Make numbers 1 - 40 using digits 2, 0, 2, 1

Question

Using only the digits in the year 2021 and basic mathematical operations (+, -, × and ÷), how many of the numbers from 1 to 40 can you make? Not all of the numbers are possible!

You may use each digit up to the number of times it appears in 2021, but not more. i.e., you can use 2 twice but the rest can be used once. You do not have to use every digit. The - operator may be used to make a negative (i.e. -1)

Also, you may concatenate (so you may make 12 by combining 1 and 2)

You don't need to use all the digits

Answer 1

Given OP's deleted answer it looks like concatenation is allowed and not all numbers must be used. Here are the numbers I was able to do (26/40)

$1 = 1$
$2 = 2$
$3 = 2 + 1$
$4 = 2 + 2$
$5 = 2 + 2 + 1$
$6 = (2+1) \times 2$
$7 = \frac{10}{2} + 2$
$8 = 10 - 2$
$9 = 10 - \frac{2}{2}$
$10 = 10$
$11 = 10 + \frac{2}{2}$
$12 = 12$
$14 = 10 + 2 + 2$
$16 = (10-2) \times 2$
$17 = 20 - 2 - 1$
$18 = 20 - 2$
$19 = 20 - 1$
$20 = 20$
$21 = 20 + 1$
$22 = 20 + 2$
$23 = 20 + 2 + 1$
$24 = 12 \times 2$
$32 = 20 + 12$
$38 = (20 - 1) \times 2$
$39 = (20 \times 2) - 1$
$40 = 20 \times 2$

Answer 2

Assuming

that "you can concatenate" means not only the individual digits, but also the results of their operations,

the available numbers are

0 to 12, 14 to 25, 28, 30, 32, 38, 39 and 40.

because of

the following Python script (sorry for such an awkward one!) which enumerate all possible binary trees with no more than 4 leaves and computes the result for expression denoted by each of the trees for all permutations of operators and operands (which are 2021 digits)

from itertools import permutations, groupby
from operator import itemgetter

# all permutations of 2021 digits
p = list(permutations([2, 0, 2, 1]))
# all operations
ops = [("+", lambda x, y: x + y), 
       ("-", lambda x, y: x - y), 
       ("*", lambda x, y: x * y),
       ("/", lambda x, y: x / y if  y != 0 else 1000000),
       ("|", lambda x, y: float(str(abs(int(x))) + str(abs(int(y)))))]

# set of values
v = set()

for x in p:
    # using 1 digit:
    v.add((f"{x[0]}", x[0]))
   # using 2 digits:
    for (s1, op1) in ops:
        v.add((f"{x[0]} {s1} {x[1]}", op1(x[0], x[1])))
    # using 3 digits:
    for (s1, op1) in ops:
        for (s2, op2) in ops:
            v.add((f"({x[0]} {s1} {x[1]}) {s2} {x[2]}", op2(op1(x[0], x[1]), x[2])))
            v.add((f"{x[0]} {s1} ({x[1]}) {s2} {x[2]})", op1(x[0], op2(x[1], x[2]))))
    # using 4 digits:
    for (s1, op1) in ops:
        for (s2, op2) in ops:
            for (s3, op3) in ops:
                v.add((f"({x[0]} {s1} {x[1]}) {s2} ({x[2]} {s3} {x[3]})", op2(op1(x[0], x[1]), op3(x[2], x[3]))))
                v.add((f"(({x[0]} {s1} {x[1]}) {s2} {x[2]}) {s3} {x[3]}", op3(op2(op1(x[0], x[1]), x[2]), x[3])))
                v.add((f"(({x[0]} {s1} ({x[1]} {s2} {x[2]})) {s3} {x[3]}", op3(op1(x[0], op2(x[1], x[2])), x[3])))
                
                v.add(( f"{x[0]} {s1} (({x[1]} {s2} ({x[2]}) {s3} {x[3]})", op1(x[0], op3(op2(x[1], x[2]), x[3]))))

results = list(r for r in v if 0 <= r[1] <= 40 and int(r[1]) == r[1])
results.sort(key=itemgetter(1))

abridged = [list(g)[0] for k, g in groupby(results, key=itemgetter(1))]

for s in abridged: print(s)

(Try it online!)

Note that all numbers not mentioned in hexomino's answer use concatenation of results (denoted here by $\#$):

$15= ((1 + 2) \# 0) / 2$
$25 = 2 \# (10 / 2)$
$28 = (1 + 2) \# 0 - 2$
$30 = (1 + 2) \# 0$

Answer 3

1 = 2 + 0 - 2 + 1
2 = (2 + 0) / 2 + 1
3 = 2^2 - 1 - 0
4 = 2^2 - 0/1
5 = 2 + 0 + 2 + 1
6 = (2 + 1)2 + 0
7 = 10/2 + 2
8 = (2 + 1)! + 2 - 0
9 = (2 + 1)^2 + 0/2
10 = (20/2) / 1
11 = 20 / 2 + 1
12 = 12 + 0/2
13 = 12 + (0/2)!
14 = 12 + 2 + 0
15 = C(10,2) / ( 2! + 0! )
16 = 2^ ( 0! + 1! + 2!)
17 = 20 - 2 -1
18 = 20 - (2/1)!
19 = 20 - 2! + 1!
20 = 20 / (2 -1)
21 = 20 + 2 - 1
22 = 20 + 2/1
23 = 20 + 2 + 1
24 = ( 12) ( 2 + 0)
25 = 122 + 0!
26 = 20 + (1 + 2)!
27 = (2 ! + 0! ) ^( 2 + 1)
28 = C(10-2, 2)
29 = [(2+0!)! / .2] -1
30 = (2 + 0!)! * (1/.2)
31 = 2^(1/.2) - 0!
32 = 10 + 22
33 = 2^(1/.2) + 0!
34 = (2^2)! + 10
35 = [ (2 + 1) 1] ^2 - 0!
36 = [ (2 + 1)! ] ^2 - 0
37 = 2! + 0!)! ^( 2) + 1
38 = (20-1) * 2
39 = 202 - 1
40 = 10 (2 + 2)
41 = 20 * 2 + 1
42 = 212 - 0
43 = 21 *2 + 0!
44 = 22(1 + 0!)
45 = p(10, 2)/2
46 = 10.2(bar)/0.2(bar)
47 = C(10,2) + 2
48 = (10/.2) - 2
49 =((2+1)! + 0!)^2
50 = 10^2/2
51 = 102/2
52 = (10/.2) + 2
53 = ( 12/.2 bar) - 0!
54 = ( 12/.2 bar) + 0
55 = (12 - 0!)/.2
56 = P(10-2,2)
57 =
58 =[ ( 2 + 0! )! / .1] - 2
59 = 12/.2 - 0!
60 = 20 ( 2 + 1)
61 = 12/.2 + 0!
62 = [( 2 + 0! )!/ .1] + 2
63 = 2^( [ 2 + 1]!) - 0!
64 = 2^([2 + 1)! ] + 0
65 = C(12,2) - 0!
66 = C(12,2) - 0
67 = C(12,2) + 0!
68 =
69 =
70 =
71 =
72 = (12) ( 2 + 0!)!
73 =
74 =
75 =
76 =
77 =
78 = C(12+0!, 2)
79 = [( 0! / .1 bar ) ^2] -2
80 = [2^( 2 + 0!)] / .1
81 =[( 2^0/.1 bar)^2]
82 =
83 = [ (0!/.1 bar)^2] +2
84 =
85 =
86 =
87 =
88 = P(10,2) - 2
89 = (20/.2 bar) - 1
90 = C(10,2) * 2
91 = ( 20/.2 bar) + 1
92 = P (10,2) +2
93 =
94 =
95 = (20 -1)/.2
96 =
97 =
98 = 10^2 - 2
99 = 20/.2 - 1
100 = 102 - 2

Answer 4

I tried to make the numbers using all the digits
I got 25 of them using all digits and 1 using only 3 digits

$1 = \frac{2}{2} \times 1 + 0$
$2 = 2 \times 1 + 2 \times 0$
$3 = 2 + 1 + 2 \times 0$
$4 = 2 + 2 \times 1 + 0$
$5 = 2 + 2 + 1 + 0$
$6 = 2 \times (2+1) + 0$
$7 = \frac{10}{2} + 2$
$8 = 20 - 12$
$9 = 10 - \frac{2}{2}$
$10 = 12 - 2 - 0$
$11 = 10 + \frac{2}{2}$
$12 = 12 + 2 \times 0$
$13 = $
$14 = 10 + 2 + 2$
$15 = $
$16 = (10 - 2) \times 2$
$17 = 20 - 2 - 1$
$18 = 20 - 2\times 1$
$19 = 20 - 2 +1 $
$20 = 10 \times 2$ //missing a 2
$21 = 20 + 2 - 1 $
$22 = 22 - 0 \times 1 $
$23 = 20 + 2 + 1 $
$24 = 12 \times 2 + 0 $
$25 = $
$26 = $
$27 = $
$28 = $
$29 = $
$30 = $
$31 = $
$32 = 22 + 10 $
$33 = $
$34 = $
$35 = $
$36 = $
$37 = $
$38 = (20 - 1) \times 2$
$39 = 20 \times 2 - 1$
$40 = (2+2) \times 10 $

Answer 5

I see one solution for 13 in the above answers using factorials, but if exponentiation is allowed, we can also do:

13 = 12 + 2^0

The answer from MATIUS using factorials and C(n,r) function has a solution for 15 using C(n,r) but here is one with only factorial:

15 = 12 + 2 + 0!

Alternate solutions for 26 and 27 using factorials:

26 = 2*(12 + 0!)
26 = (2+2)! + 0! + 1
27 = 21 + (2+0!)!

Answer 6

My try with all digits in order. (Used factorial and Exponents)

1 = (2 + 0 - 2) + 1

2 = ((2 + 0) ÷ 2) + 1

3 = (2 * 0) + 2 + 1

4 = (2 + 0 + 2) ÷ 1

5 = 2 + 0 + 2 + 1

6 = (2 + 0) * (2 + 1)

7 = (2 + 0!)! + (2 - 1)

8 = (2 + 0!)! + (2 ÷ 1)

9 = (2 + 0!)! + 2 + 1

10 = 20 ÷ (2 ÷ 1)

11 = (20 ÷ 2) + 1

12 = (2 + 0!)! * (2 ÷ 1)

13 = ((2 + 0!)! * 2) + 1

14 = 20 - (2 + 1)!

15 = ?

16 = 2 ˆ (0! + 2 + 1)

17 = 20 - 2 - 1

18 = (20 - 2) + 1

19 = 20 - (2 - 1)

20 = 20 ÷ (2 - 1)

21 = 20 + (2 - 1)

22 = 20 + (2 ÷ 1)

23 = 20 + 2 + 1

24 = (2 + 0 + (2 ÷ 1))!

25 = (2 + 0 + 2)! + 1

26 = 20 + (2 + 1)!

27 = (2 + 0!) ˆ (2 + 1)

28 = ?

29 = ?

30 = ?

31 = ?

32 = 2 ˆ ((- 0!) + (2 + 1)!)

33 = ?

34 = ?

35 = (((2 + 0!)!) ˆ 2) - 1

36 = (((2 + 0!)!) ˆ 2) * 1

37 = (((2 + 0!)!) ˆ 2) + 1

38 = ?

39 = (20 * 2) - 1

40 = (20 * 2) ÷ 1