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# Is (N^x)-2N divisible by 3 ? a. x=3 b. N=5

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Is (N^x)-2N divisible by 3 ? a. x=3 b. N=5 [#permalink]

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05 Apr 2006, 09:06
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Is (N^x)-2N divisible by 3 ?
a. x=3
b. N=5
Manager
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05 Apr 2006, 09:33
Statement 1: Insufficient, because we don't know anything about x

Statement 2: Insufficient. 5^2 - 10 = 15 which is divisible by 3, but 5^3 -10 = 115, which is not divisible by 3.

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05 Apr 2006, 18:21
laxieqv wrote:
Is (N^x)-2N divisible by 3 ?
a. x=3
b. N=5

seems C works.

from i, if n= 2, [(N^x)-2N] is not divisible by 3. if n= 3, [(N^x)-2N] is divisible by 3.
from ii, if x= 2, [(N^x)-2N] is divisible by 3. if x= 3, [(N^x)-2N] is not divisible by 3.

so C.
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05 Apr 2006, 18:35
a) x = 3,

Then we have N^3 - 2N = N(N^2-2)

If n = 5, 5(25-2) --> Not divisible by 3
If n = 3, 3(7) --> Divisible by 3

Insufficient.

b) N = 5

Then we have 5^x -10
If x = 2, 5^x-10 = 15 ---> divisible by 3
If x = 3, 5^x-10 = 115 ---> not divisible by 3

Insufficient.

Using a) and b)

N^x-2N = 5^3-10 --> can decide if it's divible by 3 or not divisible by 3

Sufficient.

Ans C
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06 Apr 2006, 02:55
hik, I'm sorry, everyone, the original question is like this:
Is 5(N^x)-2N divisible by 3 ?
a. x=3
b. N=5

I'm really sorry for the error
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06 Apr 2006, 04:17
laxieqv wrote:
hik, I'm sorry, everyone, the original question is like this:
Is 5(N^x)-2N divisible by 3 ?
a. x=3
b. N=5

I'm really sorry for the error

Well, then it's A; nice construction.

5(N^x)-2N

statement 1)

5*N^3 - 2N

Now we can plug in numbers:

5 - 2 = 3
40 - 4 = 36
135 - 6 = 129
320 - 8 = 312
625 - 10 = 615
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06 Apr 2006, 04:32
I agree with the new eqn, it shud be A
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06 Apr 2006, 06:23
Is there a "rule" here? or just plugging numbers shows the pattern?
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06 Apr 2006, 07:17

the result that i get by placing both N and X in the equation is not divisible by 3
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06 Apr 2006, 07:25
the answer is C now that i read the thread and discovered that the posted question was wrong

A) X=3 => 5(3^3)-2N
45-2N.
Now, 45 is divisible by 3 but 2N?? huh ... N can be 1 so it makes indivisible
N=2 makes indivisible .. but N=3 makes it divisible(39) so I can't say from A

B) N=5 => 5(5^x)-10
5(5^x-2)
let take x = 1 => 5(5^1 - 2) => 13 not divisible by 3
x = 2 => 5(5^2 - 2) => 38 not divisible
x = 3 => not divisible
x= 4 => divisible!!

I can't really say from B as well !!

plugin both values of x and N i get 615 => Divisible!! so I need both conditions!!
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06 Apr 2006, 07:26
Any other shortcut to deal with these kinda guys??

Man o man .. I keen thinkin if my way of doing things is right?
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06 Apr 2006, 08:54
C for first problem

&

A for second problem
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07 Apr 2006, 09:15
laxieqv wrote:
Is 5(N^x)-2N divisible by 3 ?
a. x=3
b. N=5

(1) x=3 ---> 5( N^3) - 2N = 3N^3 + 2N^3 - 2N
We have: 3N^3 is divisible by 3, now consider
2N^3-2N = 2 N( N-1) ( N+1) . The product of three consecutive number must be divisible by 3 ---> 2N^3- 2N is divisible by 3

---> 5 N^3 - 2N is divisible by 3 --> answer to the question : YES!
--->suff

(2) not suff as proved by others.

---> A it is.

BUT, it's the case in which N is integer. How's if N is not an integer?!!

Anyways, A is the OA.
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07 Apr 2006, 11:17
Agree

C for the first equation

A for the second equation
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25 Apr 2006, 04:22
I too have the same question. What if N is not an integer?
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26 Apr 2006, 17:11
laxieqv, you said that the product of 3 consecutive integers is always divisible by 3.
what if N=0 ?then we have (N-1)N(N+1)=0
if N=1, we have: (N-1)N(N+1)=0
If N=-1, which is also an integer, (N-1)N(N+1)=0

I know that the sum of 3 consecutive integers will always be divisible by 3, but it's not so obvious that the product of 3 consecutive integers will be too, unless we exclude the 0 among our consecutive integers.
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26 Apr 2006, 18:23
laxieqv wrote:
laxieqv wrote:
Is 5(N^x)-2N divisible by 3 ?
a. x=3
b. N=5

(1) x=3 ---> 5( N^3) - 2N = 3N^3 + 2N^3 - 2N
We have: 3N^3 is divisible by 3, now consider
2N^3-2N = 2 N( N-1) ( N+1) . The product of three consecutive number must be divisible by 3 ---> 2N^3- 2N is divisible by 3

---> 5 N^3 - 2N is divisible by 3 --> answer to the question : YES!
--->suff

(2) not suff as proved by others.

---> A it is.

BUT, it's the case in which N is integer. How's if N is not an integer?!!

Anyways, A is the OA.

That OA seems to assume N and x would take on integer values and not fractional. But nothing in the stem suggests that this should be the case....
Re: DS divisibility   [#permalink] 26 Apr 2006, 18:23
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