What language is the question written in?

Maybe English Google translate maybe can't translate it because of latex but don't know...

Are you sure it is English? I can't understand it.

Sure ;-)

mihajlon wrote: Two different $3$ digit numbers are picked and then for every of them is calculated sum of all $5$ numbers which are getting when digits of picked number change place (etc. if one of the number is $707$, the sum is $2401=770+77+77+770+707$). Do the given results must be different? Yes, they must. If $\overline{abc},\overline{a'b'c'}$ are the given numbers then we have $\overline{acb}+\overline{bac}+\overline{bca}+\overline{cab}+\overline{cba}=\overline{a'c'b'}+\overline{b'a'c'}+\overline{b'c'a'}+\overline{c'a'b'}+\overline{c'b'a'}$ that is $122a+212b+221c=122a'+212b'+221c'.$ Taking $\pmod 9$ we get $|x-y|=0,9,18$ where $x=a+b+c, \ y=a'+b'+c'.$ Taking $\pmod {11}$ we get $a+3b+c\equiv a'+3b'+c'\pmod {11}.$ Now, we distinguish the cases $x-y=0,9,-9,18,-18$ e.g. if $x=y+9$ then $9+2b\equiv 2b' \pmod {11}$ so $b'\equiv b-1 \pmod{11}$ so $b'=b-1.$ Thus $122A+212B+221C=0$ and $A+C=8$ and $B=1$ where $A=a-a', \ B=b-b', \ C=c-c'.$ So $A=20, C=-12. $ This is impossible. Similarly we work out the other cases...

Let $N$ be a three digit number, i.e. $N = \overline{abc}$. Permutating the three digit of $N$, we obtain six three digit numbers (including $N$) which sum is $\overline{abc} + \overline{acb} + \overline{bac} + \overline{bca} + \overline{cab} + \overline{cba} = 2(100 + 10 + 1)(a + b + c) = 222(a + b + c) = 222T$, where $T$ is the sum of digits of $N$. Hence the sum $S(N)$ of the five numbers (excluding $N$) which are digital permutations of $N$ are $(1) \;\; S(N) = 222T - N$. Assume there are two different three digit numbers $N_1$ and $N_2$ s.t. $S(N_1) = S(N_2)$. Let $T_1$ and $T_2$ be the sum of digits of $N_1$ and $N_2$ respectively. According to (1) we have $222T_1 - N_1 = 222T_2 - N_2$, i.e. $(2) \;\; N_1 - N_2 = 222(T_1 - T_2)$. The fact that $N_k \equiv T_k \pmod{9}$ combined with (2) give us $9 | 221(T_1 - T_2)$, implying $9 | T_1 - T_2$ since $GCD(221,9)=1$. Seeing that $T_1=T_2$ implies $N_1=N_2$ by (2), a contradiction which means $T_1 \neq T_2$. Therefore $|T_1 - T_2| \geq 9$, which according to (2) give us $|N_1 - N_2| \geq 222 \cdot 9 > 1000$, which is impossible since $100 \leq N_1,N_2 < 1000$. Conclusion: $S(N_1) = S(N_2)$ iff $N_1 =N_2$.