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M=O+P+Q, where O, P, and Q are consecutive positive integer;

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Manager
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M=O+P+Q, where O, P, and Q are consecutive positive integer; [#permalink] New post 12 Sep 2006, 19:09
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A
B
C
D
E

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M=O+P+Q, where O, P, and Q are consecutive positive integer; M=R*S*T, where R, S and T are positive consecutive integers. What is the remainder when M is divided by 5?
1). When O is divided by 5, the remainder is 1
2). When R is divided by 5, the remainder is 1


Q: If O, P, and Q are consecutive positive integer, is it safe to assume O < P < Q ?
VP
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 [#permalink] New post 12 Sep 2006, 19:14
answer is D.

it does not matter whether they are consecutively increasing or not.
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 [#permalink] New post 12 Sep 2006, 19:58
tennis_ball wrote:
answer is D.

it does not matter whether they are consecutively increasing or not.

I think it matters.
Lets say if O = 6 , P = 5 and Q = 7 then remainder is 3
if O = 6 P = 7 and Q = 8 then remainder is 1.

But I think it is safe to assume O<P<Q.

St1:
Let O = x then
M = x+x+1 + x+2 = 3x+3
x/5 has a remainder of 5. 3x will have a remainder of 3 and 3x+3 will have a remainder of 1: SUFF

St2: Let R = x
Then M = x(x+1)(x+2) = x^3 + 3x^2 + 2x
x/5 has a remainder of 1. This means last digit of x is either 1 or 6.
So last digit of x^3 will be either 6 or 1. i.e Remainder = 1
Last digit of x^2 will be either 6 or 1. Last digit of 3x^2 will be either 8 or 3. i.e remainder = 3
Last digit of 2x will be 2 - i.e remainder = 2

Total remainder = 1+3+2 = 6
Hence final remainder will be 6-5 = 1: SUFF

Hence answer is D
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 [#permalink] New post 13 Sep 2006, 01:09
Think it is E)
O-5,Q-6,R-7 them M=18 rem-3
O-6,Q-5,R-4 then M=15 rem is 0
A Is NOT suff

B)
Numbers may be 5,6,7 where the product is divisible by 5 without reminder or 6,7,8 where there will be a remainder
B is NOT suff

E should be it

So it is not safe to assume anything, IMO
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 [#permalink] New post 13 Sep 2006, 06:21
Does anyone know for sure whether we can assume an increasing order?
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 [#permalink] New post 13 Sep 2006, 07:16
WHAT is the OA? then we know whether we can assume.
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 [#permalink] New post 13 Sep 2006, 13:04
tennis_ball wrote:
WHAT is the OA? then we know whether we can assume.



We can assume that they are ascending order- that's what consecutive means!
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 [#permalink] New post 15 Sep 2006, 00:27
kevin am lost here... as the numbers are consecutive we still can never know the order

what am i missing here

Plz Help
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 [#permalink] New post 15 Sep 2006, 05:56
This can be E or D depending upon whether we consider O < P < Q.

PS,
On the GMAT can we assume O < P < Q? Or is it usually specified?

Thanks/
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Re: DS:consecutive integers [#permalink] New post 15 Sep 2006, 07:41
I found a solution here
C it is
(1) alone we have O divided by 5 remainder is 1 so O,P,Q divided by 5 must have remainder is one of these sets (0,1,2);(1,2,3);(4,0,1).So M diveded by 5 could have remainder of 3 or 1(1+2+3=6) or 0. This tells us nothing
(2) alone R,S,T divided by 5 must have remainder is one of these sets (0,1,2);(1,2,3);(4,0,1). So M diveded by 5 could have remainder of 2 or 1 or 4 . This also tells nothing
But(1)(2) together we can say that M divided by 5 must have remainder of 1
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 [#permalink] New post 15 Sep 2006, 08:44
I'd say D:

1)reminders (o,p,q) = (1,2,3) 1+2+3=6 6/5 = reminder 1

2)reminders (r,s,t) = (1,2,3) 1*2*3=6 6/5 = reminder 1
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 [#permalink] New post 15 Sep 2006, 08:55
I say this E..

M=O+P+Q; M=R*S*T

we are told (o, p, q) (r, s, t) they are consecutive numbers

(1)when o is divided by 5 remainder is 1

so say o is 6, then p=7, q=8...remainder will be 1

however if o=6, p=5, q=4; then remainder will be 0....

Insuff

(2) same thing if R=6, s=7, t=8 remainder is given..however if r=6, s=5 and t=4..then remainder is 0...

together its insuff too...

E it is...

if however it is an ascending order...then its a D..
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 [#permalink] New post 15 Sep 2006, 10:06
D for me

From stem:

m = x+(x+1)+(x+2) = 3x+3 so m is divisible by 3

also

m = y.(y+1).(y+2)...so m is divisible by 6 ok...

From 1:
O=5k+1
So O+P+Q=15k+6 which will leave a remainder of 1

Suff

From 2:
R=5k+1
So RST=(5k+1)(5k+2)(5k+3) = 125k^3 + 150k^2 +55k + 6...again remainder is alway 1...SUFF
  [#permalink] 15 Sep 2006, 10:06
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M=O+P+Q, where O, P, and Q are consecutive positive integer;

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