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M=O+P+Q, where O, P, and Q are consecutive positive integer; [#permalink]
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12 Sep 2006, 20:09
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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
Joined: 25 Jun 2006
Posts: 1166

answer is D.
it does not matter whether they are consecutively increasing or not.



CEO
Joined: 20 Nov 2005
Posts: 2894
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD  Class of 2008

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 65 = 1: SUFF
Hence answer is D
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Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

Think it is E)
O5,Q6,R7 them M=18 rem3
O6,Q5,R4 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



Intern
Joined: 04 May 2006
Posts: 40

Does anyone know for sure whether we can assume an increasing order?



VP
Joined: 25 Jun 2006
Posts: 1166

WHAT is the OA? then we know whether we can assume.



GMAT Instructor
Joined: 04 Jul 2006
Posts: 1262
Location: Madrid

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!



SVP
Joined: 05 Jul 2006
Posts: 1747

kevin am lost here... as the numbers are consecutive we still can never know the order
what am i missing here
Plz Help



VP
Joined: 02 Jun 2006
Posts: 1260

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/



Manager
Joined: 01 Jun 2006
Posts: 139

Re: DS:consecutive integers [#permalink]
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15 Sep 2006, 08:41
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



Senior Manager
Joined: 15 Jul 2006
Posts: 381

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



Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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..



Director
Joined: 28 Dec 2005
Posts: 752

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










