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I decided to ask myself how many different images my laptop's screen could display. I came up with (number of colors)^(number of pixels) so assuming 32768 colors I'm trying to get my head around the number, but I have a feeling it's too big to actually calculate.

Am I right that it's too big to calculate? If not, then how? If so then how would you approach grasping the magnitude?

Update: I realized a simpler way to get the same number is 2^(number of bits of video RAM) or "all the possible configurations of video RAM" - correct me if I'm wrong.

Aaron Anodide
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    That’s roughly $10^{4.5\times10^6}$, which is a $1$ followed by four and a half million zeroes. – Brian M. Scott Mar 21 '13 at 20:02
  • thanks, it's disturbing how much I forget from the days of going to school – Aaron Anodide Mar 21 '13 at 20:04
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    24-bit graphics allows for $2^{24} = 16777216$ colors, not $32768$. – Thomas Andrews Mar 21 '13 at 20:06
  • thanks for the corrections - i'll think a little longer before posting next time.. i appreciate the help – Aaron Anodide Mar 21 '13 at 20:07
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    However, the vast majority of these images will just be noise. – Raskolnikov Mar 21 '13 at 20:12
  • I think that means that there are more than $8*10^{7579366}$ and there are only $10^80$ atoms in the visible universe. – kaine Mar 21 '13 at 20:19
  • Yet it's easy to calculate the right-most digit! – DJohnM Mar 21 '13 at 20:44
  • Somewhere in that collection of images is a proof (or counterexample) to the Riemann hypothesis, all the unwritten music and unwritten books, pictures of distant planets, pictures of hypothetical planets and dictionaries of never spoken languages. You could find cures for diseases and the deepest secrets of every person alive. The only trouble is putting it all together. –  Mar 21 '13 at 22:23
  • Where do you get 1049088 pixels from? That number looks like it probably has a typo in it somewhere, since it's $2^9\cdot 3 \cdot 683$, and 683 is unlikely to divide either the x or y resolution of your display - but $3\cdot 683=2049$, and so the original number is also $2^9\cdot(2^{11}+1)$... – Steven Stadnicki Feb 04 '14 at 19:15

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Your original number is $2^{15*2^{20}} <2^{2^{24}} < 10^{2^{21}}< 10^{3*10^6} $ which is certainly computable since it has less than 3,000,000 digits.

The new, larger number is $2^{24*2^{20}} <2^{2^{25}} < 10^{2^{22}}< 10^{6*10^6} $ which is still computable since it has less than 6,000,000 digits.

wythagoras
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marty cohen
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  • I fixed a small typo in the MathJax (a pair of braces). It is not true that $$2^{2^{24}} = 2^{8 \cdot 2^{21}} = (2^8)^{2^{21}} = 256^{2^{21}} < 10^{2^{21}}$$ as you claim. Also, technically, you are abusing the term "computable". – wythagoras Oct 23 '16 at 14:31
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Using the fact that $24$ bit color allows $2^{24}$ colors in a pixel, you get $(2^{24})^{1049088}=2^{24\cdot 1049088}=2^{25178112}$ If you like powers of $10$ better, this is about $10^{25178112\cdot \log_{10}2}\approx 10^{7.58\cdot 10^6}$

Ross Millikan
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