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:

\( 1 = -2 - 0 - 1 + 4 \)

\( 2 = - 2 - 0 \times 1 + 4 \)

\( 3 = 2 \times 0 - 1 + 4 \)

\( 4 = 2 \times 0 \times 1 + 4 \)

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

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## Comments

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TopNewest5 = 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?

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

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

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\(10=(2+0)\times(1+4)\)

Pero no puedo obtener 11...

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\(13=-2^0+14\)

\(14=2\times 0+14\)

\(15=2^0+14\)

\(16=2+0+14\)

\(17=2+0!+14\)

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

\(19=20-1^4\)

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

\(21=-2-0-1+4!\)

\(22=-2+0\times 1+4!\)

\(23=2\times 0-1+4!\)

\(24=2\times 0\times 1+4!\)

\(25=2\times 0+1+4!\)

\(26=2+0\times 1+4!\)

\(27=2+0+1+4!\)

\(28=2+0!+1+4!\)

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

\(30=(2+0!)!\times 1+4!\)

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

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

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

\(34=20+14\)

Is combining digits allowed? How about totient function?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\( 50 = \left \lfloor 201 \div 4 \right \rfloor \)

\( 51 = \left \lceil 201 \div 4 \right \rceil \)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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What if square roots were not allowed?

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

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the formula for the totient function.

First, I'll assume you knowWhen 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.

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

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

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

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

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

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

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

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

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

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

\(46=-2+(0!+1)\times 4!\)

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

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

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

\(50=2+(0!+1)\times 4!\)

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

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

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

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

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

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

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

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

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

\(76 = (20-1) \times 4 \)

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You're smart lol

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[(2+0)x4]+1

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Well, this is in the wrong order.

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

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\(100=20\times(1+4)\)

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\(1,638,400,000,000,000,000=20^{14}\)

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

Noone told me me anything not being able to:

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

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maybe every number can be formed

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5 = 2+0-1 + 4

6 = 2 + 0 X 1 + 4

7 = 2 + 0 + 1 + 4

8 = (2 + 0 X 1) X 4

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2 raised to 0 +1 raised to 4

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

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

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