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# n could be the sum of consecutive integers r, s, and t

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Director
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n could be the sum of consecutive integers r, s, and t [#permalink]

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06 Mar 2005, 11:17
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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]
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06 Mar 2005, 11:28
Consider (1)

n = r + s + t = r + (r +1) + (r + 2) [since r,s,t are consecutive integers]
= 3(r+1)

When r is divided by 5, remainder is 1, hence r = 5p + 1

n = 3(5p+1 + 1) = 3(5p+2) = 15p + 6

When n will be divided by 5, it should give remainder as '1'. Hence (1) alone is sufficient

Consider (2)

n = x * y * z = x * (x+1) * (x+2) [since x,y,z are consecutive integers]

When x is divided by 5, remainder is 1, hence x = 5p + 1

x/5 --> Remainder is 1
(x+1)/5 --> Remainder is 2
(x+2)/5 --> Remainder is 3

Hence remainder when 'n = x * (x+1) * (x+2)' is divided by 5 is

1 * 2 * 3 = 6 /5 ==> 1 remainder

Hence (2) alone is sufficient

Both are independently sufficient

Ketan
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06 Mar 2005, 12:22
Can you assume this: r <s<t or x<y<z ?
Or can you assume this: n = rst in (1) but not n = xyz (They are all different)?
VP
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06 Mar 2005, 15:54
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................
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06 Mar 2005, 16:04
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
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06 Mar 2005, 17:15
When one says consecutive integers r,s,t....I think we r supposed to read it as r<s<t.....just my opinion....anyone with official definition.
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09 Mar 2005, 03:17
I will follow the similar logic as above. let s not be trickier than GMAT s writers.
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09 Mar 2005, 06:51
Yeah, OA is (D)
But I think (E) must be correct.
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09 Mar 2005, 07:31
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?
Director
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09 Mar 2005, 08:15
it is asking for the remainder so how are we saying if 1 is sufficient or 2 is sufficient. Honghu can u please explain this one?
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Re: DS - Divided by 5 [#permalink]

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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).
Director
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09 Mar 2005, 09:54
great explanation. thanks...once again..
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09 Mar 2005, 11:29
D it is

I just picked numbers assuming r is the smallest and x is the smallest

for statement 1
if r divided by 5 remainder is 1, u can pick r=6 for example you
6+7+8=21=N

N/5 remainder is 1
we can check by adding 10 to each which is really like adding 30 to the above! you still get 1 as remainder

statement 2
I just foucesed on the unit digit, it truns out to be 0 and therefore the remainder is 0

sufficient

D it is..
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09 Mar 2005, 12:03
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
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09 Mar 2005, 13:17
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.
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09 Mar 2005, 14:12
It is D. Good question. And a very good explanation by KetanM
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09 Mar 2005, 14:40
Caspace wrote:
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.
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09 Mar 2005, 15:05
one more D. great explanations.
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Re: DS - Divided by 5 [#permalink]

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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.
Re: DS - Divided by 5   [#permalink] 09 Mar 2005, 18:30
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