1 In an exam, there are 30 multiple-choice questions. Correct answers earn 5 points each, incorrect answers earn 0 points, and unanswered questions earn 1 point each. If person A scores more than 80 points, and tells B the score, B can deduce how many questions A answered correctly. If A's score is slightly lower but still greater than 80, B cannot deduce the number of correct answers. What score did A get in this exam? (2nd American Mathematical Invitational Competition Question)
1. Jia's score $S$ is determined by his answer situation, so we can introduce parameters: Let the number of questions Jia answered correctly, unanswered, and incorrectly be $x$, $y$, and $z$, respectively, with $x+y+z=30$. Then Jia's score $S=5x+y+0z=5x+y$. The phrase "if Jia's score is lower but still greater than 80, then Yi cannot deduce it" means we need to find $S>80$ such that the corresponding $x$ is unique and $S$ is the smallest. Since $S$ can be guessed as $(x, y)$, it can also be guessed as $(x-1, y+5)$, or $(x+1, y-5)$. To ensure the guess can only be $(x, y)$, $(x-1, y+5)$ and $(x+1, y-5)$ must not exist. Because $(x-1, y+5)$ does not exist, $x-130$. But $x \geqslant 1$, so it must be that $(x-1)+(y+5)>30$. Thus, $x+y \geqslant 27$. Because $(x+1, y-5)$ does not exist, $y-530$. But $x+y+z=30$ implies $x+y \leqslant 30$, so it must be that $y-5<0$. Thus, $y \leqslant 4$. Therefore, $S=5x+y=5(x+y)-4y \geqslant 5 \times 27-4y \geqslant 5 \times 27-4 \times 4=119$. The equality holds when $x=23, y=4$. Hence, $S_{\text{min}}=119$, meaning Jia scored 119 points in this exam.