Problems:

If x + y + xy = 1, where x and y are non-zero real numbers, what is xy + 1/xy - y/x - x/y? (The answer is 4 but needing simple algebraic manipulation.)

The quartic polynomial P(x) satisfies P(1) = 0 and attains its maximum value of 3 at both x = 2 and x = 3. Compute P(5). More appreciated if the solution does not require calculus at least.

Let S(X) be the sum of elements of a nonempty finite set X, where X is a set of numbers. Calculate the sum of all numbers S(X) where X ranges over all nonempty subsets of the set {1, 2, 3, ..., 16}. Please show quick method.

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## Comments

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TopNewestThe polynomial is $P(x) = 3 - \tfrac34(x-2)^2(x-3)^2$, and so $P(5) = -24$.

For simplicity, let $S(\varnothing) = 0$, so that $X$ can range over all subsets of $\{1,2,\ldots,n\}$. Each number $1 \le j \le n$ occurs in precisely half of the $2^n$ subsets of $\{1,2,\ldots,n\}$, and so contributes a total of $j \times 2^{n-1}$ to the total sum $S_\mathsf{total} \; = \; \sum_{X \subseteq\{1,2,\ldots,n\}} S(X)$ Thus $S_\mathsf{tot} \; = \; 2^{n-1}\sum_{j=1}^n j \; = \; n(n+1)2^{n-2}$

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Hello.. I just noticed that For x = 1, the value is not 0... contradicting the given...

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So, OK, if $x=1$ then $y=0$. This just means that for $x$ and $y$ to be nonzero

andsatisfy the identity $x+y+xy=1$, they also have to be not equal to $1$. That just means that there is an additional "hidden" restriction on the possible values of $x$ and $y$, but nothing more. If you think about it, $x$ and $y$ cannot be equal to $-1$, either.Log in to reply

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$\tfrac34$ instead of $\tfrac14$.

Good point. I have corrected the solution, which neededLog in to reply