Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
M=O+P+Q, where O, P, and Q are consecutive positive integer; [#permalink]
12 Sep 2006, 19:09
0% (00:00) correct
0% (00:00) wrong based on 1 sessions
HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
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 ?
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.
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
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