# Puzzles

## Four integers

\(a\), \(b\), \(c\) and \(d\) are four positive (and non-zero) integers.

$$abcd+abc+bcd+cda+dab+ab+bc+cd+da+ac+bd\\+a+b+c+d=2009$$

What is the value of \(a+b+c+d\)?

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$$abcd+abc+bcd+cda+dab+ab+bc+cd+da+ac+bd\\+a+b+c+d=(a+1)(b+1)(c+1)(d+1)-1$$

So:

$$(a+1)(b+1)(c+1)(d+1)=2010\\=2\times 3\times 5\times 67$$

Therefore \(a+b+c+d=1+2+4+66=73\).

#### Extension

Which numbers could 2009 be replaced with so that the problem still has a unique solution?

## Times roamin'

What is the product of this series?

$$(x-a)(x-b)(x-c)...(x-z)$$

## x to the power of x again

Let \(y=x^{x^{x^{x^{...}}}}\) [\(x\) to the power of (\(x\) to the power of (\(x\) to the power of (\(x\) to the power of ...))) with an infinite number of \(x\)s]. What is \(\frac{dy}{dx}\)?

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\(y=x^{x^{x^{x^{...}}}}\) so \(y=x^y=e^{y\ln{x}}\).

By the chain and product rules, \(\frac{dy}{dx}=e^{y\ln{x}}(\frac{dy}{dx}\ln{x}+\frac{y}{x})\).

Rearranging, we get \(\frac{dy}{dx}=\frac{ye^{y\ln{x}}}{x(1-e^{y\ln{x}}\ln{x})}\).

This simplifies to \(\frac{dy}{dx}=\frac{x^{x^{x^{x^{...}}}}x^{x^{x^{x^{...}}}}}{x(1-x^{x^{x^{x^{...}}}}\ln{x})}\).

#### Extension

What would the graph of \(y=x^{x^{x^{x^{...}}}}\) look like?

## x to the power of x

If \(x^{x^{x^{x^{...}}}}\) [\(x\) to the power of (\(x\) to the power of (\(x\) to the power of (\(x\) to the power of ...))) with an infinite number of \(x\)s] is equal to 2, what is the value of \(x\)?