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If x is a positive integer and z is a non-negative integer such that (2,066)z is a divisor of, 176,793, what is the value of zx - xz?

A) -81 B) -1 C) 0 D) 1 E) It Cannot Be Determined

An odd number is never divisible by an even number. What value of z could make the expression (2,066)z equal to an odd number that is a factor of any number?

pratikbais please format the questions properly. Thank you. The question should read:

If x is a positive integer and z is a non-negative integer such that 2,066^z is a divisor of 3,176,793, what is the value of z^x - x^z? A. -81 B. -1 C. 0 D. 1 E. It Cannot Be Determined

3,176,793 is an odd number. The only way it to be a multiple of 2,066^z (even number in integer power) is when \(z=0\), in this case \(2,066^z=2,066^0=1\) and 1 is a factor of every integer. Hence \(z=0\) --> \(z^x-x^z=0^x-x^0=0-1=-1\).

Answer: B.

Must know for the GMAT: \(a^0=1\), for \(a\neq{0}\) - any nonzero number to the power of 0 is 1. Important note: the case of 0^0 is not tested on the GMAT.

Re: If x is a positive integer and z is a non-negative integer [#permalink]

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02 Aug 2014, 04:33

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Also, 2066^z = a number that ends with digit 6. If you multiply 6 *6*6*..... , it always end with 6 at the end. So 3176793 cannot be divided by 2066, then z= 0 -> B is the answer
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Re: If x is a positive integer and z is a non-negative integer [#permalink]

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02 Nov 2015, 01:24

The number 3.176.793 is a multiple of 3 becasue 3+1+7+6+7+9+3 = 36, which is a multiple of 3. The number 2.066 is not a multiple of three because 2+0+6+6 = 14, which is not a multiple of 3.

A quick tip: If the question says "non-negative integer", alarms should go off in your head and you should immediately consider the case of 0. It is easier to say positive integers. When the test maker goes for "non negative integer", it usually means that 0 will have some role to play in the solution. Also considering the huge numbers the question deals with, it is a good idea to try 0 and 1 (if needed). The moment you put z = 0, you get the answer as -1. Since it is a PS question, you are done.
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