13
$\begingroup$

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: $\sqrt[n]{x} = x^{\frac{1}{n}}$.

$\endgroup$
  • $\begingroup$ I see from your example $x^{\frac{1}{n}}$ that you allow the number $1$. I assume that $x$ and $n$ has to be created from Φ's? $\endgroup$ – Stig Hemmer Jan 22 '18 at 11:12
  • $\begingroup$ @StigHemmer $\frac\Phi\Phi=1$ ;-) also no, I don't think $1$ is allowed, it just explains how roots should be represented. $\endgroup$ – EKons Jan 22 '18 at 12:57
  • $\begingroup$ And although $\Phi$ itself is an interesting number, none of the solutions posted use any of its properties besides $\Phi \neq 0$. $\endgroup$ – ivan100sic Jan 22 '18 at 16:02
  • $\begingroup$ @ivan100sic When I created this puzzle I acutally thought of a way that would use the properties of $\Phi$, but unfortunately the simple solutions outperform it regarding the number of $\Phi$s used. But the linked question doesn't have this loophole. $\endgroup$ – A. P. Jan 22 '18 at 16:18
  • $\begingroup$ Fun fact: $$2008=7^4-7^3-7^2-1.$$ That is pretty cool. $\endgroup$ – Feeds Jul 31 '18 at 11:34
10
$\begingroup$

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) $

$\endgroup$
  • 3
    $\begingroup$ Not that I could come up with anything better but it’s slightly disappointing that the golden ratio could really be replaced with “any positive number”. Not just for this answer but all/most answers $\endgroup$ – ferret Jan 22 '18 at 16:31
12
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Now I feel silly. Well done! $\endgroup$ – hexomino Jan 21 '18 at 22:00
  • 1
    $\begingroup$ The one who should feel silly is me. I should have checked for simpler solutions than mine. Hopefully you still had fun. Here is a follow-up that is very unlikely to be solved that easy. $\endgroup$ – A. P. Jan 21 '18 at 22:39
7
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ I literally just posted a comment with the same sentiment. +1 $\endgroup$ – ferret Jan 22 '18 at 16:39
  • $\begingroup$ The "at the loss of accuracy" is akin to saying "if we can use a ceiling function", which was explicitly outlawed, so I don't think this should then go on to say "my current final score of...". $\endgroup$ – Jonathan Allan Jan 25 '18 at 18:38
  • $\begingroup$ @JonathanAllan Disagree, because it's referring to the accuracy of the result, not an intermediate step. If I tried to use something like ceiling(phi) in the calculation I'd say that is against the rules. $\endgroup$ – corsiKa Jan 25 '18 at 20:41
  • $\begingroup$ Huh? When push comes to shove you did not make 2008 using that many phis, you made something close which would require an unavailable operation at the end. You could say "at the expense of acuraccy I can do it in one" (or zero maybe!). $\endgroup$ – Jonathan Allan Jan 25 '18 at 20:57
  • $\begingroup$ @JonathanAllan If you want to feel that way, sure. I could add a few more powers of phi and make it even closer but it still wouldn't be exact. I would still argue that this is in the spirit of the competition. And you can't deny the elegance of it - I mean, c'mon, base phi is pretty darned cool. So go pee in someone else's soup... $\endgroup$ – corsiKa Jan 25 '18 at 23:34
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Guessing you used a script for that, but still.. impressive! $\endgroup$ – Beastly Gerbil Jan 21 '18 at 21:43
  • $\begingroup$ Actually, I didn't need to. Everything I've written here can be derived on pen and paper, in just a few lines, from the equation $\Phi^2 = \Phi + 1$. I'm not even sure this answer is optimal. $\endgroup$ – hexomino Jan 21 '18 at 21:53
  • $\begingroup$ Oh ok thats even more impressive! $\endgroup$ – Beastly Gerbil Jan 21 '18 at 21:58
  • $\begingroup$ Reduces to 40 by using my version of 2. $\endgroup$ – Jonathan Allan Jan 21 '18 at 21:58
  • $\begingroup$ ...or 28 using my newer method. $\endgroup$ – Jonathan Allan Jan 21 '18 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.