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Try to make all numbers 1-40 using the digits 2, 0, 2, 6.

Rules:

Use all four digits exactly once.

Allowed operations: +,−,×,÷,! (factorial), $x^y$ (exponentiation), √ (square root).

Parentheses and concatenation are allowed. (e.g. 20+2×6).

Squaring uses the digit 2.

The modulus operator is not allowed.

Rounding is not allowed.

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    $\begingroup$ Welcome to Puzzling Stack Exchange. Friendly tip: this kind of question is better received if you're not asking for a giant list. For example, you could try to solve it yourself, and then ask for one particular number that you know to be tricky. $\endgroup$ Commented Aug 27 at 15:20
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    $\begingroup$ Why is there a $4$ in the example? And must the digits be ordered to make it more challenging? $\endgroup$ Commented Aug 27 at 19:11
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    $\begingroup$ Did you mean "sqrt consumes the digit 2"? The $x^{y}$ is allowed which would consume y=2 so why mention 2 specifically? $\endgroup$ Commented Aug 27 at 21:01
  • $\begingroup$ @WeatherVane - Looks like they took it directly from this - puzzling.stackexchange.com/questions/124968/… $\endgroup$ Commented Aug 28 at 14:01
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    $\begingroup$ @MacGyver88 I see – the wording is identical except for being 2024 and allowing concatentation, with the 4 copy/pasted too. $\endgroup$ Commented Aug 28 at 14:11

2 Answers 2

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Here are my answers:

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

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    $\begingroup$ Nice... also the simpler $32 = 20 + 2 \times 6$ which is in order :) $\endgroup$ Commented Aug 27 at 21:06
  • $\begingroup$ @WeatherVane. Yes, but where’s the fun in that? ;) $\endgroup$ Commented Aug 27 at 21:15
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    $\begingroup$ Also $ 40 = 6! \div (20 - 2) $ $\endgroup$ Commented Aug 27 at 21:32
  • $\begingroup$ @WeatherVane. Ah, that's a good one! I tried a lot to make 40 out of 5! or 6!, but didn't think of this. $\endgroup$ Commented Aug 27 at 21:39
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    $\begingroup$ @WeatherVane. It's also nice that, with your expression for 40, you don't need the sqrt to make these numbers. Occasional exponentiation and mostly just arithmetic and factorial. $\endgroup$ Commented Aug 27 at 21:47
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Here are my answers (just pattern matching, no specific method except for consecutive numbers sometimes):

1 = 2/2+0x6
2 = 2+0x2x6
3 = 6/2+0x2
4 = 2x2+0x6
5 = 6-2/2+0
6 = 6+0x2x2
7 = 6+2/2+0
8 = 6+2+0x2
9 = 6+2+2^0
10 = 6+2+2+0
11 = 6x2-2^0
12 = 6x2+0x2
13 = 6x2+2^0
14 = 6x2+2+0
15 = 6x2+2+0!
16 = (6+2)x2+0
17 = (6+2)x2+0!
18 = 6^2/2+0
19 = 6^2/2+0!
20 = (2+2)x(6-0!)
21 = 22-6^0
22 = 22+0x6
23 = 22+6^0
24 = 26-2+0
25 = 26-2^0
26 = 26+0x2
27 = 26+2^0
28 = 26+2+0
29 = 26+2+0!
30 = 62/2-0!
31 = 62/2+0
32 = 60/2+2
33 = 6^2-2-0!
34 = 6^2-2+0
35 = 6^2-2^0
36 = 6^2+0x2
37 = 6^2+2^0
38 = 6^2+2+0
39 = 6^2+2+0!
40 = 20x2 (not sure how to incorporate 6 here)

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    $\begingroup$ The instructions say "Use all four digits exactly once." which makes that answer for 40 invalid, unfortunately. $\endgroup$ Commented Aug 27 at 15:21

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