How much caffeine is left in your system? The half-life of caffeine in your system is 6 hours. If you drink 100mg of caffeine every day, what percentage is left in your system after a week?
If you drink 100mg caffeine every day for a month (30d), how long does it take for the level of caffeine in your system to drop below 1mg?
 A: Hint: If you drink $x$ coffee after one day you get $\frac{x}{2^4}$ left in your system. The rest is writing the proper sum.
For the first question, you will have:
$$
\frac{x}{16}+\frac{x}{16^2}+...+\frac{x}{16^7}=x\frac{1}{16}\frac{1-\frac{1}{16^7}}{1-\frac{1}{16}}
$$
For the last question, you get: 
$$
\frac{x}{16}+\frac{x}{16^2}+...+\frac{x}{16^{30}}=x\frac{1}{16}\frac{1-\frac{1}{16^{30}}}{1-\frac{1}{16}}=y
$$
You want to find $d$ such that:
$$
\frac{y}{16^d}\leq \frac{x}{100}
$$
A: The answer to both depends on when (and how, since one rarely drinks the coffee instantaneously) the coffee is consumed during the day.
For the purposes of this problem, however, I will assume that all the coffee is consumed at exactly $t_{\text{coffee}}$ every day and that we are looking for the quantity remaining at the end of the $7$th day.
At the end of a day, the contribution from that day is $100 ( \frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}}$, after $n$ days, this contribution  decreases to 
$100 ( \frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}} ( \frac{1}{2^{4} } )^n$ (since $4 = \frac{24}{6}$). Summing up, and dividing by $700$ to get the ratio gives:
$$\frac{100}{700}  (\frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}} 
\left( ( \frac{1}{2^{4} } )^6 + ... + ( \frac{1}{2^{4} } )^1 + 1\right)  = 
\frac{1}{7}  (\frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}} 
\left( \frac{1-(\frac{1}{2^{4}})^7 }{1-\frac{1}{2^{4} }} \right)$$
So, doing the computation shows that the percentage left varies between $0.95 \% -15.24 \%$ (taking $t_{\text{coffee}} \in [0,24)$). Note that this is the percentage of the total caffeine consumed during the week, which is a bit misleading, since if you took the average over $n$ days, and let $n \to \infty$, then the percentage would decrease to zero.
The other computation follows similar lines (but is less misleading):
I am assuming that you drink coffee every day for 30 days and then go cold turkey.
Then you want to find the minimum $n$ (whole days) such that
$$100  (\frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}} 
\left( ( \frac{1}{2^{4} } )^{29} + ... + ( \frac{1}{2^{4} } )^1 + 1\right) (\frac{1}{2^4} )^{ n}  \le 1$$
A quick calculation shows that the answer is $n=1$ or $n=2$ (depending on $t_{\text{coffee}}$).
