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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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?

Log in to reply

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

Log in to reply

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

Log in to reply

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

Pero no puedo obtener 11...

Log in to reply

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?

Log in to reply

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Log in to reply

What if square roots were not allowed?

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

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.

Log in to reply

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Log in to reply

Log in to reply

Log in to reply

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

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

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

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

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

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

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

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

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

You're smart lol

Log in to reply

[(2+0)x4]+1

Log in to reply

Well, this is in the wrong order.

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

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

Noone told me me anything not being able to:

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

Log in to reply

maybe every number can be formed

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

Log in to reply

2 raised to 0 +1 raised to 4

Log in to reply

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

Log in to reply

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

Log in to reply