188. Among the first 100000001 Fibonacci numbers (see problem 185), is there a number that ends with four zeros?
188. Let's leave in each member of the Fibonacci series, written with five or more digits, only the last four digits. We will get a sequence of numbers, each of which is less than $10^{4}$. Let's denote by $a_{k}$ the member of this sequence standing at the $k$-th place. Note that if we know $a_{k+1}$ and $a_{k}$, then we can calculate $a_{k-1}$, because the $(k-1)$-th member of the Fibonacci series is equal to the difference between the $(k+1)$-th and $k$-th members, and the last four digits of the difference can be determined by the last four digits of the minuend and subtrahend. From this it follows that if for some numbers $k$ and $n$ it turns out that $a_{k}=a_{n+k}, a_{k+1}=a_{n+k+1}$, then $a_{k-1}=a_{n+k-1}, a_{k-2}=a_{n+k-2}, \ldots, a_{1}=a_{n+1}$. But since $a_{1}=0$, this will mean that $a_{n+1}=0$, i.e., that in the Fibonacci series, the $(n+1)$-th place is occupied by a number ending in four zeros.

It remains to show that among $10^{8}+1$ pairs

| $a_{1}$, |  | $a_{2}$, |
| :---: | :---: | :---: |
| $a_{2}$, |  | $a_{3}$, |
| $\cdots . . . .$. |  |  |
| $a_{10^{8}}$ |  | $a_{10^{8}+1}$, |
| $a_{10^{8}+1}$, |  | $a_{10^{8}+2}$ |

there will be at least two identical ones. But this will certainly be the case, because all numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{10^{8}+2}$ do not exceed $10^{4}$, and from $10^{4}$ numbers $0,1,2,3,4, \ldots, 9999$ only $10^{4} \cdot 10^{4}=10^{8}$ different pairs can be formed (because the first number can take $10^{4}$ different values and the second number can take $10^{4}$ different values).

Note. It is even possible to precisely indicate the first number in the Fibonacci series ending in four zeros; the number of this number is 7501 (see problem 174 of the book [5] and the corollaries from it).