# Tag Info

114

Does this count?

110

If you are allowed to use decimals, then

94

91

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:

73

72

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

69

68

65

I believe that this is the smallest:

58

53

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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52

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 + ... 52 Here's one way I found: Or, using just the characters explicitly allowed in the question: 51 No rules? Looks like 88 to me if I squint. 51 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 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 46 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$$Which ... 43 ... -3: MINUS ONE MINUS ONE MINUS ONE -2: MINUS ONE MINUS ONE -1: MINUS ONE 0: ONE MINUS ONE 1: ONE 2: ONE MINUS MINUS ONE 3: ONE MINUS MINUS ONE MINUS MINUS ONE ... two letters fewer since we do not use PLUS any more. 43 Here's another solution: Where: 43 The answer is 42 And three more à la Perry 41 First off, latest edit - just for fun, how to get 5 from just 0 and 1: Before rule change posted: With the changed rules: And while we're at it, here's 0 to 28: And here's how to get 5 from just 0, 0 and 1: And how to get 5 from just 0, 0 and 0: 40 The simplest one is: 39 If √ can mean nth root:$$\sqrt[1234567]{-8+9} 3 operators. Obviously...

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\$\begin{align} 0 & = (1 + 2 - 3) \times (4 + 5) \\ 1 & = 1 + (2+3-5) \times 4 \\ 2 & = 2 + (1+3-4) \times 5 \\ 3 & = 1 -2+3-4+5 \\ 4 & = 1 \times (2+3-5) + 4 \\ 5 & = 1-2-3+4+5 \\ 6 & = 5+1 \times (2+3-4) \\ 7 & = 5+2 \times (4-3) \times 1 \\ 8 & = 5+3 \times (4-2-1) \\ 9 & = 5+3 + (4-2-1) \\ 10 & = 5+4 + (3-2) \...

38

If the double factorial is allowed, then I propose WolframAlpha agrees that the result is 19.

37

Perhaps involving only subtracting, For example,

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