GRE Test Question Homework: Is (n^*)^* = n?

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The discussion revolves around a GRE practice question involving the operation defined as n^* = 32 - n. The user is confused about the equality (n^*)^* = n, as stated in the answer key. It is clarified that the asterisk represents an operation that subtracts a number from 32, not an exponent. By applying this definition, the user realizes that evaluating m^* leads to the conclusion that the two values are indeed equal. This clarification resolves the initial confusion regarding the operation's meaning.
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Homework Statement



Hi,
I was doing some GRE practice tests and came across this question:

for all number n, n^*=32-n (apparently where the asterisk is an exponent)

They then give me two values, which are (n^*)^* and n, and I am to say whether choice 1 is a greater value than choice 2, choice 2 is a greater value than choice 1, the two values are equal, or there is not enough info to determine. The answer key says they are equal. This means given the above equation, (n^*)^* = n

I can't see how this is true... Anyone care to explain? Thanks
 
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Let n^*=32-n=m. Then, (n^*)^*=m^*. Can you evaluate m^*?
 
cristo said:
Let n^*=32-n=m. Then, (n^*)^*=m^*. Can you evaluate m^*?

I'm not sure. Is there some way you want me to rewrite this? m^*=(32-n)^*
 
Xkaliber said:
I'm not sure. Is there some way you want me to rewrite this? m^*=(32-n)^*

Yes. The star is shorthand for the operation that "subtracts a given number from 32." In the case of m^*, the given number is 32-n. What is the result when you apply * to that?
 
lol, that was easy. I had in my mind that * was some sort of exponential value, not a more general operator. Thanks
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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