# Make numbers 1-50 using $\pi$ and its digits (but with some penalties) [closed]

In this problem, you will be allowed to use some operations and additional digits from the basic approximation of $$\pi=3.14$$ having some penalties, as follows:

Operations:

• Using basic operations and parenthesis ($$+, -, *, /, (),$$) gives you 4 points per number:

$$3+1+4 = 8$$

• Using additional operations without numbers (! - factorial, $$\sqrt{}$$ - square root, unary minus,...) gives you 3 points $$-3 + \sqrt{1+4!} = 22$$

• Using additional operations with numbers ($$x^2, x^3,...$$ - exponentiation, other roots, etc) gives you 2 points $$\sqrt{3^2 - 1} + 4 = 6$$

• Concatenation of original digits gives you 1 point.

$$cat(3,1)+4 = 35$$

this results in a partial score $$S_p$$:

$$S_p = \sum_{i=0}^{50}{S_i}$$

Number of decimal digits of pi:

The previous partial score must be divided by the number of decimal digits used, starting from $$3.14$$, which contains 2 decimal digits. For example, if $$\pi = 3.14$$ is used, the total score $$S_T$$ is $$S_p/2$$. But if $$\pi = 3.14159$$ is used, $$S_T$$ is $$S_p/5$$.

Valid options are (from 2 to 5 digits): 3.14, 3.142, 3.1416, 3.14159.

Rules:

• All digits should be used once and in their order
• Exponents must be integers (positive and negative ;))
• Operations can be done on groups with parenthesis, e.g. $$(3+1)!+4$$ or $$3+(1+4)^2$$
• Operations/operators must be defined before expanding, this is, it is not allowed to add operators after solving partially. e.g. This is not allowed:
• $$concat((3+1),4) = 44)$$
• $$3^4+1+4 = 81+1+4 = 8+1+1+4 = 14$$
• Stacked operators are valid (Double factorial, multiple exponents, etc)
• Unary minus $$-$$ is allowed, e.g. $$-3+1+4 = 2$$

Question:

Having a theoretical max score of 100, How close can you get to it?

What is the highest score you can achieve?

P.D.

• Making $$0$$ is valid for bonus points ;)

*
• Is the "$n$ points per number" scoring system the same for all numbers or it's defined for each number? For example, if I write $8=3+1+4$ but $35=31+4$, what's the score for 8 - 4 (since I've used only basic operations for 8) or 1 (since I've used concatenation for some other number)? Also, what types of rounding are allowed (i.e. can $\pi\approx3.142$ be used instead of $\pi\approx3.141$)? Jan 6, 2020 at 15:20
• FIrst you have to choose the number of decimal digits to use for making all the numbers. In your example, you have 4 points from "8" and 1 point from "35". I will update the rounding options. Thanks. Jan 6, 2020 at 15:29
• I'm afraid this sort of open-ended question isn't considered on-topic here. As you'll see if you follow that link, I have a few misgivings about the rule, but this particular question does seem like it fits Deusovi's "more game than puzzle" description, and doesn't really have such a thing as a correct answer as opposed to a potentially endless stream of small improvements to the score. So I'm going to have to close it as too broad. My apologies. Jan 6, 2020 at 20:24
• (The specific close reason here -- "needs more focus" -- isn't really quite right. We only have a finite number of those reasons available. I think that one is the nearest one to meaning the right thing.) Jan 6, 2020 at 20:26

Current score: 50

Using: 3.1415, for a divisor 4
Basic operations only - 4 points per number, 200 partial score.

Working:

1=1
2=1+1
3=3
4=4
5=5
6=5+1
7=3+4
8=4*(3-1)
9=3*(4-1)
10=5*(3-1)
11=(3*4)-1
12=3*4
13=3*4+1
14=(3+4)*(1+1)
15=3*5
16=3*5+1
17=3*5+1+1
18=3*(5+1)
19=(4*5)-1
20=4*5
21=(3+4)*(5-1-1)
22=(4*5)+1+1
23=(4*5)+3
24=(5+1)4
25=5
(1+4)
26=(5*(1+4))+1
27=3*(4+5)
28=(3+4)(5-1)
29=(3+4)
(5-1)+1
30=3*(4+1+5)
31=(3*(4+1+5))+1
32=4*(3+5)
33=3*(1+4+1+5)
34=((3+4)*5)-1
35=(3+4)*5
36=3*4*(5-1-1)
37= ((3+4)5)+1+1
38=(3-1)
((4*5)-1)
39=((3+1+4)*5)-1
40=4*5*(3-1)
41=4*5*(3-1)+1
42=3*(((4-1)*5)-1)
43=(4*5*(1+1))+3
44=((3*(5-1)-1)4
45=3
(4-1)5
46=(3
(4-1)*5)+1
47=(3*4*(5-1))-1
48=3*4*(5-1)
49=(3+4)(5+1+1)
50=5
((3*4)-1-1)

Bonus:

0=3-1+4-1-5

• I seem to be able to either separate my lines of working onto separate lines or spoilerbar them - this formatting is taking some getting used to. My * multipliers are getting eaten, too. Can anyone fix that or give me a hand fixing it myself? Jan 6, 2020 at 12:06
• If you want to add newlines inside the spoiler, append double spaces at the end of the line :) Jan 6, 2020 at 12:14
• Hi @LizWeir, it's a nice work, but I didn't add the rule that all digits should be used in their order. Sorry for that, it's my bad. I will clarify the question. Jan 6, 2020 at 12:18
• Ah! That changes things, yeah. ;-) Jan 6, 2020 at 12:53

My solution

using an approximation $$\pi\approx3.14$$ with the score of 70.5 ($$=(4\times7+3\times27+2\times16)/2$$)

Note

I believe that operators like $$[x]$$ (integer part of $$x$$) are allowed and classified as "additional operations without numbers" (note the "..." and "etc" in the question, so nothing disallows using it).

Here we go (numbers in parentheses indicate point count):

$$1=(3+1)/4\ (4)$$
$$2=-3+1+4\ (3)$$
$$3=3!+1-4\ (3)$$
$$4=3-1+\sqrt4\ (3)$$
$$5=3\times1+\sqrt4\ (3)$$
$$6=3-1+4\ (4)$$
$$7=3\times1+4\ (4)$$
$$8=3+1+4\ (4)$$
$$9=3!-1+4\ (3)$$
$$10=3!\times1+4\ (3)$$
$$11=3!+1+4\ (3)$$
$$12=3\times1\times4\ (4)$$
$$13=3^2\times1+4\ (2)$$
$$14=3^2+1+4\ (2)$$
$$15=3\times(1+4)\ (4)$$
$$16=(3+1)\times4\ (4)$$
$$17=-3!-1+4!\ (3)$$
$$18=-3!\times1+4!\ (3)$$
$$19=-3!+1+4!\ (3)$$
$$20=-3-1+4!\ (3)$$
$$21=-3\times1+4!\ (3)$$
$$22=-3!+1+4!\ (3)$$
$$23=-[\sqrt3]\times1+4!\ (3)$$
$$24=[\sqrt3]-1+4!\ (3)$$
$$25=[\sqrt3]\times1+4!\ (3)$$
$$26=3-1+4!\ (3)$$
$$27=3\times1+4!\ (3)$$
$$28=3+1+4!\ (3)$$
$$29=3!-1+4!\ (3)$$
$$30=3!\times1+4!\ (3)$$
$$31=3!+1+4!\ (3)$$
$$32=(3+1+4)!!!!\ (3)$$(quadruple factorial means $$8\times4$$)
$$33=[\sqrt{3-1}\times4!]\ (3)$$
$$34=3^2+1+4!\ (2)$$
$$35=3\times1+\sqrt{4^5}\ (2)$$
$$36=3!\times(-(1-4))!\ (3)$$
$$37=-3^3\times1+4^3\ (2)$$
$$38=-3^3+1+4^3\ (2)$$
$$39=3!+1+\sqrt{4^5}\ (2)$$
$$40=3^2-1+\sqrt{4^5}\ (2)$$
$$41=[\sqrt3\times1\times4!]\ (3)$$
$$42=3^3-1+4^2\ (2)$$
$$43=3^3\times1+4^2\ (2)$$
$$44=3^3+1+4^2\ (2)$$
$$45=3^2\times(1+4)\ (2)$$
$$46=-3^4-1+\sqrt{4^7}\ (2)$$
$$47=-3^4\times1+\sqrt{4^7}\ (2)$$
$$48=(3-1)\times4!\ (3)$$
$$49=(3\times1+4)^2\ (2)$$
$$50=3^3-1+4!\ (2)$$

Bonus:

$$0=3+1-4$$