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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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  • 1
    $\begingroup$ do we have to use every digit 1-9 exactly once? $\endgroup$ – JonMark Perry Jan 30 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 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 at 11:43
  • $\begingroup$ That will teach me to read the question properly. Did not realise floor and ceil were allowed. Also if using 1 on its own, you could apply sqrt abitrarily often to it and it still works $\endgroup$ – hexomino Jan 30 at 11:46
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    $\begingroup$ This has many similarities with an earlier puzzle. $\endgroup$ – Bass Jan 30 at 13:18
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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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  • 9
    $\begingroup$ +1 quadratic scaling beats linear scaling every time. $\endgroup$ – hexomino Jan 30 at 12:53
  • $\begingroup$ nice job! @ThomasBlue $\endgroup$ – Omega Krypton Jan 30 at 13:16
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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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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 at 12:14

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