http://olympiads.mccme.ru/vmo/

Do you mean $m(n) \ne n$ ?

yes-yes, $m(n)<n$.

What do you mean "great natural divisor" ?

$n=p_{1}^{\alpha _{1}}...p_{k}^{\alpha _{k}},p_{1}< p_{2}< ...< p_{k}\Rightarrow m(n)=p_{1}^{\alpha _{1}-1}...p_{k}^{\alpha _{k}}\Rightarrow 10^{\alpha }=p_{1}^{\alpha _{1}-1}...p_{k}^{\alpha _{k}}(p_{1}+1)$ case 1: if $k$ in greater than 3: $p_{i}\neq 2,5$ which is imposible. case 2: if $k=1$:if $\alpha=1$ $p+1=10^{\beta }$,$9\mid 10^{\alpha }-1=p$ , if $\alpha>1$ $p^{\alpha -1}(p+1)=10^{\beta }$ if $p=2$ $2^{\alpha }=10^{\beta }$ which is not true if $p=5$ $6.5^{\alpha -1}=10^{\beta }$ which is not true. case 3: k=2 $p^{\alpha -1}q^{\beta }(p+1)=10^{m}$ if $\alpha =1$ ($p<q$) $q^{\beta }(p+1)=10^{m}$ if q is not 2 ($p<q$) $\Rightarrow q=5,p< q,p\neq 2\Rightarrow p=3\Rightarrow 4.5^{\beta }=10^{m}\Rightarrow \beta =m=2\Rightarrow n=75$ , $\alpha >1$: $\Rightarrow p=2\Rightarrow 3.2^{\alpha }.q^{\beta }=10^{n}$ which is not true. if $k=3$: $\Rightarrow \alpha =1\Rightarrow q^{a}.r^{b}(p+1)=10^{m},q,r\neq 2(p< q< r)\Rightarrow q=5\Rightarrow (p\neq 2)p=3\Rightarrow q=2,5$ which is not true (p< q< r) so answer only is $n=75$

Let $n+m(n)=10^k$ with $k \in \mathbb{N}$. It is easy to see that $n \ge 2$. If $n$ is even then $m(n)= \frac n2$. Hence $3m(n)=10^k$, contradiction. Thus, $n$ is odd. Let $n=l \cdot m(n)$ with $l \in \mathbb{N}, l \ge 2$. Hence we get $(l+1) \cdot m(n)=10^k$. Since $n$ is odd then $m(n)$ is also odd. Therefore $m(n)=5^x,l+1=5^y \cdot 2^k$ with $x,y \in \mathbb{N},x+y=k,x \ge 1$. If $l>5$ then $m(n)$ is not the greatest divisor of $n$. It follows that $l \le 5$ or $5^y \cdot 2^k \le 6$. We get $y=0$ and $k \in \{1,2 \}$. With $k=1$ then $l=1$, contradiction. With $k=2$ then $l=3$ and $x=2$. We find $n= 5^2 \cdot 3=75$. The answer is $\boxed{n=75}$.

Supongamos que $n$ cumple las condiciones del problema, entonces $n$ no puede ser un número primo, pues se tendría que $n+1$ es potencia de $10$, y esto obliga a que $9 \mid n$, que sería un absurdo, por lo tanto, $n$ es compuesto. En efecto, sea $p$ el menor número primo que divide a $n$, entonces $m(n)=\frac{n}{p}$, entonces $n+\frac{n}{p}=10^{t}$, para algún entero positivo $t$. Note que $p$ no puede ser $2$, pues se tendría que $\frac{3n}{2}=10^{t}$, pero como $n$ es par, entonces $3 \mid 10^{t}$, lo cual es absurdo, por lo tanto $p$ es impar. Luego, se tiene que $(\frac{n}{p}).(p+1)=10^{t}$, y como los números $\frac{n}{p}$ y $p+1$ son entero, se deduce que $\frac{n}{p}$ y $p+1$ son divisores de $10^{t}$, y como $n$ es impar, entonces $n$ es múltiplo de $5$. Por lo tanto, $p \leq 5$, pero si $p=5$, entonces $p+1=6$ debería ser divisor de $10^{t}$, por lo tanto $2<p<5$, lo cual obliga a que $p=3$. En consecuencia tenemos que $4 \times \frac{n}{3}=10^{t}$, pero $\frac{n}{3}$ es impar, por lo tanto $\frac{n}{p}=25$, para que pueda cumplir las condiciones de que ese producto sea una potencia de $10$. Finalmente se deduce que $n=75$.

Another solution Let $p$ be the least prime divisor of $n$. If $p=2$ then this implies $\frac{3n}{2} = 10^x$ or $3|10$ which is bad and doesn't work If $p=3$ then this implies $\frac{4n}{3} = 10^x$ or $n = \frac{3*2^x*5^x}{4}$. Since $3$ is n's least prime divisor this means $x=2$ and $n=75$ If $p=5$ then $\frac{6n}{5} = 10^x$ implying $6|10$ which is bad and doesn't work. Now assume $p > 5$. Then we have $\frac{(p+1)n}{p} = 10^x$, or $n = \frac{2^x*5^x*p}{p+1}$. Since $p$ is n's least prime divisor and $p > 5$ this implies that $p+1 = 2^x*5^x$, or $p = 10^x-1$. Taken mod 3, we see that this means $p \equiv 1^x-1 \equiv 0 \pmod{3}$. Since $p>3$ this means p is not prime, which is a contradiction. Thus, the only possibility is $n = 75$

n + m(n) = n + n/k where k is the smallest prime that divides n. Now, n + n/k + 10p =>n = k*5p*2p /(k+1) Now if k+1 does not divide k thus it divides 10p . Now on dividing 5p*2p by k+1, we must not get any prime less than k as then n would get prime factor less than k which is against our considerations. There either k+1 is 3 or primes which may eliminate 5p*2p from numerator on division. But as none of 10p - 1 is not a prime, the only option is k = 3. k is not 2 or 5 as 3 or 6 does not divide 10p. Thus k = 3. and it has to eliminate 2p , thus p=2. Therefore 10p = 102 = 100. Therefore on solving for n by putting values of 10p as 100 and k as 3 we get 75 as the only solution. Thank you.

5p = 5^p, 2p = 2^p, 10p = 10^p, 102 = 10^2

mathuz wrote: Define $m(n)$ to be the greatest proper natural divisor of $n\in \mathbb{N}$. Find all $n \in \mathbb{N} $ such that $n+m(n) $ is a power of $10$. N. Agakhanov Let $n+m(n)=10^a$. If $n$ is a prime then $m(n)$ is clearly $1$. In that case, \begin{align*} p+1 & = 10^a\\ p & = 10^a-1 \end{align*}Right side is divisible by $10-1=9$, so $p$ can not be prime. Now, if $n>1$ and not a prime, then $n$ has a smallest prime divisor. Then the greatest proper divisor of $n$ will be $\dfrac{n}{p}$, let's say $n=pk$. Be careful here, $p$ is the smallest prime does not mean that $k$ is not divisible by $p$. For example, $12=2^2\cdot3$ so $k=6$. Take $k=p^rl$ where all prime factors of $l$ must be greater than $p$ and $r\geq0$. \begin{align*} n+m(n) & = 10^a\\ pk+k & = 10^a\\ p^rl(p+1) & = 10^a \end{align*}If $r\geq1$ then $p$ divides $10$. So $p=2$ or $p=5$. If $p=2$, then $p+1=3$ divides $10^a$, contradiction. If $p=5$, $p+1=6$ divides $10^a$, again contradiction. So $r=0$ and $l(p+1)=10^a$. It is clear that $l$ is odd, otherwise $2|l$ and hence $p<2$ since all prime factors of $l$ are greater than $p$. This also provides with $p+1\leq l$. Since $l$ is odd, $l=5^x$ for some $1\leq x\leq a$. Since $p$ is less than all prime factors of $l$, we must have $p=3$. \begin{align*} 5^x\cdot4 & = 2^a5^a \end{align*}which immediately gives $a=2$, so $x=a=2$. Thus, $n=pk=3\cdot5^2=75$.

your solution is much same as mine

Let $p$ be the smallest prime factor of $n$. Then $(p+1)\cdot \dfrac{n}{p}=10^x\implies \dfrac{n}{p}\mid 10^x$ so $n=2^i\cdot 5^j \cdot p$. Since $p+1\mid 10^x$ then $p=3$ or $p > 5$. If $p=3$ then $2\nmid n$ and $4n=3\cdot 10^x\implies x=2\implies n=75$. If $p>5$ then $n=p$ and $p+1=10^x$ impossible since $10^x-1\equiv 0\pmod{3}$. Thus the only solution is $\boxed{n=75}$.

Cute problem Clearly $n$ is not even. Let $p$ be the smallest prime factor of $n$. $$n + \frac np = 10^k \implies n(p+1)=p \cdot 2^k \cdot 5^k \implies p+1 = 2^k \cdot a$$where $a$ is odd. $$\implies n \cdot a = p \cdot 5^k$$If $p>5$ then $5^k$ cannot be a factor of $n \implies a = 5^k$ and $n=p$, now it's easy to check that this does not work. The only possible case left is $p=3$. $$n \cdot 4 = 3 \cdot 10^k \implies n = 3 \cdot 2^{k-2} \cdot 5^k \implies k-2 = 0 \implies k = 2 \implies n=75$$which is the only possible solution.

Finally doing little Number Theory after sometime. Let $p$ be the smallest prime dividing $n$. \[ n+\frac{n}{p}=10^m \tag{$\star$}\]for some $m \in \mathbb{N}$. Now we rule out the possibility of $p=2$ in $(\star)$. Substituting the value, $p=2$ we get that $3n=2\cdot 10^m$ which is a contradiction due to$\pmod{3}$ reasons. So $p$ must be an odd prime.

. Now we have deduced $n=5^m \cdot p$ but as $5$ is a prime dividing $n$ and $p$ must be smaller than $5$, we have the only possibility that $p=3$. Putting everything back, we get $n=75$. Just for sake of satisfaction, we verify our answer: \[75+m(75)=75+25=100=10^2\]Now, we are finally done. $\blacksquare$