer such that 12^n is a divisor of 3,176,793, what is the value of n^12-12^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.

-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 non-negative integer such that 12n is a divisor of [#permalink]

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28 Mar 2013, 21:12

nave81 wrote:

If n is a non-negative 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.

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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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.

Re: If n is a non-negative integer such that 12^n is a divisor [#permalink]

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05 Jul 2014, 16:43

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Re: If n is a non-negative integer such that 12^n is a divisor [#permalink]

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15 Aug 2015, 11:25

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Re: If n is a non-negative integer such that 12^n is a divisor [#permalink]

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09 Mar 2016, 10:36

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.

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&^12-12^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 non-negative integer such that 12^n is a divisor
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09 Mar 2016, 10:36

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