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.

-From PMO

No vote yet

2 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 \[ S_\mathsf{tot} \; = \; 2^{n-1}\sum_{j=1}^n j \; = \; n(n+1)2^{n-2} \]

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply