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 \( \sqrt[3]{ \, } \).

## 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? – Ajala Singh · 3 years, 8 months ago

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– Jorge Tipe · 3 years, 8 months ago

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

– Bob Krueger · 3 years, 8 months ago

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

Pero no puedo obtener 11... – Kenny Lau · 3 years, 8 months ago

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– Jorge Tipe · 3 years, 8 months ago

\(11=(2+0!)!+1+4\)Log in to reply

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

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

\(15=2^0+14\)

\(16=2+0+14\)

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

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

\(19=20-1^4\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(34=20+14\)

Is combining digits allowed? How about totient function? – Daniel Chiu · 3 years, 8 months ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\( 66 = \left ( \phi ( \phi ( ((2+0!)!)!)) \right ) \times 1 + \sqrt{4} \) – Pi Han Goh · 3 years, 8 months ago

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What if square roots were not allowed? – Calvin Lin Staff · 3 years, 8 months ago

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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. – Daniel Chiu · 3 years, 8 months ago

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– Evan Robinson · 3 years, 8 months ago

That's pretty cool :) Of course, the next question is, what happens if the totient function is not allowed?Log in to reply

– Daniel Chiu · 3 years, 8 months ago

Well, without the totient function, I can't find a way to make 33.Log in to reply

– Michael Tang · 3 years, 8 months ago

Why does totient eventually make a power of 2?Log in to reply

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. – Daniel Chiu · 3 years, 8 months ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(51=(2+0!)\times(1+\phi^9((\phi(4!)!)))\) – Daniel Chiu · 3 years, 8 months ago

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– Abdur Rehman Zahid · 2 years, 8 months ago

\(48=\phi(20)\times\phi(14)\)Log in to reply

– Pi Han Goh · 3 years, 8 months ago

I honestly don't know, been trying hard to continue after \(34\) without the use of square root. Turns out it works very handy with floor/ceiling function and factorials. This is because when we combine \(2\) and \(0\) as such: \( ((2+0!)!)! = 720 \), and apply the functions in different orders, we can get many (possibly infinite) natural numbers.Log in to reply

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

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

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

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

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

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

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

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

\(76 = (20-1) \times 4 \) – Fahim Shahriar Shakkhor · 3 years, 8 months ago

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– Sharky Kesa · 3 years, 8 months ago

64 can also equal \(4^{(2+1)} \times 0!\)Log in to reply

– Bogdan Simeonov · 3 years, 8 months ago

33 is also \(\frac{\varphi(201)}{4}\)Log in to reply

– Evan Robinson · 3 years, 8 months ago

This is probably a better solution since \( \phi ((2+0!+1)!) +4! = \phi (24) + 24 = 8 + 24 = 32 \) not \( 33 \).Log in to reply

– Daniel Chiu · 3 years, 8 months ago

True enough I messed up, nice catch!Log in to reply

You're smart lol – Kenny Lau · 3 years, 8 months ago

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– Christal Juntilla · 3 years, 8 months ago

[(2+0)x4]+1Log in to reply

– Daniel Chiu · 3 years, 8 months ago

Well, this is in the wrong order.Log in to reply

– Kenny Lau · 3 years, 8 months ago

True, we always depend on other functions, disregarding the basic stuffs.Log in to reply

\(100=20\times(1+4)\) – Abdur Rehman Zahid · 2 years, 9 months ago

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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:

\(2,432,902,008,176,640,000^{87,178,291,200}! = (20!^{14!})!\) – John Muradeli · 3 years, 7 months ago

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maybe every number can be formed – Yash Gupta · 3 years, 8 months ago

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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 – Angelica Ermino · 3 years, 8 months ago

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2 raised to 0 +1 raised to 4 – Chaitu Sakhare · 3 years, 8 months ago

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(0!+0!+0!+0!+0!+0!)!=720 – Anirudha Nayak · 3 years, 8 months ago

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9 = [(2+0)x4]+1 – Christal Juntilla · 3 years, 8 months ago

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