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=−2−0−1+4 1 = -2 - 0 - 1 + 4
2=−2−0×1+4 2 = - 2 - 0 \times 1 + 4
3=2×0−1+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]{ \, } .

#NumberTheory #Numbers

Note by Chung Kevin
7 years, 5 months ago

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Comments

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 - 7 years, 5 months ago

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

Jorge Tipe - 7 years, 5 months ago

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

Bob Krueger - 7 years, 5 months ago

10=(2+0)×(1+4)10=(2+0)\times(1+4)

Pero no puedo obtener 11...

Kenny Lau - 7 years, 5 months ago

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

Jorge Tipe - 7 years, 5 months ago

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@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=20−1419=20-1^4

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

21=−2−0−1+4!21=-2-0-1+4!

22=−2+0×1+4!22=-2+0\times 1+4!

23=2×0−1+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 - 7 years, 5 months ago

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@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!⌋)+1−ϕ(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!)!)!))−1−4 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!)!)!))+1−4 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!)!)!))+1−4 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 - 7 years, 5 months ago

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@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 - 7 years, 5 months ago

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@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=log⁡2(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=log⁡2(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 - 7 years, 5 months ago

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

Michael Tang - 7 years, 5 months ago

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@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 - 7 years, 5 months ago

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

Sadie Robinson - 7 years, 5 months ago

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

Daniel Chiu - 7 years, 5 months ago

@Calvin Lin – 35=ϕ(ϕ(ϕ((2+0!)!)))−1−435=\phi(\phi(\phi((2+0!)!)))-1-4

36=ϕ(ϕ(ϕ((2+0!)!)))×1−436=\phi(\phi(\phi((2+0!)!)))\times 1-4

37=ϕ(ϕ(ϕ((2+0!)!)))+1−437=\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!)!)))+1−ϕ(4)39=\phi(\phi(\phi((2+0!)!)))+1-\phi(4)

40=ϕ(ϕ(ϕ((2+0!)!)))+1−ϕ(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 - 7 years, 5 months ago

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

Abdur Rehman Zahid - 6 years, 5 months ago

@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 - 7 years, 5 months ago

@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=(20−1)×476 = (20-1) \times 4

Fahim Shahriar Shakkhor - 7 years, 5 months ago

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

Sharky Kesa - 7 years, 5 months ago

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

Bogdan Simeonov - 7 years, 5 months ago

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@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 .

Sadie Robinson - 7 years, 5 months ago

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@Sadie Robinson – True enough I messed up, nice catch!

Daniel Chiu - 7 years, 5 months ago

@Jorge Tipe – Factorial.....

You're smart lol

Kenny Lau - 7 years, 5 months ago

[(2+0)x4]+1

Christal Juntilla - 7 years, 5 months ago

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Well, this is in the wrong order.

Daniel Chiu - 7 years, 5 months ago

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

Kenny Lau - 7 years, 5 months ago

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

Anirudha Nayak - 7 years, 5 months ago

2 raised to 0 +1 raised to 4

Chaitu Sakhare - 7 years, 5 months ago

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 - 7 years, 5 months ago

maybe every number can be formed

Yash Gupta - 7 years, 5 months ago

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 M. - 7 years, 3 months ago

100=20×(1+4)100=20\times(1+4)

Abdur Rehman Zahid - 6 years, 6 months ago

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

Krishna Karthik - 2 years, 6 months ago

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

Christal Juntilla - 7 years, 5 months ago
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