15-10 A bag of peanuts contains 1988 peanuts. A monkey takes one peanut on the first day, and from the second day onwards, the number of peanuts it takes each day is the total of all the peanuts taken in previous days. If on a certain day the number of peanuts left in the bag is less than the total number of peanuts already taken, on that day it starts over by taking one peanut and follows the original rule for a new cycle. If this continues, on which day will the monkey have taken all the peanuts from the bag?
(Science Experimental Class Mathematics Admission Test, 1989)
［Solution］Suppose the monkey used $k$ days in the first round, meaning that on the $k+1$-th day, it started taking one peanut again.
At this point, in the first $k$ days, a total of
$$
1+1+2+2^{2}+2^{3}+\cdots+2^{k-2}=2^{k-1}
$$

peanuts were taken, and it satisfies $1988-2^{k-1}<2^{k-1}$.
If in the second round it used $t$ days, then in the second round, it took $2^{t-1}$ peanuts, and
$$
1988-2^{k-1}-2^{t-1}<2^{t-1}, \quad t<k .
$$

Therefore, we only need to decompose 1988 into the sum of powers of 2, i.e.,
$$
1988=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+2^{2} .
$$

So, the total number of days used is
$$
\begin{aligned}
& (10+1)+(9+1)+(8+1)+(7+1)+(6+1)+(2+1) \\
= & 11+10+9+8+7+3 \\
= & 48 \text { (days). }
\end{aligned}
$$