117

A lateral thinking answer:


112

Does this count?


110

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


109

If you are allowed to use decimals, then


94

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:


94

How about


76

because


76

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


74

Here's an answer which


70

Answer: How?


69

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


67

I believe that this is the smallest:


67

Another answer could be


63

Lateral thinking!


63

I think...


59

What about this where


58

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 ...


57

The equation is equivalent to: As $A$ to $I$ are $1$ to $9$:


55

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} ...


53

How about such a variant?


53

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


52

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


52

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 ...


51

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 + ...


48

I tried to make a digital clock. $0 = (7 + 1 + 2) \times 0$ $1 = (2 + 7 + 1) ^ 0$ $2 = (7 + 1) \times 0 + 2$ $3 = 7 \times 0 + 2 + 1$ $4 = 2 \times 7 - 10$ $5 = 7 - 2 + 1 \times 0$ $6 = 7 - 1 + 2 \times 0$ $7 = 7 + 1 * 2 \times 0$ $8 = 7 + 1 + 0 \times 2$ $9 = 7 + 2 + 1 \times 0$ $10 = 1 + 2 + 7 + 0$ $11 = 12 - 7^0$ $12 = 12 + 7 \times 0$ $13 = 12 + 7 ^ 0$ $...


48

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


47



44

And three more à la Perry


43

... -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.


43

Here's another solution: Where:


Only top voted, non community-wiki answers of a minimum length are eligible