$n^3= {8S(n) }^3 + 6S(n)n + 1 $ $(2S(n))^3+1^3+(-n)^3-3*(2S(n))(1)(-n)=0$ $2S(n)+1-n =0$ if $n$ is $k$-digit. $0 = 2S(n)+1-n \leq 18k+1-10^{k-1}$ $k\leq 2$ For $k=1$ there are not solutions. For $k=2$ $2a+2b+1-10a-b=0$ $b=8a-1$ $a=1,b=7$ $n=17$

Where do you guys get the paper If anyone of you gave the exam then please post the paper

Only n=17 satisfy condition

RagvaloD wrote: $n^3= {8S(n) }^3 + 6S(n)n + 1 $ $(2S(n))^3+1^3+(-n)^3-3*(2S(n))(1)(-n)=0$ $2S(n)+1-n =0$ if $n$ is $k$-digit. $0 = 2S(n)+1-n \leq 18k+1-10^{k-1}$ $k\leq 2$ For $k=1$ there are not solutions. For $k=2$ $2a+2b+1-10a-b=0$ $b=8a-1$ $a=1,b=7$ $n=17$ Here you used the fact that $a^{3}+b^{3}+c^{3}=abc \Longleftrightarrow a+b+c=0$

Okay it's pretty cool problem .But the $P4$ was nothing for RMO . We are given $n^3 =8S(n)^3+6S(n)n +1 $. $\implies (2(S(n))^3+(-n)^3+(1)^3 -3.1.(-n)(2S(n))=0$. Which Implies either $n=2S(n)+1$ or $2S(n)=n=1$. Easily second case is not possible . So our concerns is to solve $n=2S(n)+1$. So we can say that ,$n=2S(n)+1 \le18([\log n]+1)+1$. Suppose ,$10^3>n>10^2$ but then we get ,$n\le 55$. Similarly we get if $10^k>n>10^{k-1}$ then we have $n<18(k+1)+1<10^{k-1}$. This contradiction shows that $n\le 10^2$. If n is one digit number then $n=S(n)$ so we don't get any solution for one digit $n$. Now we should look into 2 digit case in this case $n\le 18(2)+1=37 \cdots (*)$. Suppose $n=\bar{ab} =10 a+b$ , $\implies 10a+b =2(a+b)+1$. from$(*)$ we get $1\le a\le 3$. See we get solution as$(a,b)=(1,7)$. So $\boxed{n=17}$ is only solution.