Make 2008 from Φ (Golden Ratio)

Question

I know that questions of the form "Use [arbitrary set of numbers] to make 2018" seem to be not very well recieved in the last days. That's why this puzzle is about making 2008.

But instead of using some randomly chosen numbers, you will use the most beautiful number I can imagine: $\Phi = \tfrac{1}{2} \left( 1 + \sqrt{5} \right)$ – the Golden Ratio.

To make 2008 you may use

• the operators $+$, $-$, $\cdot$, $/$
• exponentiation
• brackets $($ $)$

and any number of instances of $\Phi$. The answer with the least number of $\Phi$s will be accepted.

You may not use operators or functions other than in this list, so don't even ask for rounding ($\lfloor$ $\rfloor$, $\lceil$ $\rceil$) or logarithm ($\log_a (x)$). If you want to use roots, this is ok as long as you express them as exponent: $\sqrt[n]{x} = x^{\frac{1}{n}}$.

Answer 1

Score:

21

The solution:

$ \frac{x+x}{x} \cdot ( \frac{x+x+x+x}{x} + (\frac{x}{x} + (\frac{x+x+x}{x})^{ \frac{x+x}{x} })^ {\frac{x+x+x}{x}})$ where $x = \phi $.

Based on

$ 2 \cdot (4 + (1 + 3^2)^3) $

Answer 2

Update...

 25  23

Using:

$N = \frac{\phi + \phi + ...}{\phi}$
and simple substitution into:
$2008 = 2048 - 40 = 8 \cdot (256 - 5) = 2^3 \cdot (2^{(2^3)} - 5)$:

$\frac{\phi + \phi}{\phi} ^{\frac{\phi + \phi + \phi}{\phi}} \cdot (\frac{\phi + \phi}{\phi} ^{(\frac{\phi + \phi}{\phi} ^{\frac{\phi + \phi + \phi}{\phi}})} - \frac{\phi + \phi + \phi + \phi + \phi}{\phi})$

---

Previous attempt (less "golfed")...

38 $\phi$s...

Using:

$1 = \frac{\phi}{\phi}$
and simple substitution into:
$2008 = 2000 + 8 = 2^4 \cdot 5^3 + 2^3$:
$(\frac{\phi}{\phi}+\frac{\phi}{\phi})^{(\frac{\phi}{\phi}+\frac{\phi}{\phi}+\frac{\phi}{\phi}+\frac{\phi}{\phi})} \cdot (\frac{\phi}{\phi}+\frac{\phi}{\phi}+\frac{\phi}{\phi}+\frac{\phi}{\phi}+\frac{\phi}{\phi})^{(\frac{\phi}{\phi}+\frac{\phi}{\phi}+\frac{\phi}{\phi})} + (\frac{\phi}{\phi}+\frac{\phi}{\phi})^{(\frac{\phi}{\phi}+\frac{\phi}{\phi}+\frac{\phi}{\phi})}$

Answer 3

For everybody who wants to laugh at my unnecessarily complicated answer (Beware! This will spoiler the right answer to this follow-up question.):

$2008 = 2207 - 199$,
where $2207 = L(16)$ and $199 = L(11)$ – the $16$th and $11$th Lucas numbers, respectively. The $n$th Lucas number can be expressed by $\Phi$ as $$L(n) = \Phi^n + \left( - \Phi \right)^{-n}.$$ With this $$\begin{align} 2008 &= \overbrace{\left( \Phi^{16} + \Phi^{-16} \right)}^{2207} - \overbrace{\left( \Phi^{11} - \Phi^{-11} \right)}^{199} = \Phi^{10} \left( \Phi^6 - \Phi \right) + \Phi^{-12} \left( \Phi + \frac{\Phi}{\Phi^5} \right) \\ &= \Phi^{\frac{\Phi + \Phi}{\Phi} \cdot \frac{\Phi + \Phi + \Phi + \Phi + \Phi}{\Phi}} \left( \Phi \Phi \Phi \Phi \Phi \Phi - \Phi \right) + \Phi^{- \frac{\Phi + \Phi + \Phi + \Phi}{\Phi} \cdot \frac{\Phi + \Phi + \Phi}{\Phi}} \left( \Phi + \frac{\Phi}{\Phi \Phi \Phi \Phi \Phi} \right). \end{align}$$ Unfortunately this takes 34 $\Phi$s – much more than the accepted answer.

Answer 4

Our first step is to construct some identities:

$1 = \Phi * \Phi - \Phi$ which has 3 $\Phi$s

As my kindergarten daughter can tell me,

$2 = 1 + 1 = \Phi * \Phi + \Phi * \Phi - \Phi - \Phi$ which has 6 $\Phi$s

Expanding further:

$5 = 2 * 2 + 1 = (\Phi * \Phi + \Phi * \Phi - \Phi - \Phi) * (\Phi * \Phi + \Phi * \Phi - \Phi - \Phi) + \Phi * \Phi - \Phi$ which has 15 $\Phi$s.

And throw it into the soup:

$2008 = ((5*5*5*2)+1)*2*2*2 = ((((\Phi * \Phi + \Phi * \Phi - \Phi - \Phi) * (\Phi * \Phi + \Phi * \Phi - \Phi - \Phi) + \Phi * \Phi - \Phi)*((\Phi * \Phi + \Phi * \Phi - \Phi - \Phi) * (\Phi * \Phi + \Phi * \Phi - \Phi - \Phi) + \Phi * \Phi - \Phi)*((\Phi * \Phi + \Phi * \Phi - \Phi - \Phi) * (\Phi * \Phi + \Phi * \Phi - \Phi - \Phi) + \Phi * \Phi - \Phi)*(\Phi * \Phi + \Phi * \Phi - \Phi - \Phi))+\Phi * \Phi - \Phi)*(\Phi * \Phi + \Phi * \Phi - \Phi - \Phi)*(\Phi * \Phi + \Phi * \Phi - \Phi - \Phi)*(\Phi * \Phi + \Phi * \Phi - \Phi - \Phi)$

For a total of

72 $\Phi$s

using only the properties of $\Phi$ - none of these operations would apply to any other real number.

For using way less $\Phi$ instances at the expense of accuracy, we can express the number 2008

In base phi as 101 001 000 000 000

Which expands to

$$ \frac{\Phi^{\Phi^{\Phi^{\Phi{^\Phi}}}}}{\Phi} + ({\frac{\Phi^{\Phi^{\Phi}}}{\Phi}})^{\Phi^{\Phi}} + ({\Phi^{\Phi^{\Phi}}}*{\Phi})^{\Phi} \approx 2008 $$

In other words

$ \Phi^{15} + \Phi^{13} + \Phi^{10} \approx 1364.00 + 521.00 + 122.99 \approx 2007.99 \approx 2008 $

Now this uses

17 $\Phi$ -

we can attempt to reduce this by factoring out

$\Phi^{9}$,

like so

$$ (\Phi*{\Phi^{\Phi^{\Phi^{\Phi}}}}) ( ({\Phi*\Phi*\Phi})^\Phi + {\Phi^{\Phi^{\Phi}}} + \Phi ) $$

Which is the same result but only uses my current final score of

13

Answer 5

As an opening bid, I think I can do it using

$44$ $\Phi$s

which is

$\left(\Phi.\Phi - \left( \frac{\Phi}{\Phi.\Phi}\right)\right)^\left(\left(\Phi.\Phi - \left( \frac{\Phi}{\Phi.\Phi}\right)\right)^\left( \Phi.\Phi + \left(\frac{\Phi}{\Phi.\Phi.\Phi} \right)\right)+\left( \Phi.\Phi + \left(\frac{\Phi}{\Phi.\Phi.\Phi} \right)\right)\right)$
$ -\left(\left(\Phi.\Phi - \left( \frac{\Phi}{\Phi.\Phi}\right)\right)^\left( \Phi.\Phi + \left(\frac{\Phi}{\Phi.\Phi.\Phi} \right)\right) * \left[\left(\Phi.\Phi - \left( \frac{\Phi}{\Phi.\Phi}\right)\right)+\left( \Phi.\Phi + \left(\frac{\Phi}{\Phi.\Phi.\Phi} \right)\right)\right]\right) = 2008$

Idea:

We use the following two equations
$\left(\Phi.\Phi - \left( \frac{\Phi}{\Phi.\Phi}\right)\right) = 2$
$\left( \Phi.\Phi + \left(\frac{\Phi}{\Phi.\Phi.\Phi} \right)\right) = 3$

together with

$2^\left(2^3+3\right) - (2^3 * (2+3)) = 2008$

although I think it should be possible with fewer.