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This is very similar to the 2, 0, 1, 8 problem. Just try to make all numbers 1-30 using the digits 2, 0, 1, 9.

Rules:

  • Use all four digits exactly once
  • Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root
  • Parentheses and concatenation are allowed (e.g. $20 + (1 * 9)$)
  • Squaring uses the digit 2 so expressions using multiple twos, like 22 or $12+92$, are not allowed
  • Keep the order "2 0 1 9" in all expressions
  • The modulus operator is not allowed
  • Rounding is not allowed (e.g. $201/9=22$)
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11
  • 51
    $\begingroup$ but.. but... how could you? we are not in 2019 yet! $\endgroup$
    – Kepotx
    Apr 17, 2018 at 17:07
  • $\begingroup$ Is "0 1 9 2" still considered in to be in order? $\endgroup$
    – infinitezero
    Apr 17, 2018 at 18:55
  • $\begingroup$ @infinitezero no $\endgroup$
    – jkl3699
    Apr 17, 2018 at 19:19
  • $\begingroup$ If squaring uses the two, why would square root not use both the one and the two? $\endgroup$
    – T James
    Apr 17, 2018 at 22:17
  • 16
    $\begingroup$ Just for fun, $(-2^{-(0!)})!^{-1+\sqrt9}=\pi$, if you use the gamma function for the factorial $\endgroup$
    – LegionMammal978
    Apr 17, 2018 at 23:24

10 Answers 10

Reset to default
17
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$30 = 20 + 1 + 9$
$29 = 20^1 + 9$
$28 = 20 - 1 + 9$
$27 =( 2 + 0 + 1)* 9$
$26 = 2 + 0 + (1 + \sqrt{9})!$
$25 = 2^0 + (1 + \sqrt{9})!$
$24 = (2 + 0 - 1 + \sqrt{9})!$
$23 = -2^0 + (1 + \sqrt{9})!$
$22 = -2 + 0 + (1 + \sqrt{9})!$
$21 = 2 + 0 + 19$
$20 = (2 + 0)(1 + 9)$
$19 = 20 - 1^9$
$18 = 2^{(0+1)} * 9$
$17 = -2 + 0 + 19$
$16 = 20 - (1 + \sqrt{9})$
$15 = (2 + 0 + 1)! + 9$
$14 = 20^1 - (\sqrt{9})!$
$13 = 20 - 1 - (\sqrt{9})!$
$12 = 2 + 0 + 1 + 9$
$11 = 2^{(0 + 1)} + 9$
$10 = 2 + 0 - (1 - 9)$
$9 = (2 + 0 + 1)! + \sqrt{9}$
$8 = (2 + 0)(1 + \sqrt{9})$
$7 = -2^{(0 + 1)} + 9$
$6 = 2 + 0 + 1 + \sqrt{9}$
$5 = 2^0 + 1 + \sqrt{9}$
$4 = 2 + 0 - 1 + \sqrt{9}$
$3 = (2 + 0 - 1) * \sqrt{9}$
$2 = (2 + 0) * 1^9$
$1 = (2 - 0 - 1)^9$

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1
  • 1
    $\begingroup$ There is also (2^0)*1*9 for nine $\endgroup$
    – MEE
    Apr 18, 2018 at 13:54
11
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$1 = 20 - 19 $
$2 = 2 + 0 * 1 * 9$
$3 = 2 + 0 + 1^9$
$4 = 2 + 0! + 1^9$
$5 = (2 + 0!)! - 1^9$
$6 = (2 + 0 + 1^9)!$
$7 = (2 + 0!)! + 1^9$
$8 = 2 * 0 - 1 + 9$
$9 = 2 * 0 * 1 + 9$
$10 = 2 * 0 + 1 + 9$
$11 = 2 + 0 * 1 +9$
$12 = 2 * 0! + 1 + 9$
$13 = 2 * (0! + 1) + 9$
$14 = (2+0!)!-1+9$
$15 = (2+0!)!+1*9$
$16 = (2+0!)!+1+9$
$17 = 20-\sqrt{1*9}$
$18 = 2 * 0! * 1 *9$
$19 = 20 - 1^9$
$20 = 20 * 1^9$
$21 = 20 + 1^9$
$22 = 20 - 1 + \sqrt{9}$
$23 = 20 + \sqrt{1 * 9}$
$24 = (2 + 0! + 1^9)!$
$25 = 20 -1 + \sqrt{9}!$
$26 = 20 + \sqrt{1 * 9}!$
$27 = (2 + 0 + 1) * 9$
$28 = 20 - 1 + 9$
$29 = 20^1 + 9$
$30 = 20 + 1 + 9 $

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0
3
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PARTIAL:

19:

$2*0+19$

21:

$2+0+19$

29:

$20+1*9$

30:

$20+1+9$

more incoming

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0
3
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00

$2 * 0 * 1 * 9$

01

$2 * 0 + 1^9$

02

$2 + 0 * 1 * 9$

03

$2 + 0 + 1 ^ 9$

04

$2 * 0 + 1 + \sqrt{9}$

05

$2 + 0 * 1 + \sqrt{9}$

06

$2 + 0 + 1 + \sqrt{9}$

07

$(20 + 1) / \sqrt{9}$

08

$2 * 0 + ( -1 + 9)$

09

$2 * 0 * 1 + 9$

10

$2 * 0 + 1 + 9$

11

$2 + 0 * 1 + 9$

12

$2 + 0 + 1 + 9$

13

$ 20 - 1 - \sqrt{9}!$

14

2$0 - 1 * \sqrt{9}!$

15

$20 + 1 - \sqrt{9}!$

16

$20 - 1 - \sqrt{9}$

17

$-2 + 0 + 19$

18

$-(2^0) + 19$

19

$2 * 0 + 19$

20

$2 ^ 0 + 19$

21

$2 + 0 + 19$

22

$20 - 1 + \sqrt{9}$

23

$20 + 1 * \sqrt{9}$

24

$20 + 1 + \sqrt{9}$

25

$20 - 1 + \sqrt{9}!$

26

$20 + 1 * \sqrt{9}!$

27

$20 + 1 + \sqrt{9}!$

28

$20 - 1 + 9$

29

$ 20 + 1 * 9$

30

$20 + 1 + 9$

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3
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  1. 20 - 19 = 1
    OR
    SQRT(20 - 19) = 1
    OR
    (20 - 19)! = 1

  2. 2 + 0 * 19 = 2
    OR
    20 / (1 + 9) = 2

  3. 2 + 0 + 1^9 = 3

  4. 2 * 0 + 1 + SQRT(9) = 4

  5. 2 + 0 * 1 + SQRT(9) = 5

  6. - 2 + 0 - 1 + 9 = 6
    OR
    2 + 0 + 1 + SQRT(9) = 6

  7. - 2 + 0 * 1 + 9 = 7
    OR (20 + 1) / SQRT(9) = 7

  8. 2 * 0 - 1 + 9 = 8

  9. 2 * 0 * 1 + 9 = 9
    OR
    2^0 - 1 + 9 = 9

  10. 20 - 1 - 9 = 10
    OR
    20 - (1 + 9) = 10
    OR
    2 * 0 + 1 + 9 = 10

  11. 2 + 0 * 1 + 9 = 11
    OR
    20^1 - 9 = 11
    OR
    2 * 01 + 9 = 11
    OR
    20 * 1 - 9 = 11

  12. 20 + 1 - 9 = 12
    OR
    2 + 0 + 1 + 9 = 12
    OR
    2 - 0 + 1 + 9 = 12

  13. 20 - 1 - (SQRT(9))! = 13

  14. 20 - 1 * (SQRT(9))! = 14

  15. 20 + 1 - (SQRT(9))! = 15

  16. 20 - 1 - SQRT(9) = 16
    OR
    20 - (1 + SQRT(9)) = 16

  17. 20 - 1 * SQRT(9) = 17

  18. 20 + 1 - SQRT(9) = 18
    OR
    (2 + 0) * 1 * 9 = 18
    OR
    (2 + 0) / (1 / 9) = 18

  19. 2 * 0 + 19 = 19

  20. 2^0 + 19 = 20

  21. 2 + 0 + 19 = 21

  22. 20 - 1 + SQRT(9) = 22

  23. 20 * 1 + SQRT(9) = 23

  24. 20 + 1 + SQRT(9) = 24

  25. 20 - 1 + (SQRT(9))! = 25

  26. 20^1 + (SQRT(9))! = 26

  27. 20 + 1 + (SQRT(9))! = 27

  28. 20 - 1 + 9 = 28

  29. 20 * 1 + 9 = 29

  30. 20 + 1 + 9 = 30

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2
  • $\begingroup$ I really like how you gave "or" options. That said, you missed a few more that make 10. $\endgroup$
    – sirjonsnow
    Apr 18, 2018 at 14:11
  • $\begingroup$ The ORs were as I found them, and not intended to be a complete solution. I have added a few for 10. $\endgroup$
    – NetJohn
    Apr 18, 2018 at 15:17
2
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Here they are, as simple and neat as I could make them!

$1-6:$

$$\small\begin{array}{c|c}1&2&3&4&5&6\\\hline2\cdot0+1^9&2+0\cdot1\cdot9&2\cdot0\cdot1+\sqrt9&2+0!+1^9&-2-0!-1+9&-2+0-1+9\end{array}$$

$7-12:$

$$\small\begin{array}{c|c}7&8&9&10&11&12\\\hline-2+0\cdot1+9&-2+0+1+9&2\cdot0\cdot1+9&2+0-1+9&2+0\cdot1+9&2+0+1+9\end{array}$$

$13-18:$

$$\small\begin{array}{c|c}13&14&15&16&17&18\\\hline2+0!+1+9&(2+0!)!-1+9&20+1-\sqrt9!&(2+0!)!+1+9&-2+0+19&2\cdot(0+1)\cdot9\end{array}$$

$19-25:$

$$\small\begin{array}{c|c}19&20&21&22&23&24&25\\\hline2\cdot0+19&2^0+19&20+1^9&2+0!+19&20\cdot1+\sqrt9&20+1+\sqrt9&(2+0!)!+19\end{array}$$

$26-30:$

$$\small\begin{array}{c|c}26&27&28&29&30\\\hline20+1\cdot\sqrt9!&20+1+\sqrt9!&20-1+9&20\cdot1+9&20+1+9\end{array}$$

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0
2
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Full solution

$20-19 = 1 $
$2 + 0 * 19 = 2 $
$2 * 0 * 1 + \sqrt{9} = 3 $
$2 * 0 + 1 + \sqrt{9} = 4$
$2 + 0 * 1 + \sqrt{9} = 5$
$-2 + 0 - 1 + 9 = 6$
$-2 + 0 * 1 + 9 = 7$
$2 * 0 - 1 + 9 = 8$
$2 * 0 * 1 + 9 = 9$
$20 - 1 - 9 = 10$
$20^1 - 9 = 11$
$20 + 1 - 9 = 12$
$20 - 1 - (\sqrt{9})! = 13$
$2 * (0 + 1 + (\sqrt{9})! = 14$
$(2 + 0 + 1)! + 9 = 15$
$2 * (0 - 1 + 9) = 16$
$-2 + 0 + 19 = 17$
$-2^0 + 19 = 18$
$2 * 0 + 19 = 19$
$20! / 19! = 20$
$2 + 0 + 19 = 21$
$20 - 1 + \sqrt{9} = 22$
$20 * 1 + \sqrt{9} = 23$
$20 + 1 + \sqrt{9} = 24$
$20 - 1 + (\sqrt{9})! = 25$
$20 + 1 * (\sqrt{9})! = 26$
$(2 + 0 + 1) * 9 = 27$
$20 - 1 + 9 = 28$
$20 ^1 + 9 = 29$
$20 + 1 + 9 = 30$

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2
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1

$20-19$

2

$2+(0*19)$

3

$(2*0*1)+√9$

4

$(2*0)+√(9)+1$

5

$2+(0*1)+√(9)$

6

$2+0+1+√(9)$

7

$-2+0+(1*9)$

8

$(2*0)-1+9$

9

$(2*0*1)+9$

10

$(2*0)+1+9$

11

$2+(0*1)+9$

12

$2+0+1+9$

13

$20-1-(√9)!$

14

$20*1-(√9)!$

15

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

16

$2^(0+1+√9)$

17

$20-(1*√9)$

18

$(2+0*1)*9$

19

$(2*0)+19$

20

$(2^0)+19$

21

$2+0+19$

22

$20-1+√9$

23

$(20*1)+√9$

24

$20+1+√9$

25

$20-1+(√9)!$

26

$(20*1)+(√9)!$

27

$(2+0+1)*9$

28

$20-1+9$

29

$(20*1)+9$

30

$20+1+9$

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2
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  • 1= ((2 + 0 + 1) / √9)!
  • 2= 2 + 0 * 1 * 9
  • 3 = 2 + (0 * 1 * 9)!
  • 4 = 2^(0 - 1 + √9)
  • 5 = 2 + 0 * 1 + √9
  • 6 = -(2 + 0 + 1) + 9
  • 7 = -2 + 0 * 1 + 9
  • 8 = -2 + 0 + 1 + 9
  • 9 = 2 * 0 * 1 + 9
  • 10 = 2 * 0 + 1 + 9
  • 11 = 2 + 0 * 1 + 9
  • 12 = 2 + 0 + 1 + 9
  • 13 = 2 + 0! + 1 + 9
  • 14 = 2 * -(0! + 1 - 9)
  • 15 = (2 + 0 + 1)! + 9
  • 16 = -2 + (0! + 1) * 9
  • 17 = -2 + 0 + 19
  • 18 = 2 * (0 * 1 + 9)
  • 19 = 2 * 0 + 19
  • 20 = 20 * 1^9
  • 21 = 20 + 1^9
  • 22 = 20 - 1 + √9
  • 23 = 20 * 1 + √9
  • 24 = 20 + 1 + √9
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1
  • 1
    $\begingroup$ Welcome to PSE! Please use spoilers '>!' to hide your answer. Happy puzzling = ) $\endgroup$
    – 19aksh
    Jul 13, 2019 at 11:27
2
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Solutions for 25-36

  • 25 = 2 - 0! + (1 + √9)!

  • 26 = 2 + 0 + (1 + √9)!

  • 27 = (2 + 0 + 1) * 9

  • 28 = 20 - 1 + 9

  • 29 = 20 + 1 * 9

  • 30 = 20 + 1 + 9

  • 31 = 2 || 0! + 1 + 9

  • 32 = 2 ^ (0! + 1 + √9)

  • 33 = (2 - 0!) || 1 * √9

  • 34 = 2 || 0! + 1 || √9

  • 35 = —not here yet—

  • 36 = (2 + 0!)! ^ (-1 + 9)

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1
  • 2
    $\begingroup$ I believe you should merge this into your previous answer of solutions from 1-24, but nice answer nevertheless. $\endgroup$
    – Mr Pie
    Jul 14, 2019 at 2:41

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