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If r☉s = rs + r + s, then for what value of s is r

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Manager
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If r☉s = rs + r + s, then for what value of s is r [#permalink]

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New post 27 Nov 2007, 16:22
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If r☉s = rs + r + s, then for what value of s is r ☉s equal to r for all values of r?
(A) –1
(B) 0
(C) 1
(D) 1/(r+1)
(E) r

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Re: Symbolism [#permalink]

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New post 27 Nov 2007, 16:51
yogachgolf wrote:
If r☉s = rs + r + s, then for what value of s is r ☉s equal to r for all values of r?
(A) –1
(B) 0
(C) 1
(D) 1/(r+1)
(E) r



rs+r+s=r

I get B. Lets say r= 5, -2, or 0. (mix up the values)

if r*s=rs+r+s. and s=0. then try rs+r+s=r. Which values work?

A: 5(-1)+5+-1=5? No Used 5 for r
B: -2(0)+-2+0=-2? Yes Used -2 for r
C: -2+-2+1=-2? No Used -2 for r
D: 5/6 +5 +1/6 =5? NO used 5 for r
E: No need for values as we can see that r^2+2r is not equal to r if r is anything but 0.


There is another way that I just found that is much easier.

rs+r+s=r ---> s(r+1)+r=r ---> s(r+1)=0 s=0 r=-1.

B.

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New post 16 Dec 2007, 09:57
i am getting A.

r?s=rs+r+r=r(s+2)

you want r?s=r, then you should set s+2 to give 1, so that you are only left with r.

s+2=1, s=-1.

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Concentration: Entrepreneurship, Other
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New post 16 Dec 2007, 10:38
pmenon wrote:
r?s=rs+r+r=r(s+2)

In original question: r?s=rs+r+s

rs+r+s=r ==> s(r+1)=0 is correct for any r only at s=0.

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  [#permalink] 16 Dec 2007, 10:38
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