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).

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  1. Pingback: Logic Puzzle: 5, 5, 7, 7 = 181 | By Way Of Contradiction

  2. 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

  3. 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.

  4. 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

  5. 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 =

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