5
$\begingroup$

In my last question I somehow tricked myself, because I wanted to make a puzzle about a number that looks like a year. But the complicated solution I thought of turned out to be outperformed by a much simpler ansatz. Here is a puzzle that (hopefully) has my intended solution as optimum:

Make the number $439204$ by using an arbitrary number of $\Phi$s and

  • the operators $+$, $−$, $\cdot$, $/$ (including unary "$-$")
  • exponentiation
  • brackets $($ $)$.

The aim is to use as few $\Phi$s as possible.

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

$\endgroup$
2
  • $\begingroup$ Could you clarify whether we are allowed unary minus ? $\endgroup$
    – Gareth McCaughan
    Jan 21, 2018 at 23:46
  • $\begingroup$ Unary minus is allowed. I edited the question to clarify this. $\endgroup$
    – A. P.
    Jan 21, 2018 at 23:48

2 Answers 2

6
$\begingroup$

I can do

38 ... no, 18,

as follows.

Write $L_n$ for the $n$th Lucas number, so $L_0=2$, $L_1=1$, and $L_{n+2}=L_{n+1}+L_n$. It happens that $L_{27}=439204$. And we have $L_n=\phi^n+(-\phi)^{-n}$. And $27=3^3$.

Therefore,

$$\begin{eqnarray}439204 &=& \phi^{27}+(-\phi)^{-27} \\ &=& \phi^{27}-\phi^{-27} \\ &=& \phi^{\left(\left(\frac{\phi+\phi+\phi}{\phi}\right)^{\left(\frac{\phi+\phi+\phi}{\phi}\right)}\right)} - \phi^{-\left(\left(\frac{\phi+\phi+\phi}{\phi}\right)^{\left(\frac{\phi+\phi+\phi}{\phi}\right)}\right)}. \end{eqnarray}$$

$\endgroup$
5
  • $\begingroup$ In that case I think it should be possible in 18 $\endgroup$ Jan 21, 2018 at 23:43
  • $\begingroup$ Ah - you saw it too :) $\endgroup$ Jan 21, 2018 at 23:44
  • $\begingroup$ Yup, seconds after posting 38. $\endgroup$
    – Gareth McCaughan
    Jan 21, 2018 at 23:45
  • $\begingroup$ I really should've done what I was going to do when I first saw the number change... an oeis search oeis.org/search?q=439204&language=english&go=Search $\endgroup$ Jan 21, 2018 at 23:48
  • $\begingroup$ Yep :-). (What I actually did when I saw the new number was to look up "Lucas numbers". But I already knew about the nice formula for those.) $\endgroup$
    – Gareth McCaughan
    Jan 21, 2018 at 23:58
0
$\begingroup$

Using a similar method to my answer to the previous question...

 43  42

Since

$N = \frac{\phi + \phi + ...}{\phi}$
and
$439204 = 2^{(2^3)}\cdot 5\cdot 7^3+2\cdot (3^4+1)$

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

$\endgroup$

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.