Problem 1801

*Answer:* 2003.

Subtracting the second equation from the first one gives
$$a_1(p-1) + \dots + a_m(p^m - 1) = 2002.$$
As the l.h.s. of the obtained equality is divisible by $p-1$, $2002 = 2 \cdot 7 \cdot 11 \cdot 13$ must also be divisible by $p-1$. Thus $p-1$ equals one of $1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001$ and $2002$. Since $p$ is prime, only $2, 3, 23$ and $2003$ remain. The first equation of the given system is the $p$-ary representation of 2013, whence the coefficients $a_i$ are uniquely determined by $p$.

Now we study all cases.
1. If $p=2$ then $m=10$ as $2^{10} < 2013 < 2^{11}$. The second equation implies that all $a_i$s must be ones, but $1+2+2^2+\dots+2^{10} = 2^{11}-1 = 2047$. Hence there is no solution in this case.

2. Let $p=3$. As $2013 = 2 \cdot 3 + 3^2 + 2 \cdot 3^3 + 2 \cdot 3^5 + 2 \cdot 3^6$ whereas $2+1+2+2+2 = 9 \neq 11$, this case gives no solution either.

3. Let $p=23$. As $2013 = 12 + 18 \cdot 23 + 3 \cdot 23^2$ while $12 + 18 + 3 > 11$, this case gives no solution either.

4. For $p=2003$, we get $2013 = 10+2003$ and $10+1=11$, so the conditions are satisfied.
Consequently, 2003 is the only prime number with the desired property.