# Logic Puzzle: One, Two, Three

Using the three digits, 1, 2, and 3, each at most once, and the combining them using addition, subtraction, multiplication, division, exponentiation, square root, factorial, unary negation, digit concatenation, decimal point, vinculum, and parenthesis, construct all the positive integers from 1 to 30. (Digit concatenation and decimal points only allowed on the original 3 digits. You do not need a 0 before the decimal point.)

The solution will eventually be posted in the comments, but if you solve it before then, feel free to show off (even with partial progress).

1. 1 = 1
2 = 2
3 = 3
4 = 3+1
5 = 3+2
6 = 3!
7 = 3!+1
8 = 3!+2
9 = 3!+2+1
10 = 3/(.2+.1)
11 = sqrt((3+2)!+1)
12 = 12
13 = 13
14 = (3/.2)-1
15 = (3/.2)
16 = (3/.2)+1
17 = (2/.1)-3
18 = 21-3
19 = ???
20 = (2/.1)
21 = 21
22 = 23-1
23 = 23
24 = 23+1
25 = (3-.2′)/.1′, where ‘ represents a vinculum on the previous digit
26 = 13*2
27 = 21+3!
28 = (3/.1)-2
29 = 31-2
30 = 3/.1

2. I’ll get the easy ones out of the way:

01=1/(3-2)
02=1+3-2
03=3/(2-1)
04=3+2-1
05=(3+2)*1
06=1+2+3
07=2*3+1
08=2^3*1
09=3^2*1
10=3^2+1
11=3!*2-1
12=3!*2*1
13=3!*2+1
14=(3!+1)*2
15=13+2
16=(1+3)^2
17=???
18=21-3
19=cat(1,3^2)
20=???
21=21 (doesn’t use ‘3’)
22=23-1
23=23*1
24=23+1
25=(3!-1)^2
26=13*2
27=(1+2)^3
28=???
29=31-2
30=???

I’ve tried to use all three digits where possible.

• Ah, according to the reddit thread we’re allowed e.g. “.2” as an abbreviation of “0.2”.
So we can do:
17=2/.1-
20=(1+3)/.2
28=3/.1 – 2
30=3/.1 (doesn’t use ‘2’)

I think I’m done. Didn’t need sqrt anywhere either, although I did find a silly way to do 19=sqrt(cat(3!^2,1))

• Yeah, that 19 doesn’t count. Sorry.

• Hmmm…
Aha!

19=sqrt(3!!/2+1)

• Good Job!

3. I got partial solution without 17,20,28 and 30: http://pastebin.com/G784scX8
I get the feeling I’m missing something obvious, because I’m not able to contruct 20 and 30.

• The digit concatenation operation is only allowed for the original 3 digits.

• You’re allowed .2=0.2, for example.

4. I’ve got all of them but the illusive 19, I’ll post my solutions if I find it.

5. Solution:
1=1
2=2
3=3
4=1+3
5=2+3
6=3!
7=3!+1
8=2^3
9=3^2
10=3^2+1
11=13-2
12=12
13=13
14=2*(3!+1)
15=12+3
16=2^(3+1)
17=2/.1-3
18=21-3
19=Sqrt((3!)!/2+1)
20=2/.1
21=21
22=23-1
23=23
24=23+1
25=(3!-2)!+1
26=2*13
27=(2+1)^3
28=3/.1-2
29=31-2
30=3/.1

6. Using all 3 digits exactly once, I’m only missing 19 and 30. Vinculum is represented by underscore following the digit.
1 = (3-1)/2
2 = (1+3)/2
3 = 3!/2!*1
4 = 3+2-1
5 = 3!-2!+1
6 = 3+2+1
7 = 3*2+1
8 = 1*2^3
9 = 3^2*1
10 = 3^2+1
11 = 13-2
12 = 3!*2!*1
13 = 3!*2!+1
14 = (3!+1!)*2
15 = 2/(.1_)-3
16 = 2^(3+1)
17 = 2/.1-3
18 = 3!*(2!+1!)
19 =
20 = (3+1)/.2
21 = 2/(.1_)+3
22 = (3+1)!-2
23 = 2/.1+3
24 = 21+3
25 = 3/(.1_)-2 or (3!-1)^2 or 3/.12
26 = 3!/(.2_)-1
27 = 3^(2+1)
28 = 3/.1-2
29 = 3/(.1_)+2
30 =

• 30 = 3/(.2-.1)
I haven’t come up with a better solution for 19 than yours, though I don’t like it because technically, a square root uses a ‘2’.