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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.

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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
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
However, the vast majority of these images will just be noise. –  Raskolnikov Mar 21 '13 at 20:12

2 Answers 2

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.

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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 $10^{25178112\cdot \log_{10}2}\approx 10^{7.58\cdot 10^6}$

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