Solution to Exercise 16 The difficulty of this exercise lies in constructing the integer n. One might first wonder where the zeros will appear in the decimal representation of $5^{n}$. If one wants to avoid finding good rational approximations of $\log _{10} 5$ with small denominators, which is not very elegant, one should avoid looking for zeros at the beginning of the decimal representation. Moreover, it is clear that $5^{\text {n }}$ will never end in a zero. From this, one could hope to have zeros among the last digits.

The idea is to note that if $k \geqslant 1$ is fixed, then for all $n>k$, we have $5^{n} \equiv 5^{k}\left(\bmod 5^{k}\right)$. Therefore, for $5^{n}-5^{k}$ to end with k zeros, it suffices that $5^{n} \equiv 5^{k}\left(\bmod 2^{k}\right)$. Since $\varphi\left(2^{k}\right)=2^{k-1}$, this is the case as soon as $n \equiv k\left(\bmod 2^{k-1}\right)$. And in this case, provided that $5^{k}$ is written with strictly fewer than $k$ digits, there will be consecutive zeros among the last $k$ digits of $5^{n}$.

More precisely, consider the integer $\mathrm{n}=20+2^{19}=524308<10^{6}$. Clearly $5^{n} \equiv 5^{20}\left(\bmod 5^{20}\right)$, and $5^{n} \equiv 5^{20}\left(\bmod 2^{20}\right)$ since $\varphi\left(2^{20}\right)=2^{19}$. Thus, the integer $5^{n}-5^{20}$ is divisible by $10^{20}$ and therefore ends with at least 20 zeros. Since

$$5^{10}=\frac{10^{10}}{2^{10}}<\frac{10^{10}}{10^{3}}=10^{7}$$

we have $5^{20}<10^{14}$. Thus, $5^{20}$ has at most 14 digits, so the last 20 digits of $5^{n}=\left(5^{n}-5^{20}\right)+5^{20}$ start with a series of 6 consecutive zeros.

Graders' Comments: This problem was quite difficult and clever, and very few students submitted an attempt at a solution. However, among those who submitted a solution, a very large number provided a (nearly) complete solution. In addition to the solution $\mathrm{n}=2^{19}+20$ given in the solution, some students proved that $\mathrm{n}=2^{19}+21$ and $\boldsymbol{n}=2^{18}+20$ also work, with the same type of reasoning. Here are some general remarks:

- It turns out that a computer program (in Python, for example) can easily show that $n=3375$ is also a solution. However, there is very likely no simple mathematical proof of this fact. The few copies that mentioned this solution without justification therefore did not receive any points.
- Some students imagined that it was possible to take $n$ sufficiently small and negative so that $5^{n}$ is strictly less than $10^{-6}$, and thus starts with 6 consecutive zeros in its decimal representation. The statement was indeed imprecise in this regard, but in a number theory problem at this position in the exam, students could have deduced that only natural numbers were intended...
- Several students used the LTE lemma in the special case $p=2$ to estimate $v_{2}\left(5^{n}-1\right)$ where $n \geqslant 1$. Note that here, we indeed have $v_{2}\left(5^{n}-1\right)=2+v_{2}(n)$, but recall that for this case $p=2$, in general, if $x$ and $y$ are two odd integers, we have $v_{2}\left(x^{n}-y^{n}\right)=v_{2}(x-y)+v_{2}(x+y)+v_{2}(\mathrm{n})-1$.
- Finally, we expected students to justify precisely the inequalities invoked concerning the powers of 2, 5, or 10 used. In particular, all students who clearly used their calculator to see, for example, that $5^{20}$ has at most 14 digits lost a point.