What is remainder of the following?

Hi henilshaht,

We're asked for the remainder when (6^35 - 6^25 - 2) is divided by 6. This is a great 'concept question', meaning that you don't actually have to do much math if you recognize the concepts involved.

To start, the GMAT will NEVER expect you to calculate that type of large value by hand, so that is NOT what this question requires.

Since we're dividing a number by 6, the only possible values for the remainder are 0, 1, 2, 3, 4 and 5. If you increase the numerator by 1, then the remainder will increase by 1 (the exception being that if the current remainder is 5, then increasing the numerator by 1 will 'cycle' the 5 back to a 0). In that same way, if you decrease the numerator by 1, then the remainder will decrease by 1 (again, the exception being if the current remainder is already 0, then it will 'cycle' the 0 to a 5).

Both 6^35 and 6^25 are MULTIPLES of 6... so dividing either of those individual numbers by 6 will give you a remainder of 0. Subtracting a multiple of 6 from a larger multiple of 6 will result in a number that is ALSO a multiple of 6... so whatever 6^35 and 6^25 actually are... the calculation (6^35 - 6^25) IS a multiple of 6.... meaning that the remainder would be 0 if there was no other math to consider.

However, that extra "-2" means that the remainder will no longer be 0. We have to 'cycle' that 0 down "2 spots" ... and the remainder will become 4.

Final Answer: E

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

henilshaht wrote:$$\frac{^{6^{35}\ -\ 6^{25}\ -\ 2}}{6}$$

A. 0
B. 1
C. 2
D. 3
E. 4[/list]

Since 6^35 and 6^25 are divisible by 6, the remainder when each is divided by 6 is 0. Therefore, the remainder is the last term “-2”. Of course, the remainder can’t be negative, and in such a case, we add 6 to -2 to make it positive (and thus the true remainder can be revealed). Therefore, the remainder is -2 + 6 = 4.

Answer: E

henilshaht wrote:$$\frac{^{6^{35}\ -\ 6^{25}\ -\ 2}}{6}$$

A. 0
B. 1
C. 2
D. 3
E. 4[/list]

$$\frac{^{6^{35}\ -\ 6^{25}\ -\ 2}}{6} = \frac{^{6^{35}\ -\ 6^{25}\ -\ 6 + 4}}{6}= \frac{^6({6^{34}\ -\ 6^{24}\ -\ 1) + 4}}{6}$$

At this point, we can see that the numerator is 4 greater than some multiple of 6,
So when we divide the numerator by 6, the remainder must be 4.

Answer: D