If n is a non-negative integer such that 12^n is a divisor

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"

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

Nice question!

Bunuel's approach is very good.

Thanks!

Thanks Bunnel's for this in depth explanation!!

-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.

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.

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=0-1=-1

B

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: new-tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029223

0^anything=0

anything^0=1

Therefore the only value for n=0.

Answer : 0-1=-1(B)

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 NON-NEGATIVE 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

It does help,
But I wonder, it could still be a divisor and have a remainder correct? Hence should be 12^6?
6^12-12^6
Please help

Posted from my mobile device

Hi billionaire999,

By definition, a 'divisor' divides evenly into a larger number (meaning that there is NO remainder).

For example, both 2 and 4 are divisors of 8 (since 8/2 = 4r0 and 8/4 = 2r0) but 5 is not a divisor of 8 (since 8/5 = 1r3).

GMAT assassins aren't born, they're made,
Rich

Given: n is non negative so it could be 0 and any positive number
12 = 2x2x3
3176793 is not divisible by 2
hence 12^n to be a divisor of 3176793 has to mean n =0
then 0^12-12^0
0-1 = -1
Hence B