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Is (N^x)-2N divisible by 3 ? a. x=3 b. N=5 [#permalink]
 05 Apr 2006, 09:06
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Is (N^x)-2N divisible by 3 ?
a. x=3
b. N=5
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Manager
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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.
Answer is C.
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Re: DS divisibility [#permalink]
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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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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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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Director
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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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VP
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I agree with the new eqn, it shud be A
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Manager
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Is there a "rule" here? or just plugging numbers shows the pattern?
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Senior Manager
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Answer is E
the result that i get by placing both N and X in the equation is not divisible by 3
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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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Senior Manager
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Any other shortcut to deal with these kinda guys??
i take about 2 mins to solve such answers 
Man o man .. I keen thinkin if my way of doing things is right?
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C for first problem 
&
A for second problem
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Re: DS divisibility [#permalink]
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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Agree
C for the first equation
A for the second equation 
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Manager
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I too have the same question. What if N is not an integer?
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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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Re: DS divisibility [#permalink]
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....
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Re: DS divisibility
[#permalink]
26 Apr 2006, 18:23
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