Bulgarian National Mathematical Olympiad

If $a=\overline{a_1{a}_2{a}_3{a}_4{a}_5{a}_6}$ satisfies the second condition, then $10^5<a<10^6$ and $\overline{a_4{a}_5{a}_6}=\overline{a_1{a}_2{a}_3}+4$. It follows that

$$a=1000\overline{a_1{a}_2{a}_3}+\overline{a_4{a}_5{a}_6}=1001\overline{a_1{a}_2{a}_3}+4.$$

So, we are looking for 3-digit numbers $A=\overline{a_1{a}_2{a}_3}$ such that $1001A+4=n^k$ for $k\ge 3$. Using that $1001=7\cdot 11\cdot 13$, we get $n^k\equiv 4(\mathrm{mod}t)$ for $t=7,11,13$.

1. Let $k=3$. Then $n^3\equiv 4(\mathrm{mod}7)$. On the other hand, $n^3\equiv \pm 1(\mathrm{mod}7)$, a contradiction.

2. Let $k=4$. Then $n^4\equiv 4(\mathrm{mod}13)$. On the other hand, $n^4\equiv 1,3,9(\mathrm{mod}13)$, a contradiction.

3. Let $k=5$. Then $n^5\equiv 4(\mathrm{mod}11)$. On the other hand, $n^5\equiv \pm 1(\mathrm{mod}11)$, a contradiction.

4. Let $k=7$. Then $n^7\equiv 4(\mathrm{mod}7)$. On the other hand, $n^7\equiv n(\mathrm{mod}7)$ and hence $n\equiv 4(\mathrm{mod}7)$. If $n=4$, then $4^7<10^5$, and if $n\ge 11$, then $n^7\ge 11^7>10^6$, a contradiction.

5. Let $k=11$. Since $4^{11}>2^{20}>10^6$, then $n<4$. But $n^{11}\equiv n(\mathrm{mod}11)$ implies $n^{11}\equiv 4(\mathrm{mod}11)$, a contradiction.

6. Let $k\ge 13$. Since $3^{13}=2187.729>2000.700>10^6$, then $n=2$. The inequality $2^{20}>10^6$ implies $k\le 19$. Moreover $2^k\equiv 4(\mathrm{mod}11)$ and so $2^{k-2}\equiv 1(\mathrm{mod}11)$. Since the order of 2 modulo 11 is equal to 10, it follows that $10|k-2$, a contradiction to $13\le k\le 19$.

Finally, if $k>4$ is an even number, then 4 divides $k$ or $k$ has an odd divisor. Thus the problem reduces to one of above cases.

Answer. There do not exist 6-digit numbers with the given properties.