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## 'Weekly Challenge 48: Quorum-sensing' printed from http://nrich.maths.org/

The bacteria divide twice an hour, so divide 48 times in 24 hours.

Thus $2^{48}$ bacteria cells after 24 hours, assuming the bacteria split in two when they divide.

X has half life of 10 mins, so there are $6\times 24$ half lives in 24 hours. We only consider the bacteria released near the end then, as very little from the beginning will be left after 24 hours.

Now $10^{11}< 2^{48}$ so if Rudolph's nose was 4mL, his nose will certainly be glowing.

Assuming instead that Rudolph's nose was 4mL, we need the concentration of bacteria to be $4\times 10^{11}$.

After 23.5 hours there were $2^{47}$ cells and $2^{47}$ have just been released. Right at 24 hours, there will be $\frac {2^{47}}{2}+2^{48}$ cells, which is greater than $4\times 10^{11}$.

So Rudolph will have a glowing nose.