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N = P + Q + R and P, Q, R is the consecutive integers. N = X [#permalink]
11 Nov 2005, 08:16

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A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 1 sessions

N = P + Q + R and P, Q, R is the consecutive integers.
N = X * Y * Z and X, Y, Z is the consecutive integers, too.
What is the remainder of N to be divided by 5 ?

(1) The remainder of P to be divided by 5 is 1
(2) The remainder of X to be divided by 5 is 1 _________________

N = P + Q + R and P, Q, R is the consecutive integers. N = X * Y * Z and X, Y, Z is the consecutive integers, too. What is the remainder of N to be divided by 5 ?

(1) The remainder of P to be divided by 5 is 1 (2) The remainder of X to be divided by 5 is 1

It's E.
Consider two cases:
case 1: N= 1+2+3 = 1*2*3=6
in this case, P=X=1 which divided by 5 has remainder of 1. (exactly what's from the two statements)
N=6 ---> the remainder of N to be divided by 5 is 1.

case 2: N= 1+0+-1 = 1*0* (-1) = 0
in this case, P=X= 1which divided by 5 has remainder of 1 ( exactly the two statements)
N= 0 ---> the remainder of N to be divided by 5 is 0.

=> we can't find exactly what the remainder is even if we combine two statements.

i would go with D.
N = P + Q + R and P, Q, R is the consecutive integers.
N = X * Y * Z and X, Y, Z is the consecutive integers, too.
What is the remainder of N to be divided by 5 ?

(1) The remainder of P to be divided by 5 is 1
take any value of P where remainder is 1 6,16,11,21...
and multiply the each next to consicutive integer like
6+7+8/5=Remainder 1
16+17+18/5=Remainder 1
21+22+23/5=Remainder 1

how do you know P is the smallest consecutive number?
for all I know P could 6, Q=5 and R =4

?? did you read my post??

cool_jonny009 wrote:

i would go with D. N = P + Q + R and P, Q, R is the consecutive integers. N = X * Y * Z and X, Y, Z is the consecutive integers, too. What is the remainder of N to be divided by 5 ?

(1) The remainder of P to be divided by 5 is 1 take any value of P where remainder is 1 6,16,11,21... and multiply the each next to consicutive integer like 6+7+8/5=Remainder 1 16+17+18/5=Remainder 1 21+22+23/5=Remainder 1

I also got D assuming that the number are in order. In questions when they say consecutive, I think, I have always come across ordered sequence (ascending)

P = 1, Q = 2, R = 3 -> N = (1+2+3)/5 -> Remainder is 1
P = 6 Q = 5, R = 7 -> N = (6+5+7)/5 -> Remainder is 3

(2) Insufficient

Let X = b]1[/b], 6, 11, 16 ......

X = 1, Y = 2, Z = 3 -> N = (1*2*3)/5 -> Remainder is 1
X = 1, Y = 0, Z = -1 -> N = (1*0*-1)/5 -> Remainder is 0

I believe this question is testing whether it is correct to assume P, Q and R and X, Y and Z are in order. I don't think this is valid assumption. If the assumption was valid, then this would be an easy question.

we don't know the order, first of all, secondly even if we know the order, we don't the whether the consequtive integers are positive or negative. _________________

Throw my vote with D too. When it says x,y,z are consecutive integers I always think I can assume x=n-1, y=n and z=n+1. In other words, I'm assuming their orders. It's a good point to notice though. It would be nice for us to see if it is a official question and if the anwer does not confirm our assumption. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

The basic requirement for remainder of a number X when divide by n is:

X = n.q + r ........ (where q is an integer quotient and r the remainder)

So, if X is negative:

-X = -n.q - r

eg. -6/5

-6 = -1(5) - 1

X = n.q + r

here the property of r is:

0<= r <q

in this example, -1 < 0

I guess, we should leave as out-of-scope as I did not find any problem where we need to think about this. ETS only talks about remainer in the context of positive integers.
The only think to remember is:
q <> 0 _________________

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