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n could be the sum of consecutive integers r, s, and t [#permalink]
06 Mar 2005, 11:17
00:00
A
B
C
D
E
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
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n could be the sum of consecutive integers r, s, and t or the product of consecutive integers x, y, and z. What is the remainder when n is divided by 5? (1) When r is divided by 5, the remainder is 1 (2) When x is divided by 5, the remainder is 1[/i]
I go with E. because,
from (1), we do not know the order of rst. r is 5k+1 but s could be 5k+2, 5k or 5k-1. similarly, t could be 5k+2, 5k or 5k-1.
from (2) also, we do not know the order of xyz. x is 5k+1 but y could be 5k+2, 5k or 5k-1 and so does z.
from 1 and 2, we do not know. guys, correct, if any................
I admit that this is a very good question...
I naturally assumed that r <s<t or x<y<z and answered D, just because it is usually the GMAT format in this kind of problem, the smallest number first...I am curious about the OA and the feeling of the other members concerning this one
I personally believe that in the term "consecutive integers r, s, and t" we could assume that r, s and t are in order. Anybody has any real proof from an real ETS question that would disapprove this?
Re: DS - Divided by 5 [#permalink]
09 Mar 2005, 09:13
ketanm already had a very nice solution to this. Basically you should try to write n out as an algebra expression using the given conditions.
n=r+s+t=r+(r+1)+(r+2) (consecutive integer)
= 3r +3 (!)
Also n=x*y*z=x*(x+1)*(x+2) (2)
(1) When r is divided by 5, the remainder is 1
r=5k+1
substite into (1)
n=3(5k+1)+3=15k+6
When 15k is divided by 5, reminder is zero. When 6 is divided by 5, reminder is 1.
Sufficient
(2) When x is divided by 5, the remainder is 1
x=5k+1
n=(5k+1)(5k+2)(5k+3) (A)
=5k((5k+2)(5k+3))+(5k+2)(5k+3)
=5k((5k+2)(5k+3))+5k*(5k+3)+2*(5k+3)
=5k((5k+2)(5k+3))+5k*(5k+3)+2*5k+6 (B)
I'm writing this out so for people who do not have a intuitive feeling about the original equation (A) can know how to access it. You don't have to multiply them all out. From (B) you can see all the items before 6 is divisible by 5. And the reminder is 1 when 6 is divide by 5.
Therefore it is also sufficient.
This is why the answer would be (D).
If we are not sure if r,s,t and x,y,z are in order, then we would not know s=r+1, t=r+2, etc, then we would not be able to arrive to our previous conclusion. In this case the answer would be (E).
I agree it should be D, in the OA most examples for consecutive integers seem to be in the increase order not decreasing...on the test i would go for D
I used a simpler approach by using the old numbes rule for 5 saying that r and x had to be either 1 or 6 (or 11 or 16 etc) and went from their with plugging.
1+2+3 and 6+7+8 and 1*2*3 and 6*7*8 and you get the same answer
I used a simpler approach by using the old numbes rule for 5 saying that r and x had to be either 1 or 6 (or 11 or 16 etc) and went from their with plugging.
1+2+3 and 6+7+8 and 1*2*3 and 6*7*8 and you get the same answer
D.
Yes, very good approach. It would require you have pretty good mathematic intuition but it will save you a lot of time.
Re: DS - Divided by 5 [#permalink]
09 Mar 2005, 18:30
qhoc0010 wrote:
n could be the sum of consecutive integers r, s, and t or the product of consecutive integers x, y, and z. What is the remainder when n is divided by 5? (1) When r is divided by 5, the remainder is 1 (2) When x is divided by 5, the remainder is 1[/i]
Guys, I disagree with those who have assumed the order of the integers.
In DS, we are not supposed to make an assumption because it distorts the core concept of DS. If we statrt assume, then there will be no question such as DATA SUFFICIENCY because with the assumption all DS questions are solvable. In DS question we basically deal with whether the information provided in the question is sufficient. therefore, in such questions, we must have to work arround facts and figures provided.
this is my personal opinion. Seek more opinion from you guys and moderators.
gmatclubot
Re: DS - Divided by 5
[#permalink]
09 Mar 2005, 18:30