Can you beat this burning 2014 game?

What is the smallest positive integer which cannot be formed using the digits 2, 0, 1, 4 in that order? You are allowed to use +,,×,÷ +, -, \times, \div, exponential, factorials, etc. Be creative, and share what you can get. Let's see how high we can go.

To get started, I have:
1=201+4 1 = -2 - 0 - 1 + 4
2=20×1+4 2 = - 2 - 0 \times 1 + 4
3=2×01+4 3 = 2 \times 0 - 1 + 4
4=2×0×1+4 4 = 2 \times 0 \times 1 + 4


For clarity, the only numbers which appear are 2,0,1,42, 0, 1, 4 in that order. You cannot use a greek symbol to present a number ( e.g. π=3.14159 \pi = 3.14159\ldots . The square root sign \sqrt{ \, } is borderline, since it actually represents 2 \sqrt[2]{ \, } , but we'd allow it. You can't use cube root, unless you have a 3, as in 3 \sqrt[3]{ \, } .

Note by Chung Kevin
5 years, 9 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

5 = 2+0-1 + 4

6 = 2 + 0 X 1 + 4

7 = 2 + 0 + 1 + 4

8 = (2 + 0 X 1) X 4

I can't get 9, maybe something like 2X4 +1?

Ajala Singh - 5 years, 9 months ago

Log in to reply

9=(2+0+1)49=\sqrt{(2+0+1)^4} is allowed?

Jorge Tipe - 5 years, 9 months ago

Log in to reply

If that isn't allowed, there's always (20!)(14)(-2-0!)\cdot (1-4).

Bob Krueger - 5 years, 9 months ago

Log in to reply

10=(2+0)×(1+4)10=(2+0)\times(1+4)

Pero no puedo obtener 11...

Kenny Lau - 5 years, 9 months ago

Log in to reply

@Kenny Lau 11=(2+0!)!+1+411=(2+0!)!+1+4

Jorge Tipe - 5 years, 9 months ago

Log in to reply

@Jorge Tipe 12=(2+0+1)×412=(2+0+1)\times 4

13=20+1413=-2^0+14

14=2×0+1414=2\times 0+14

15=20+1415=2^0+14

16=2+0+1416=2+0+14

17=2+0!+1417=2+0!+14

18=2+(0!+1)418=2+(0!+1)^4

19=201419=20-1^4

20=(2+0!)!+1420=(2+0!)!+14

21=201+4!21=-2-0-1+4!

22=2+0×1+4!22=-2+0\times 1+4!

23=2×01+4!23=2\times 0-1+4!

24=2×0×1+4!24=2\times 0\times 1+4!

25=2×0+1+4!25=2\times 0+1+4!

26=2+0×1+4!26=2+0\times 1+4!

27=2+0+1+4!27=2+0+1+4!

28=2+0!+1+4!28=2+0!+1+4!

29=(2+0!)!1+4!29=(2+0!)!-1+4!

30=(2+0!)!×1+4!30=(2+0!)!\times 1+4!

31=(2+0!)!+1+4!31=(2+0!)!+1+4!

32=2((0!+1)4)32=2((0!+1)^4)

33=ϕ((2+0!+1)!)+4!33=\phi((2+0!+1)!)+4!

34=20+1434=20+14

Is combining digits allowed? How about totient function?

Daniel Chiu - 5 years, 9 months ago

Log in to reply

@Daniel Chiu 35=((2+0!)!+1)!÷4 35 = \left \lfloor \sqrt{ \left ( (2+0!)! + 1 \right )! } \right \rfloor \div \sqrt {4}

36=(2+0!)!(1×4) 36 = \sqrt {\left ( 2+0! \right )!^{(1 \times 4)} }

37=ϕ((((2+0!)!)!)+(1×4!) 37 = \left \lfloor \sqrt{ \phi ( ( ((2+0!)!)!) } \right \rfloor + (1 \times 4!)

38=ϕ((((2+0!)!)!)+(1×4!) 38 = \left \lceil \sqrt{ \phi ( ( ((2+0!)!)!) } \right \rceil + ( 1 \times 4!)

39=ϕ((((2+0!)!)!)+1+4! 39 = \left \lceil \sqrt{ \phi ( ( ((2+0!)!)!) } \right \rceil + 1 + 4!

40=ϕ(ϕ((2+0!)!))×(1+4) 40 = \sqrt { \phi \left ( \phi \left ( (2+0!)! \right ) \right ) } \times (1 + 4)

41=((2+0!)!)!+14 41 = \left \lceil \sqrt{((2+0!)!)!} \right \rceil + 14

42=ϕ(((2+0!)!)!)+1×4! 42 = \phi \left ( \left \lceil \sqrt{((2+0!)!)!} \right \rceil \right ) + 1 \times 4!

43=ϕ(((2+0!)!)!)+1+4! 43 = \phi \left ( \left \lceil \sqrt{((2+0!)!)!} \right \rceil \right ) + 1 + 4!

44=2014 44 = \left \lfloor \sqrt{2014} \right \rfloor

45=2014 45 = \left \lceil \sqrt{2014} \right \rceil

46=ϕ((ϕ(((2+0!)!)!))!1+4!) 46 = \phi \left ( \left ( \left \lceil \sqrt {\sqrt { \phi ( ((2+0!)!)! )} } \right \rceil \right ) ! - 1 + 4! \right )

47=(ϕ(((2+0!)!)!))!1+4! 47 = \left ( \left \lceil \sqrt {\sqrt { \phi ( ((2+0!)!)! )} } \right \rceil \right ) ! - 1 + 4!

48=(2+0!+1)!+4! 48 = (2+0!+1)! +4!

49=(ϕ(((2+0!)!)!))!+1+4! 49 = \left ( \left \lceil \sqrt {\sqrt { \phi ( ((2+0!)!)! )} } \right \rceil \right ) ! + 1 + 4!

50=201÷4 50 = \left \lfloor 201 \div 4 \right \rfloor

51=201÷4 51 = \left \lceil 201 \div 4 \right \rceil

52=ϕ(((2+0!)!)!)×1×4 52 = \left \lfloor \sqrt { \phi \left ( ((2+0!)!)! \right ) } \right \rfloor \times 1 \times 4

53=ϕ(20!)+1ϕ(4!) 53 = \phi \left ( \left \lfloor \sqrt { \sqrt { \sqrt{20!}}} \right \rfloor \right ) + 1 - \phi (4!)

54=ϕ(((2+0!)!)!)×1×ϕ(4!) 54 = \phi \left ( \left \lceil \sqrt{ ((2+0!)!)! } \right \rceil \right ) \times 1 \times \left \lfloor \sqrt {\phi (4!) } \right \rfloor

55=ϕ(ϕ(ϕ(((2+0!)!)!)))1+4! 55 = \phi ( \phi ( \phi ( ((2+0!)!)!))) - 1 + 4!

56=ϕ(((2+0!)!)!)×1×ϕ(ϕ(4!)) 56 = \left \lceil \sqrt{ \phi ( ((2+0!)!)! ) } \right \rceil \times 1 \times \left \lceil \phi ( \phi (4!)) \right \rceil

57=ϕ(ϕ(ϕ(((2+0!)!)!)))+1+4! 57 = \phi ( \phi ( \phi ( ((2+0!)!)!))) + 1 + 4!

58=ϕ(ϕ(((2+0!)!)!))ϕ((1+4)!) 58 = \phi ( \phi ( ((2+0!)!)!)) - \left \lceil \sqrt{ \phi ( ( 1+4)! ) } \right \rceil

59=ϕ(ϕ(((2+0!)!)!))14 59 = \phi ( \phi ( ((2+0!)!)!)) - 1 - 4

60=ϕ(ϕ(((2+0!)!)!))(1×4) 60 = \phi ( \phi ( ((2+0!)!)!)) - ( 1 \times 4)

61=ϕ(ϕ(((2+0!)!)!))+14 61 = \phi ( \phi ( ((2+0!)!)!)) + 1 - 4

62=ϕ(ϕ(((2+0!)!)!))(1×4) 62 = \phi ( \phi ( ((2+0!)!)!)) - ( 1 \times \sqrt{4})

63=ϕ(ϕ(((2+0!)!)!))+14 63 = \phi ( \phi ( ((2+0!)!)!)) + 1 - \sqrt{4}

64=(ϕ(ϕ(((2+0!)!)!)))×(14) 64 = \left ( \phi ( \phi ( ((2+0!)!)!)) \right ) \times (1^4)

65=ϕ(ϕ(((2+0!)!)!))+1+4 65 = \phi ( \phi ( ((2+0!)!)!)) + 1 + \sqrt{4}

66=(ϕ(ϕ(((2+0!)!)!)))×1+4 66 = \left ( \phi ( \phi ( ((2+0!)!)!)) \right ) \times 1 + \sqrt{4}

Pi Han Goh - 5 years, 9 months ago

Log in to reply

@Pi Han Goh Oh wow. It seems possible that with factorial, totient, floor/ceiling and roots, you can get wriggle around a lot.

What if square roots were not allowed?

Calvin Lin Staff - 5 years, 9 months ago

Log in to reply

@Calvin Lin Oh actually I can construct every positive integer. Take ((((((0+1+4)!)!)!)!)!)!((((((0+1+4)!)!)!)!)!)!, or an arbitrarily large number of factorials, then apply the totient function a number of times such that the number is the desired power of 2, then take log 2\log\ 2. n=log2(ϕ(ϕ((ϕ(ϕ(((((0+1+4)!)!)!)!))))n=\log_2 (\phi(\phi(\cdots(\phi(\phi(((\cdots((0+1+4)!)!\cdots)!)!)\cdots))) This works since repeatedly applying the totient function eventually gives a power of 2, and the totient of a power of 2 is the previous power of 2.

For example, 50=log2(ϕ501(((0+1+4)!)!))50=\log_2(\phi^{501}(((0+1+4)!)!)) where ϕn\phi^n denotes applying the totient function nn times. Calculated via Mathematica.

Daniel Chiu - 5 years, 9 months ago

Log in to reply

@Daniel Chiu Why does totient eventually make a power of 2?

Michael Tang - 5 years, 9 months ago

Log in to reply

@Michael Tang First, I'll assume you know the formula for the totient function.

When applying totient to an odd prime, another 2, along with possibly some other odd primes, is generated. With each totient, the power on 2 increases/stays the same unless the number is a power of 2. Eventually, all odd primes are gone, and there is a large power of 2 remaining.

Daniel Chiu - 5 years, 9 months ago

Log in to reply

@Daniel Chiu That's pretty cool :) Of course, the next question is, what happens if the totient function is not allowed?

Evan Robinson - 5 years, 9 months ago

Log in to reply

@Evan Robinson Well, without the totient function, I can't find a way to make 33.

Daniel Chiu - 5 years, 9 months ago

Log in to reply

@Calvin Lin 35=ϕ(ϕ(ϕ((2+0!)!)))1435=\phi(\phi(\phi((2+0!)!)))-1-4

36=ϕ(ϕ(ϕ((2+0!)!)))×1436=\phi(\phi(\phi((2+0!)!)))\times 1-4

37=ϕ(ϕ(ϕ((2+0!)!)))+1437=\phi(\phi(\phi((2+0!)!)))+1-4

38=ϕ(ϕ(ϕ((2+0!)!)))1×ϕ(4)38=\phi(\phi(\phi((2+0!)!)))-1\times\phi(4)

39=ϕ(ϕ(ϕ((2+0!)!)))+1ϕ(4)39=\phi(\phi(\phi((2+0!)!)))+1-\phi(4)

40=ϕ(ϕ(ϕ((2+0!)!)))+1ϕ(ϕ(4))40=\phi(\phi(\phi((2+0!)!)))+1-\phi(\phi(4))

41=ϕ(ϕ(ϕ((2+0!)!)))+1×ϕ(ϕ(4))41=\phi(\phi(\phi((2+0!)!)))+1\times\phi(\phi(4))

42=ϕ(ϕ(ϕ((2+0!)!)))+1×ϕ(4)42=\phi(\phi(\phi((2+0!)!)))+1\times \phi(4)

43=ϕ(ϕ(ϕ((2+0!)!)))+1+ϕ(4)43=\phi(\phi(\phi((2+0!)!)))+1+\phi(4)

44=ϕ(ϕ(ϕ((2+0!)!)))+1×444=\phi(\phi(\phi((2+0!)!)))+1\times 4

45=ϕ(ϕ(ϕ((2+0!)!)))+1+445=\phi(\phi(\phi((2+0!)!)))+1+4

46=2+(0!+1)×4!46=-2+(0!+1)\times 4!

47=ϕ(2)+(0!+1)×4!47=-\phi(2)+(0!+1)\times 4!

48=(2+0×1)×4!48=(2+0\times 1)\times 4!

49=ϕ(2)+(0!+1)×4!49=\phi(2)+(0!+1)\times 4!

50=2+(0!+1)×4!50=2+(0!+1)\times 4!

51=(2+0!)×(1+ϕ9((ϕ(4!)!)))51=(2+0!)\times(1+\phi^9((\phi(4!)!)))

Daniel Chiu - 5 years, 9 months ago

Log in to reply

@Daniel Chiu 48=ϕ(20)×ϕ(14)48=\phi(20)\times\phi(14)

Abdur Rehman Zahid - 4 years, 9 months ago

Log in to reply

@Calvin Lin I honestly don't know, been trying hard to continue after 3434 without the use of square root. Turns out it works very handy with floor/ceiling function and factorials. This is because when we combine 22 and 00 as such: ((2+0!)!)!=720 ((2+0!)!)! = 720 , and apply the functions in different orders, we can get many (possibly infinite) natural numbers.

Pi Han Goh - 5 years, 9 months ago

Log in to reply

@Pi Han Goh 67=ϕ(ϕ((((2+0!)!)!)))1+467 = \phi(\phi((((2+0!)!)!))) - 1 + 4

68=ϕ(ϕ((((2+0!)!)!)))×1+468 = \phi(\phi((((2+0!)!)!))) \times 1 + 4

69=ϕ(ϕ((((2+0!)!)!)))+1+469 = \phi(\phi((((2+0!)!)!))) + 1 + 4

70=ϕ(ϕ((((2+0!)!)!)))+(1+4)!70 = \phi(\phi((((2+0!)!)!))) + (1 + |\sqrt{4}|)!

71=ϕ(ϕ((((2+0!)!)!)))1+ϕ(4!)71 = \phi(\phi((((2+0!)!)!))) - 1 + \phi(4!)

72=ϕ(ϕ((((2+0!)!)!)))×1+ϕ(4!)72 = \phi(\phi((((2+0!)!)!))) \times 1 + \phi(4!)

73=ϕ(ϕ((((2+0!)!)!)))+1+ϕ(4!)73 = \phi(\phi((((2+0!)!)!))) + 1 + \phi(4!)

74=ϕ(ϕ((((2+0!)!)!)))+ϕ(ϕ(1+4!))74 = \phi(\phi((((2+0!)!)!))) + \phi(\phi(-1+4!))

75=(2+0!)×(1+4!)75 = (2+0!) \times (1+4!)

76=(201)×476 = (20-1) \times 4

Fahim Shahriar Shakkhor - 5 years, 9 months ago

Log in to reply

@Pi Han Goh 64 can also equal 4(2+1)×0!4^{(2+1)} \times 0!

Sharky Kesa - 5 years, 9 months ago

Log in to reply

@Daniel Chiu 33 is also φ(201)4\frac{\varphi(201)}{4}

Bogdan Simeonov - 5 years, 9 months ago

Log in to reply

@Bogdan Simeonov This is probably a better solution since ϕ((2+0!+1)!)+4!=ϕ(24)+24=8+24=32 \phi ((2+0!+1)!) +4! = \phi (24) + 24 = 8 + 24 = 32 not 33 33 .

Evan Robinson - 5 years, 9 months ago

Log in to reply

@Evan Robinson True enough I messed up, nice catch!

Daniel Chiu - 5 years, 9 months ago

Log in to reply

@Jorge Tipe Factorial.....

You're smart lol

Kenny Lau - 5 years, 9 months ago

Log in to reply

[(2+0)x4]+1

Christal Juntilla - 5 years, 9 months ago

Log in to reply

Well, this is in the wrong order.

Daniel Chiu - 5 years, 9 months ago

Log in to reply

True, we always depend on other functions, disregarding the basic stuffs.

Kenny Lau - 5 years, 9 months ago

Log in to reply

(0!+0!+0!+0!+0!+0!)!=720

Anirudha Nayak - 5 years, 9 months ago

Log in to reply

2 raised to 0 +1 raised to 4

Chaitu Sakhare - 5 years, 9 months ago

Log in to reply

5 = 2+0-1 + 4

6 = 2 + 0 X 1 + 4

7 = 2 + 0 + 1 + 4

8 = (2 + 0 X 1) X 4

angelica ermino - 5 years, 9 months ago

Log in to reply

maybe every number can be formed

Yash Gupta - 5 years, 9 months ago

Log in to reply

1,638,400,000,000,000,000=20141,638,400,000,000,000,000=20^{14}

2,432,902,008,176,640,00087,178,291,200=20!14!2,432,902,008,176,640,000^{87,178,291,200} = 20!^{14!} is a possibility as well if you can use double digits

Noone told me me anything not being able to:

2,432,902,008,176,640,00087,178,291,200!=(20!14!)!2,432,902,008,176,640,000^{87,178,291,200}! = (20!^{14!})!

John Muradeli - 5 years, 7 months ago

Log in to reply

100=20×(1+4)100=20\times(1+4)

Abdur Rehman Zahid - 4 years, 10 months ago

Log in to reply

Are there any people from 2018 that are commenting on this apart from me?

Kao Cen Darach - 10 months, 4 weeks ago

Log in to reply

9 = [(2+0)x4]+1

Christal Juntilla - 5 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...