Waste less time on Facebook — follow Brilliant.
×

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


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

Note by Chung Kevin
3 years, 6 months ago

No vote yet
1 vote

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 · 3 years, 6 months ago

Log in to reply

@Ajala Singh \(9=\sqrt{(2+0+1)^4}\) is allowed? Jorge Tipe · 3 years, 6 months ago

Log in to reply

@Jorge Tipe If that isn't allowed, there's always \((-2-0!)\cdot (1-4)\). Bob Krueger · 3 years, 6 months ago

Log in to reply

@Jorge Tipe \(10=(2+0)\times(1+4)\)

Pero no puedo obtener 11... Kenny Lau · 3 years, 6 months ago

Log in to reply

@Kenny Lau \(11=(2+0!)!+1+4\) Jorge Tipe · 3 years, 6 months ago

Log in to reply

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

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

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

\(15=2^0+14\)

\(16=2+0+14\)

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

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

\(19=20-1^4\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(34=20+14\)

Is combining digits allowed? How about totient function? Daniel Chiu · 3 years, 6 months ago

Log in to reply

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\( 66 = \left ( \phi ( \phi ( ((2+0!)!)!)) \right ) \times 1 + \sqrt{4} \) Pi Han Goh · 3 years, 6 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 · 3 years, 6 months ago

Log in to reply

@Calvin Lin Oh actually I can construct every positive integer. Take \(((((((0+1+4)!)!)!)!)!)!\), 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 \(\log\ 2\). \[n=\log_2 (\phi(\phi(\cdots(\phi(\phi(((\cdots((0+1+4)!)!\cdots)!)!)\cdots)))\] 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, \[50=\log_2(\phi^{501}(((0+1+4)!)!))\] where \(\phi^n\) denotes applying the totient function \(n\) times. Calculated via Mathematica. Daniel Chiu · 3 years, 6 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 · 3 years, 6 months ago

Log in to reply

@Evan Robinson Well, without the totient function, I can't find a way to make 33. Daniel Chiu · 3 years, 6 months ago

Log in to reply

@Daniel Chiu Why does totient eventually make a power of 2? Michael Tang · 3 years, 6 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 · 3 years, 6 months ago

Log in to reply

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(51=(2+0!)\times(1+\phi^9((\phi(4!)!)))\) Daniel Chiu · 3 years, 6 months ago

Log in to reply

@Daniel Chiu \(48=\phi(20)\times\phi(14)\) Abdur Rehman Zahid · 2 years, 6 months ago

Log in to reply

@Calvin Lin 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. Pi Han Goh · 3 years, 6 months ago

Log in to reply

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

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

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

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

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

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

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

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

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

\(76 = (20-1) \times 4 \) Fahim Shahriar Shakkhor · 3 years, 6 months ago

Log in to reply

@Pi Han Goh 64 can also equal \(4^{(2+1)} \times 0!\) Sharky Kesa · 3 years, 6 months ago

Log in to reply

@Daniel Chiu 33 is also \(\frac{\varphi(201)}{4}\) Bogdan Simeonov · 3 years, 6 months ago

Log in to reply

@Bogdan Simeonov This is probably a better solution since \( \phi ((2+0!+1)!) +4! = \phi (24) + 24 = 8 + 24 = 32 \) not \( 33 \). Evan Robinson · 3 years, 6 months ago

Log in to reply

@Evan Robinson True enough I messed up, nice catch! Daniel Chiu · 3 years, 6 months ago

Log in to reply

@Jorge Tipe Factorial.....

You're smart lol Kenny Lau · 3 years, 6 months ago

Log in to reply

@Ajala Singh [(2+0)x4]+1 Christal Juntilla · 3 years, 6 months ago

Log in to reply

@Christal Juntilla Well, this is in the wrong order. Daniel Chiu · 3 years, 6 months ago

Log in to reply

@Christal Juntilla True, we always depend on other functions, disregarding the basic stuffs. Kenny Lau · 3 years, 6 months ago

Log in to reply

\(100=20\times(1+4)\) Abdur Rehman Zahid · 2 years, 7 months ago

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:

\(2,432,902,008,176,640,000^{87,178,291,200}! = (20!^{14!})!\) John Muradeli · 3 years, 5 months ago

Log in to reply

maybe every number can be formed Yash Gupta · 3 years, 6 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 · 3 years, 6 months ago

Log in to reply

2 raised to 0 +1 raised to 4 Chaitu Sakhare · 3 years, 6 months ago

Log in to reply

(0!+0!+0!+0!+0!+0!)!=720 Anirudha Nayak · 3 years, 6 months ago

Log in to reply

9 = [(2+0)x4]+1 Christal Juntilla · 3 years, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...