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The Story:

Today, the Maths quizzes are handed out by the teacher.

You got 100/100. Everyone got 100/100.

But you're not happy with it. Everyone's not happy with it.

You want to score lower. Everyone wants to score lower.

What? Why? Yes, that's the truth.

So, one of your classmates, proposed an idea. You are all going to compete for the lowest mark.

But, everyone knows each other's mark.

So, why the challenge? Because... everyone has a chance (or more)

The challenge: The teacher sees that you are all unsatisfied, and decided to give you all one more chance. This would be your final result.

The Quiz:

The Maths teacher is queer. The quiz consists of only one question scoring 100% of the quiz.

The Question:

Use $1,2,3,4,5,6,7,8,9$ to form $100$, each digit once and only once.

Criteria:

For each use of a new kind of operator, $+10$ marks.

For each concatenation of digits, $+5$ marks.

For each operator:

Count the use of that operator. Let that be $n$. $-n(n-1)/2$ marks.

If the Maths teacher find any unnecessary operators, +50 for each one.

Can you win the challenge? What's your score?

List of allowed operators:

$+, -, *, /$ [as in division]$, \sqrt{} , \text{^}$ (as in exponentiation), $floor(), ceil()$

List of operators not allowed:

$!, \log()$

For any inquiries about operators, feel free to ask in the comments. Happy Puzzling ;)

Sorry for the horrible explanation by the Maths teacher. There was a mess during the requiz. Here is a dialogue:

JonMark Perry: Do we have to use every digit 1-9 exactly once?

Teacher: Yes.

Thomas Blue: Do we actually have to compete? I mean, wouldn't $123456789 = 100$ give us the lowest score, since it will also be wrong?

Teacher: Yes. The equation needs to be valid.

Athin: What is unnecessary operators?

Teacher: operators that can be directly removed without changing the result. Sorry for all this chaos as this is the first time I created such a puzzle...

Teacher: Puzzles belong to the intelligent, and those who find loopholes always have the wit to win... Perhaps the teacher should become the student and the student the teacher... [sigh]...

Hope this helps!

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    $\begingroup$ do we have to use every digit 1-9 exactly once? $\endgroup$
    – JMP
    Jan 30, 2019 at 11:30
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    $\begingroup$ Do we actually have to compete? I mean, wouldn't 123456789 = 100 give us the lowest score, since it will also be wrong? $\endgroup$
    – Thomas Blue
    Jan 30, 2019 at 11:38
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    $\begingroup$ Also, in this conditions can score be made arbitrarily low by adding floor(floor(floor(floor(...)))) to the final answer, so that it exploits the last criteria to the great extent? $\endgroup$
    – Thomas Blue
    Jan 30, 2019 at 11:43
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    $\begingroup$ This has many similarities with an earlier puzzle. $\endgroup$
    – Bass
    Jan 30, 2019 at 13:18
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    $\begingroup$ @OmegaKrypton fair enough, no problem - so to summarize; there must be at least one operator between each individual digit, which could be a concatenation operator. Thanks for clearing up that point and I am glad it does not affect the nice answer from Athin $\endgroup$
    – tom
    May 19, 2019 at 21:57

3 Answers 3

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28
+100
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Tricky tricks: Loophole battle with the teacher

Flooring trick ver.2 (copied the nice formulas from hexomino, tyvm)

We can use floor or ceil to reduce the score as far as we want e.g, $$floor(floor(floor(\ldots(floor(1+3-4+5+6+72+8+9))\ldots))) = 100 $$ Since each of the unary floors is unnecessary, we will get a +50 bonus for each of those. Well then, let us use more to compensate for that. Taking the example provided, and using 1001 floor operators, we get score of:
$+ 3*10 + 5 - (6*5/2) - (1001*1000/2) + 50*1001 = -450430$ points (and immediate detention)

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    $\begingroup$ +1 quadratic scaling beats linear scaling every time. $\endgroup$
    – hexomino
    Jan 30, 2019 at 12:53
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    $\begingroup$ nice job! @ThomasBlue $\endgroup$
    – Omega Krypton
    Jan 30, 2019 at 13:16
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+50
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This answer

$4 \times 5^{2^{1^{3^{6^{7^{8^9}}}}}} = 100$

has a score of

$20 - \frac{7 \times 6}{2} = -1$

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    $\begingroup$ Congrats on 10k! :D $\endgroup$
    – Mr Pie
    May 19, 2019 at 15:57
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    $\begingroup$ @MrPie I love this answer and upvote it, but disagree with the score... if some of the ^ are removed after the 1 then the answer remains the same and so we should perhaps add +250 or something for the unnecessary ^ - so I think the final score should be 249 - and if unnecessary operators are removed then it should be 18... but then maybe it depends on what is considered to be an unnecessary operator... and if removal of an operator to give concatenation is considered removal of unnecessary operator or not. $\endgroup$
    – tom
    May 19, 2019 at 16:59
  • $\begingroup$ oh got score wrong forgot concatenation - maybe 43 $\endgroup$
    – tom
    May 19, 2019 at 17:05
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    $\begingroup$ @MrPie - seems like I was wrong to worry about that - see comments below the question above, concatenation is an operator so the ^ operators between digits cannot be removed $\endgroup$
    – tom
    May 19, 2019 at 22:00
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    $\begingroup$ @athin - looks like all is fine with your answer, see comments below question. the 'extra' ^ cannot be removed $\endgroup$
    – tom
    May 19, 2019 at 22:02
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How about this to start

$1+3-4+5+6+72+8+9 = 100$

For a score of

$10 = $$10$ (for $+$) $ + 10$ (for $-$) $ + 5$ (for concatenation) $- 15$ (for using $+$ six times)

Edit: However, as Thomas Blue points out in the comments

We can use floor or ceil to reduce the score as far as we want e.g, $$floor(floor(floor(\ldots(floor(1+3-4+5+6+72+8+9))\ldots))) = 100 $$ $$ceil(ceil(ceil(\ldots(ceil(1+3-4+5+6+72+8+9))\ldots))) = 100 $$ and can use a similar trick with the square root operator and $1$ $$ \sqrt{\sqrt{\sqrt{\ldots \sqrt{1}}}}+3-4+5+6+72+8+9 = 100 $$ so perhaps these operators should be disallowed?
Update: the question has now been edited with the intention of addressing this scenario.

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    $\begingroup$ edited the rule to prevent this situation $\endgroup$
    – Omega Krypton
    Jan 30, 2019 at 12:14

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