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# Symbolism Q (got stuck!)

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Symbolism Q (got stuck!) [#permalink]  18 Sep 2003, 06:58
Hi everyone!

I'm new to this forum- what a great place!
Can't figure this thing out. I'd appreciate your help very mucho!

-If the operation @ is defined for all a[i] and [i]b by the equation
a @b[i] =(a^b)/3, then 2@(3@-1)=

A) 4
B) 2
C) -4/3
D)-2
E) -4

Thank you!
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Symbolism Q corrections [#permalink]  18 Sep 2003, 07:03
Pls disregard all those [i]'s -misprint.
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typo in question?? [#permalink]  18 Sep 2003, 12:31
well, i get

2@(1/9) which equals [2^(1/9)]/3 ...so i don't get any of the answer choices b/c this is not an integer nor negative...
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it was a typo [#permalink]  18 Sep 2003, 14:54
sorry...

The question reads: If the operation @ is defined for all a and b by the equation a@b=((a^2)*b)3, then 2@(3@-1)=

A) 4
B) 2
C) -4/3
D) -2
E) -4

thanks!
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Re: it was a typo [#permalink]  18 Sep 2003, 15:11
pete300 wrote:
sorry...

The question reads: If the operation @ is defined for all a and b by the equation a@b=((a^2)*b)3, then 2@(3@-1)=

A) 4
B) 2
C) -4/3
D) -2
E) -4

thanks!

I think the question should be a@b=((a^2)*b)/3 instead of a@b=((a^2)*b)3. If thats the case, u get the answer -4.
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you're right, it's division...
but could you be a little more specific with solution?
thanks!
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Formula: a@b=((a^2)*b)/3
Question: 2@(3@-1)= ?

work within the parenthesis first so solve (3@-1) first

(3@-1)= ((3^2)*-1)/3 = (9*-1)/3= -9/3=-3
now take -3 plug back into equation and solve the rest

2@(-3)=((2^2)*-3)/3 = (4*-3)/3= -12/3= -4

so -4 is the answer....this question is merely testing order of operations

remember PEMDAS
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thx! [#permalink]  23 Sep 2003, 18:36
Thanks a lot, guy123!
Good treat for my B.Day
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Hey Pete,
Many happy returns of the day !
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Thanks [#permalink]  02 Oct 2003, 15:33
Hey Soumala,

Thanks a lot! Be well.
Thanks   [#permalink] 02 Oct 2003, 15:33
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