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Re: Divisor of 3,176,793 [#permalink]
18 Sep 2010, 19:11

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Expert's post

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Orange08 wrote:

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. _________________

Re: If n is a non-negative integer such that 12n is a divisor of [#permalink]
28 Mar 2013, 21:12

Expert's post

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.

Re: Divisor of 3,176,793 [#permalink]
05 Jul 2013, 08:08

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? _________________

Re: Divisor of 3,176,793 [#permalink]
05 Jul 2013, 08:17

Expert's post

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.

Re: If n is a non-negative integer such that 12^n is a divisor [#permalink]
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]
15 Aug 2015, 11:25

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