Here's a possible intended solution to show that $30031 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ is composite without factoring it. Recall the Fermat primality test: if $a^{n-1} \not \equiv 1 \bmod n$, then $n$ cannot be prime. It turns out that $2^{30030} \equiv 21335 \bmod 30031$, so $30031$ must indeed be composite. There is a well-known algorithm called binary exponentiation that is reasonably fast to implement by hand and that could conceivably be done on an exam. (I am not totally convinced that this would be faster than trial division until $p = 59$, though. And if you followed Leonid's suggestion in the comments your life would be even easier.)
Edit: Here is the solution by trial division. It suffices to check the primes from $17$ up. Note that the fact that we know the prime factorization of $30030$ helps a lot. All of the arithmetic necessary beyond what I wrote down was mental.
For $17$ write $30031 = 30 \cdot 77 \cdot 13 + 1 \equiv -4 \cdot 9 \cdot -4 + 1 \equiv -8 \bmod 17$.
For $19$ write $30031 \equiv -8 \cdot 1 \cdot -6 + 1 \equiv 11 \bmod 19$.
For $23$ write $30031 \equiv 7 \cdot 8 \cdot -10 + 1 \equiv -99 \bmod 23$.
For $29$ write $30031 \equiv 1 \cdot -10 \cdot 13 + 1 \equiv -100 \bmod 29$.
For $31$ write $30031 \equiv 30000 \bmod 31$.
For $37$ write $30031 \equiv -7 \cdot 3 \cdot 13 + 1 \equiv -13 \bmod 37$.
For $41$ write $30031 \equiv -11 \cdot 5 \cdot 13 + 1 \equiv -14 \cdot 13 + 1 \equiv -140 \bmod 41$.
For $43$ write $30031 \equiv -13 \cdot -9 \cdot 13 + 1 \equiv -26 \bmod 43$.
For $47$ write $30031 = 210 \cdot 143 + 1 \equiv 25 \cdot 2 + 1 \equiv 4 \bmod 47$.
For $53$ write $30031 \equiv -2 \cdot -16 + 1 \equiv 33 \bmod 53$.
Finally, for $59$ write $30031 \equiv 33 \cdot 25 + 1 \equiv 11 \cdot 16 + 1 \equiv 0 \bmod 59$.