Since $a^{10} < 1000(a+1)$, we have $a \le 2$. (i) $a=2$. $2(b+2)(b^2+4)(2b^6+64)<100b+2100$. Therefore $b=0,1$. (i-i) $b=1$. \[(3+c+d)(5+c^2+d^2)(66+3c^6+4d^6) < 1100\]Either $c \ge 2$ or $d \ge 2$, the LHS exceed $2200$. Hence $c \le 1$, $d \le 1$. Substituting yields no solution. (i-ii) $b=0$. Similarly \[ (c+d+2) \left(c^2+d^2+4\right) \left(3 c^6+4 d^6+64\right) < 1050 \]and we have $c \le 1$, $d \le 1$. Substituting yields $\overline{abcd}= \fbox{2010}$. (ii) $a=1$. We have $(1+b)(1+b^2)(1+2b^6) < 100b+1100$. We have $b \le 1$. (ii-1) $b=1$. $(2+c+d)(2+c^2+d^2)(3+3c^6+4d^6) \le 1200$. Hence $c,d \le 1$. Substituting yields no solutions. (ii-2) $b=0$. $(c+d+1) \left(c^2+d^2+1\right) \left(3 c^6+4 d^6+1\right)< 1100$. Hence, as before $c \le 1$, $d \le 1$. Substituting yields no solutions at all. Therefore the above boxed solution is the only solution.

seoneo wrote: Since $a^{10} < 1000(a+1)$, we have $a \le 2$. (i) $a=2$. $2(b+2)(b^2+4)(2b^6+64)<100b+2100$. Therefore $b=0,1$. (i-i) $b=1$. \[(3+c+d)(5+c^2+d^2)(66+3c^6+4d^6) < 1100\]Either $c \ge 2$ or $d \ge 2$, the LHS exceed $2200$. Hence $c \le 1$, $d \le 1$. Substituting yields no solution. (i-ii) $b=0$. Similarly \[ (c+d+2) \left(c^2+d^2+4\right) \left(3 c^6+4 d^6+64\right) < 1050 \]and we have $c \le 1$, $d \le 1$. Substituting yields $\overline{abcd}= \fbox{2010}$. (ii) $a=1$. We have $(1+b)(1+b^2)(1+2b^6) < 100b+1100$. We have $b \le 1$. (ii-1) $b=1$. $(2+c+d)(2+c^2+d^2)(3+3c^6+4d^6) \le 1200$. Hence $c,d \le 1$. Substituting yields no solutions. (ii-2) $b=0$. $(c+d+1) \left(c^2+d^2+1\right) \left(3 c^6+4 d^6+1\right)< 1100$. Hence, as before $c \le 1$, $d \le 1$. Substituting yields no solutions at all. Therefore the above boxed solution is the only solution. Sorry, could you explain why $a^{10}<1000(a+1)$ I mean surely it's a digit $(1 \le a \le 9)$ but why exactly is that true? could you explain more?

Mustafa.Elq wrote: Sorry, could you explain why $a^{10}<1000(a+1)$ I mean surely it's a digit $(1 \le a \le 9)$ but why exactly is that true? could you explain more? Since all the four numbers are digits $a \le (a + b +c + d)$, $a^2 \le (a^2 + b^2 + c^2 + d^2)$ and $a^6 \le (a^6 + 2b^6 + 3c^6 + 4d^6)$ Also $\overline{abcd} < 1000\cdot(a + 1)$ Therefore \[ 1000\cdot(a+1) > \overline{abcd} = a(a + b + c + d)(a^2 + b^2 + c^2 + d^2)(a^6 + 2b^6 + 3c^6 + 4d^6) \ge a^{10} .\]