In a mathematics competition, there are 30 problems. Each correctly solved problem is worth 4 points, a wrong solution results in -1 point. If someone does not attempt a problem, they get 0 points for it. How many different total scores can a contestant achieve?
If a contestant solves $j$ problems correctly, then depending on how many of the remaining $30-j$ problems they fail, they can score any number of points between $4 j-(30-j)=5 j-30$ and $4 j$. With one more correct problem, $4 j+4$ points can also be achieved, and clearly no point value is missed between $4 j$ and $4 j+4$, as long as the number of problems that can still be failed is at least 3.

This means that starting from the minimum of -30 points when $j=0$, every integer point value is continuously achievable up to $4 j$ points, as long as $30-j \geqq 3$, i.e., $j \leqq 27$.

If $j=28$, then only two problems can be failed, so $27 \cdot 4+1=109$ points cannot be achieved. If $j=29$, then two point values are missed, $28 \cdot 4+1=113$ and $28 \cdot 4+2=114$, and if $j=30$, then all three point values are missed, where $29 \cdot 4<p<30 \cdot 4$.

Thus, out of the 151 possible numbers between -30 and 120, 6 cannot be achieved, meaning a contestant's total score can be 145 different values. (Sz. Cs.)