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# Tag Info

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Does this count?

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If you are allowed to use decimals, then

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I think that This is because This works and is valid because

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Note: This answer only applies prior to the edit that clarifies that the expression on the left must evaluate to 100, rather than simply the equation being true. If you allow exponents, you can get away with just two:

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because

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Probably not the intended answer, but, I propose: Explanation:

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Since the puzzle oddly and specifically mentions the symbol for the square root, I used this: but rotated and reflected it giving:

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I believe that this is the smallest:

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Lateral thinking!

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I think...

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4 ops = 1.9934200404 points: Off by 0.00108199. 5 ops = 2.2864604146 points: Off by 0.0000340537. 6 ops = 2.7136051067 points: Off by only 0.000000266764(!) Now we can keep taking square roots of 1 in this expression to get a lower score bound for $n$ operations where $n \geq 6$, namely: $$s_n = -\ln \left( \frac{ 355/113 } \pi - 1 \right) / n$$ Which ...

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As rand al’thor points out in the solution built upon here, there must be a way to formulate a $\small 3$ with only two $\small 0 \kern1mu$s.   How promising that...                         ...uses only two $\small 1$s. And ...

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Here's one way I found: Or, using just the characters explicitly allowed in the question:

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With the digits in order:  \begin{align} 1 &= 2 + 0 - 1 ^ 7 \\ 2 &= 2 + 0 \times 1 \times 7 \\ 3 &= 2 + 0 + 1 ^ 7 \\ 4 &= -2 - 0 - 1 + 7 \\ 5 &= 2 \times (0 - 1) + 7 \\ 6 &= 2 \times 0 - 1 + 7\\ 7 &= 2 \times 0 \times 1 + 7 \\ 8 &= 2 \times 0 + 1 + 7 \\ 9 &= 2 + 0 \times 1 + 7 \\ 10 &= 2 + 0 + 1 + 7 \\ 11 &= 2 + ...

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No rules? Looks like 88 to me if I squint.

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Also, you can use any operation. Ok then. $\begin{array}{c|c} 0 & \log_{\frac1 2} \left( \log_{4!!-3} 5 \right) \\ 1 & \log_{\frac1 2} \left( \log_{4!!-3} \sqrt 5 \right) \\ 2 & \log_{\frac1 2} \left( \log_{4!!-3} \sqrt{\sqrt 5\,} \right) \\ 3 & \log_{\frac1 2} \left( \log_{4!!-3} \sqrt{\sqrt{\sqrt 5\,}\,} \right) \\ 4 & \log_{\frac1 2} ... 48 I tried to make a digital clock.$0 = (7 + 1 + 2) \times 01 = (2 + 7 + 1) ^ 02 = (7 + 1) \times 0 + 23 = 7 \times 0 + 2 + 14 = 2 \times 7 - 105 = 7 - 2 + 1 \times 06 = 7 - 1 + 2 \times 07 = 7 + 1 * 2 \times 08 = 7 + 1 + 0 \times 29 = 7 + 2 + 1 \times 010 = 1 + 2 + 7 + 011 = 12 - 7^012 = 12 + 7 \times 013 = 12 + 7 ^ 0...

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As Jo has already shown, this can be accomplished in To help visualize this problem, we can imagine: Proving minimality:

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... -3: MINUS ONE MINUS ONE MINUS ONE -2: MINUS ONE MINUS ONE -1: MINUS ONE 0: ONE MINUS ONE 1: ONE 2: ONE MINUS MINUS ONE 3: ONE MINUS MINUS ONE MINUS MINUS ONE ... two letters fewer since we do not use PLUS any more.

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Here's another solution: Where:

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