IMO Team Selection Contest

Answer: 2003.

Subtracting the second equation from the first one gives $$a_1(p-1)+\cdots +a_m(p^m-1)=2002.$$ 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 $1+2+2^2+\cdots +2^{10}=2^{11}-1=2047$. 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 $2+1+2+2+2=9\ne 11$, 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.