possible value of n

Question

If 2^201-n is divisible by 3, which of the following could be the value of n?
I 2
II 5
III 8

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I,II and III

Oski · Answer

If one of the answer is true, they all are (since there is just multiples of 3 between any of them)

So it is either "I, II and III" or "none"

Answer should be (E) since "none" is not a choice

pmenon · Answer

I get E as well, but that might just be a fluke .... would like to see someone explain their method

Oski · Answer

It IS a method. As I said, either the all work, or none of them do. And "none" is not a choice

Oski · Answer

If you want a proof per se (but you don't need it in GMAT):

First, you know (or can verify easily) that (the reminder of a*b by 3) is the reminder of (the reminder of a by 3)*(the reminder of b by 3) by 3

2^{201} = 2*(2^2)^{100}

The reminder of  2^2  by 3 being 1, we can say that the reminder of  (2^2)^{100}  by 3 is 1 as well (since  1^{100}=1 )

Therefore the reminder of  2^{201}  by 3 is 2.

If you substract 2, 5 or 8 to that number, you will find 0, -3 or -6, whom reminder by 3 are all 0.

scthakur · Answer

Hi,

I tried a lengthier option here.

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
and so on.

Thus, we see that the units digit repeats for every fifth power. i.e. 2^201 should have the unit digit as 2.

Now, any number above having unit digit as 2 can be divisible by 3 if you subtract either 2 or 5 or 8 from it (i.e. sum of digits is divisible by 3). Going by this analogy, 2^201 - n will be divisible by 3 for all three values of n.

Oski · Answer

I see that the unit digit repeats every 4 powers, doesn't it ? You have a 2 at the end of 2^1, 2^5, 2^9,...

Still: unit digit of 2^201 should be 2 (since 201 = 4*50 + 1)

But: 
I don't see how knowing just the digit unit can give you any information on the sum of digits.

And no, "any number having unit digit as 2 can be divisible by 3 if you subtract either 2 or 5 or 8 from it" is not true.

Try 12 for instance.

scthakur · Answer

Oski,

In the series of 2^n, the possible numbers with 2 as unit digit are 2, 32, 512, etc.

Now for number 2, 2-2 = 0, divisible by 3. 2-5 = -3 -> divisible by 3. 2-8 = -6 -> divisible by 3.

Similarly, 32-2 = 30 -> divisible by 3, 32-5 = 27 -> divisible by 3, 32-8 = 24 -> divisible by 3.

Similarly, 512-2 = 510 -> divisible by 3, 512-5 = 507 -> divisible by 3, 512-8 = 504 -> divisible by 3.

Thus, my analogy is that any number in this series with unit digit as 2 will be divisible by 3 if 2, 5 or 8 is subtracted from it.

Oski · Answer

I understand that.

But it is a BIG assumption.

possible value of n

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