11
$\begingroup$

Inspired by the last year's "2018 four 4s challenge", I thought it's time to welcome 2019 by a similar challenge. This time you have to use the digits 2, 0, 1, 9 in this particular order to create the numbers 1 - 30.

The rules haven't changed:

  • Use all four digits exactly once in the order 2-0-1-9.
  • Allowed operations: $+, -, \cdot, \div, !$ (factorial), $!!$ (double factorial), square root, exponentiation.
  • Parentheses and grouping (e.g. "19") are also allowed.
  • Squaring uses the digit 2, so expressions using multiple 2's, e. g. $2^2$ or $1^2+2^9$, are not allowed.
  • The modulus operator $(\%, \mod)$ is not allowed.
  • Rounding (e.g. 201/9=22) is not allowed.

I'm curious to see your creative solutions!

May each day of 2019 bring happiness, good cheer, and sweet surprises to you and all your dear ones!

Happy New Year and greetings from Germany! André

$\endgroup$
2
  • $\begingroup$ Will you be giving a green check to someone? $\endgroup$
    – flashstorm
    Commented Dec 31, 2018 at 22:46
  • $\begingroup$ Of course I'll do. $\endgroup$
    – NAMELESS
    Commented Jan 1, 2019 at 2:40

4 Answers 4

7
$\begingroup$

1

1 = 2^(0*19)

2

2 = 2 + (0*19)

3

3 = 2 + 0!^19

4

4 = 2 ^ (0! + 1 ^ 9)

5

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

6

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

7

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

8

-((2 - 01) - 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 - (1 * sqrt(9))

18

(2 + (0 * 1)) * 9

19

20 - 1^9

20

20 * 1^9

21

20 + 1^9

22

2 + 0! + 19

23

20 + 1 * sqrt(9)

24

20 + 1 + sqrt(9)

25

2 || (0! + 1 + sqrt (9))

explained:

Ok, this one needs explanation. The operation for "grouping" is known as concatenation and represented by ||. Basically this means push the digits together: 2 || 0 = 20. However, just like any operation, you can represent either side not by a number but by an equation on its own. So 0! + 1 + sqrt(9) = 5, meaning the above represents 2 || 5, qed.

26

2 || ((0! + 1) * sqrt(9))

27

(2 + 0 + 1) * 9

28

((2 + 0!) || 1) - sqrt(9)

29

(2 + (0 * 1)) || 9

30

20 + 1 + 9

$\endgroup$
4
  • $\begingroup$ edited for clarity $\endgroup$
    – flashstorm
    Commented Dec 31, 2018 at 18:59
  • $\begingroup$ @flashstorm Um... $3\ne 2+0^{19}$ $\endgroup$ Commented Dec 31, 2018 at 19:13
  • 1
    $\begingroup$ was missing an ! $\endgroup$
    – flashstorm
    Commented Dec 31, 2018 at 19:19
  • $\begingroup$ Ding! Fries are done :) $\endgroup$
    – flashstorm
    Commented Dec 31, 2018 at 19:31
11
$\begingroup$

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

DONE!

$\endgroup$
7
  • $\begingroup$ All finished now! :D $\endgroup$ Commented Dec 31, 2018 at 19:48
  • $\begingroup$ Great :) But Spoiler-Tags would be nice ;) $\endgroup$
    – NAMELESS
    Commented Dec 31, 2018 at 21:23
  • $\begingroup$ @André Oh, sorry! I can't figure out how to get the mathjax to work inside of a spoiler tag. D:< $\endgroup$ Commented Dec 31, 2018 at 21:26
  • 1
    $\begingroup$ @André Also: I've been trying to do 31, but it actually seems to be much harder than any of the previous ones! Looks like you picked just the right number to stop on! XD $\endgroup$ Commented Dec 31, 2018 at 21:27
  • 1
    $\begingroup$ Your answer for 6 is incorrect; that expression comes out to 2. However, adding parentheses around part and another factorial will make it work. $\endgroup$ Commented Dec 31, 2018 at 21:41
5
$\begingroup$

I wrote a program to determine all representable numbers between 1 and 1,000,000 following the rules, so this should be a comprehensive list.

0 through 30:

$$0 = 2\times0\times\left(1+9\right)$$ $$1 = 20-19$$ $$2 = 2+0\times\left(1+9\right)$$ $$3 = 2+0+1^{9}$$ $$4 = 2-\left(0+1-\sqrt{9}\right)$$ $$5 = \frac{20}{1+\sqrt{9}}$$ $$6 = -2-\left(0+1-9\right)$$ $$7 = 2\times\left(0-1\right)+9$$ $$8 = 2\times0-\left(1-9\right)$$ $$9 = 2\times0+1\times9$$ $$10 = 2-\left(0+1-9\right)$$ $$11 = 2+0+1\times9$$ $$12 = 2+0+1+9$$ $$13 = 20-\left(1+\left(\sqrt{9}\right)!\right)$$ $$14 = 20-1\times\left(\sqrt{9}\right)!$$ $$15 = 20+1-\left(\sqrt{9}\right)!$$ $$16 = 2\times\left(0-\left(1-9\right)\right)$$ $$17 = -2+0+19$$ $$18 = 2\times\left(0+1\times9\right)$$ $$19 = 2\times0+19$$ $$20 = 2\times\left(0+1+9\right)$$ $$21 = 2+0+19$$ $$22 = 20-\left(1-\sqrt{9}\right)$$ $$23 = 20+1\times\sqrt{9}$$ $$24 = 20+1+\sqrt{9}$$ $$25 = 20-\left(1-\left(\sqrt{9}\right)!\right)$$ $$26 = 20+1\times\left(\sqrt{9}\right)!$$ $$27 = \left(2+0+1\right)\times9$$ $$28 = 20-\left(1-9\right)$$ $$29 = 20+1\times9$$ $$30 = 20+1+9$$

31 through 100. Concatenation of expressions (denoted with ||) is only used when necessary. Interestingly enough, 31 is the smallest number that cannot be done without concatenation of expressions, and 75 is the smallest number that cannot be done in general.

$$31 = 2+\left(\left(0!+1\right)||9\right)$$ $$32 = \sqrt{2^{0+1+9}}$$ $$33 = \left(2+0!+1\right)!+9$$ $$34 = \left(2+0!\right)||\left(1+\sqrt{9}\right)$$ $$35 = 20+\left(-1+\left(\sqrt{9}\right)!\right)!!$$ $$36 = 2\times\left(-0!+19\right)$$ $$37 = \left(2+0!\right)||\left(1+\left(\sqrt{9}\right)!\right)$$ $$38 = 2\times\left(0+19\right)$$ $$39 = 20+19$$ $$40 = 20\times\left(-1+\sqrt{9}\right)$$ $$41 = \left(\left(2+0!\right)!\right)!!-\left(1+\left(\sqrt{9}\right)!\right)$$ $$42 = \left(2+0!\right)!\times\left(1+\left(\sqrt{9}\right)!\right)$$ $$43 = \left(\left(2+0!\right)!\right)!!+1-\left(\sqrt{9}\right)!$$ $$44 = 20+\left(1+\sqrt{9}\right)!$$ $$45 = \left(\left(2+0!\right)!-1\right)\times9$$ $$46 = 2\times\left(-0!+\left(1+\sqrt{9}\right)!\right)$$ $$47 = 2\times0-\left(1-\left(\left(\sqrt{9}\right)!\right)!!\right)$$ $$48 = 2\times\left(0+\left(1+\sqrt{9}\right)!\right)$$ $$49 = 2-\left(0+1-\left(\left(\sqrt{9}\right)!\right)!!\right)$$ $$50 = 2\times\left(0!+\left(1+\sqrt{9}\right)!\right)$$ $$51 = 2+0+1+\left(\left(\sqrt{9}\right)!\right)!!$$ $$52 = 2+0!+1+\left(\left(\sqrt{9}\right)!\right)!!$$ $$53 = \left(2+0!\right)!-\left(1-\left(\left(\sqrt{9}\right)!\right)!!\right)$$ $$54 = \left(2+0!\right)!\times1\times9$$ $$55 = \left(2+0!\right)!+1+\left(\left(\sqrt{9}\right)!\right)!!$$ $$56 = \left(\left(2+0!\right)!\right)!!-\left(1-9\right)$$ $$57 = \left(20-1\right)\times\sqrt{9}$$ $$58 = \left(\left(2+0!\right)!\right)!!+1+9$$ $$59 = \left(\left(2+0!\right)!-1\right)||9$$ $$60 = 20\times1\times\sqrt{9}$$ $$61 = \left(2+0!\right)!||\left(1^{9}\right)$$ $$62 = -2+\left(0!+1\right)^{\left(\sqrt{9}\right)!}$$ $$63 = \left(20+1\right)\times\sqrt{9}$$ $$64 = 2^{0+1\times\left(\sqrt{9}\right)!}$$ $$65 = \left(2+0!\right)!||\left(-1+\left(\sqrt{9}\right)!\right)$$ $$66 = 2+\left(0!+1\right)^{\left(\sqrt{9}\right)!}$$ $$67 = \frac{201}{\sqrt{9}}$$ $$68 = 20+1\times\left(\left(\sqrt{9}\right)!\right)!!$$ $$69 = -2+\sqrt{0!+\left(1+\left(\sqrt{9}\right)!\right)!}$$ $$70 = \left(\left(2+0!\right)!||1\right)+9$$ $$71 = \sqrt{2^{0}+\left(1+\left(\sqrt{9}\right)!\right)!}$$ $$72 = \left(2+0!\right)\times\left(1+\sqrt{9}\right)!$$ $$73 = 2+\sqrt{0!+\left(1+\left(\sqrt{9}\right)!\right)!}$$ $$74 = 2\times\left(\left(-0!||1\right)+\left(\left(\sqrt{9}\right)!\right)!!\right)$$ $$76 = \left(\left(2+0!\right)!+1\right)||\left(\sqrt{9}\right)!$$ $$78 = \left(2+0!\right)!\times\left(1||\sqrt{9}\right)$$ $$79 = \left(\left(2+0!\right)!+1\right)||9$$ $$80 = 20\times\left(1+\sqrt{9}\right)$$ $$81 = \left(2+0!\right)^{1+\sqrt{9}}$$ $$83 = \left(2+0!+1\right)!!||\sqrt{9}$$ $$84 = \left(2||0!\right)\times\left(1+\sqrt{9}\right)$$ $$85 = -20+\left(1+\left(\sqrt{9}\right)!\right)!!$$ $$86 = \left(2+0!+1\right)!!||\left(\sqrt{9}\right)!$$ $$89 = \left(2+0!+1\right)!!||9$$ $$90 = \frac{\left(\left(2+0!\right)!\right)!}{-1+9}$$ $$92 = 2\times\left(-0!-\left(1-\left(\left(\sqrt{9}\right)!\right)!!\right)\right)$$ $$93 = \left(\left(2+0!\right)||1\right)\times\sqrt{9}$$ $$94 = 2\times\left(0-\left(1-\left(\left(\sqrt{9}\right)!\right)!!\right)\right)$$ $$95 = \left(\left(2+0!\right)!\right)!!-\left(1-\left(\left(\sqrt{9}\right)!\right)!!\right)$$ $$96 = 2\times\left(0+1\times\left(\left(\sqrt{9}\right)!\right)!!\right)$$ $$97 = \left(\left(2+0!\right)!\right)!!+1+\left(\left(\sqrt{9}\right)!\right)!!$$ $$98 = 2\times\left(0+1+\left(\left(\sqrt{9}\right)!\right)!!\right)$$ $$99 = \left(\left(2+0!\right)!+1\right)!!-\left(\sqrt{9}\right)!$$ $$100 = 20\times\left(-1+\left(\sqrt{9}\right)!\right)$$

The rest of the numbers from 101 to 1,000,000 can be found here.

$\endgroup$
1
  • $\begingroup$ Nice. I'm curious whether your program finds anything for 28 other than $20 - (1-9)$ and $20 - 1 +9$, etc. $\endgroup$
    – user33097
    Commented Jan 2, 2019 at 13:37
4
$\begingroup$

1-16 were done as of the time I started this, so I wanted to focus only on 17-30, using answers different from ones that I've already seen. Others will be included if I find solutions that I like.

8:

$8 = \sqrt{(2^{0!+1}!!)!! / (\sqrt9)!}$

16:

$16 = ((2 + 0 + 1)!)!! / \sqrt9 = 20 - 1 - \sqrt9$

17:

$17 = (2 + 0! + 1)!! + 9$

18:

$18 = (2^0 + 1) \cdot 9 = 2 \cdot (0+1)\cdot 9 = 2^{0+1} \cdot 9$

19:

$19 = 2 \cdot 0 + 19$

20:

$20 = 2^0 + 19 = 20! / 19!$

21:

$21 = 20 + 1^9$

22:

$22 = 20 - 1 + \sqrt9$

23:

$23 = 20 + 1 \cdot \sqrt9$

24:

$24 = 2^{0! + 1} \cdot (\sqrt9)! = 2^{0! + 1}!! \cdot \sqrt9 = (2+0!+1)!! \cdot \sqrt9 = ((2 + 0!)! - 1)!! + 9$

25:

$25 = (2 + 0!)! + 19$

26:

$26 = 20 + 1 \cdot (\sqrt9)!$

27:

$27 = 2^{0! + 1}! + \sqrt9$

28: Can't get one different from what I've already seen in other answers. Will maybe try later.

29:

$29 = 20 + 1 \cdot 9$

30:

$30 = 2^{0! + 1}! + (\sqrt9)!$

I know we're supposed to stop at 30, but I accidentally found this fun one:

32:

$32 = \sqrt{20!! / (1 + 9)!}$

$\endgroup$
3
  • $\begingroup$ Well, there's a way to construct the number 31 satisfying the rules above (at least if we allow the subfactorial) :) $\endgroup$
    – NAMELESS
    Commented Jan 3, 2019 at 1:43
  • $\begingroup$ @André do you mean without concatenation? $\endgroup$
    – user33097
    Commented Jan 3, 2019 at 1:44
  • $\begingroup$ Yes, without concatenation! Hint: Try using the "!!! faculty". $\endgroup$
    – NAMELESS
    Commented Jan 3, 2019 at 2:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.