# How can I calculate or think about the large number 32768^1049088?

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.

• 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

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