You can use any Mathematical expression,any method suitable...stupidity is on.
But the challenge is you cannot use any other numbers,you can use 5 and 4 only once and you cannot use '-'(minus) sign.
I know 4 possible methods...Can you find still more ?
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The four solutions that i know are:
FIVE in Roman = V; FOUR in Roman = IV.So, 54=VIV=I=1
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5μ(4) is also posible, where μ defines the MOBIUS function.
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Yes...thats right👍👍 Good use of MOBIUS function
even cotan(45)=1
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yup..right.👍👍
even cot(45)=1
5 (mod 4) = 1
5-4=1
dxd(x+54)=1
Woahh, allaww
Same thing as 5-4 but looks different ∫45dx=1 Lol. X'D
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yess....even this is possible👍👍
Nicely done, didn't see that.
4!Gamma(5)=1
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Could u explain what the gamma() function's purpose
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The factorial n! is originally defined only for positive integers n, so 18th century mathematicians came up with an integral Gamma(t)=Γ(t)=∫0∞xt−1e−xdx that has properties similar to the factorial, but defined for all real and complex numbers t (except 0 and negative integers). It works out that Γ(t)=(t−1)!. A very versatile function, the gamma function is one of the most widely used functions in mathematics.
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Aw yes. :-)
yeah...never thought of this...👌👍
Can you please tell me what is it's actual meaning? Like 8! means 8(8-1)(8-2)......(8-7) so what is the meaning of -2! ?? @Michael Mendrin
⌊45⌋ or ⌈54⌉
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In a similar way, ⌊54⌋,⌊45⌋ etc will also work.
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Or ⌊45⌋ and ⌈54⌉
Thas possible...greatest integer function..👍
5 and 4 are used twice
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No, they're not. They are used only once in each expression. I have written them separately.
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5⊕4=1
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this made my day....ahahahah
Whats the sign between 5 and 4???
You need to describe the modulo number i.e.5⨁84 for which an extra 8 is required.
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It's not modulo, but an XOR operation, :)
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ln(exp(4)exp(5))=1
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Wow....superb !!Gud one👌👌
This one's right.
Edit it to Ln instead of log...
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I wrote it ln then changed it to log, thought that in english countries, log refers to the natural logarithm. Anyway I'm editing it again :)
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loge is also used.
Yeah...or elseHowever Ln seems better...:)
5 mod 4 =1
Greatest integer(5/4)
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Yes..this one is possible.
3 more ways are possible.
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Signum(5+4), Signum(5×4) Signum(5÷4)
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Again,find the other ways...
gcd(4,5)=1
∣cis(45°)∣=1
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yeah....cool solution...gud one ✌👍
4ϕ(5)=1
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On a similar note of numerical functions, d(4)−d(5)=1 where d(n)=the number of positive divisors of n.
yup...another solution..
v4(5!)=1
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Speechless.......
What is ν function ???
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Highest power of a prime that divides something.In this case it is not used properly, because 4 is not a prime.
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v5(⌈4!⌉)=1.
I believe the function can still be used... This also works:Log in to reply
Gcd(5,4)=1 The easiest one
signum(45),or anything which invlove 4 and 5(but real)
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yes...signum function is possible...It is a function which always gives 1 for any constant.
⎢⎢⎢⎡ζ(4)ζ(5)⎥⎥⎥⎤
totient funtion of 5 / 4
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yeah...its possible
(Totient of 5)/4
tan(45)=1
integer(5/4)=1 (we use in C language int(a/b) to get integer quotient)
5 modulo 4=1
Is this possible 5 + 4i^2
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This gives 1 but extra 2 cant be used as in i^2.
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Just write it like this: 5+4i×i=1
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i while any number other than 5 or 4 cannot be used.
Even like that is not correct; used the imaginary numberd(σ(σ(σ(σ(σ(4))))))d(σ(σ(σ(σ(σ(5))))))=1
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Lawl is that too much... =="
Where d(n) is the number of divisors of n and σ(n) is the sum of divisors of n.
In programming languages (int)5/4=1
(5/4) × (4/5) = 1.25 x 0.8 = 1 QED
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The condition is you cannot use 5 and 4 more than once.
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Ah fair enough
small omega of 5 / small omega of 4 ( note small omega counts the number of distinct prime factors )
The most obvious gcd(5,4)=1
Also, 4φ(5)=1
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yep..The gcd(5,4) is the most obvious.
No offence to maths but 4 = IV and 5 = V in roman numerals. So IV/V = I ( after cancelling V on both sides) = 1
5%4
gcd (4,5)=1
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what does ceil and floor mean
ceil(4/5),floor(5/4)
(1) 5%4
5%4 (mod)=1
5%4.
5%4
sum to infinity of i=1 (4*5^-i) //sorry but i don't know how to write the notation
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∑i=1∞4×5−i
The expression gives the value 1 but extra 1 and ∞ cannot be used.
(5+4)/(5+4)
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you've broken the rules
Hi Vinay this discussion is a good food for thought....Next
Try to express 0 to 9 numerals using the digit 3 only. One hint is using factorial may come in handy...
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Thank you...
Generating 0 to 100 will be more interesting..I think you should post a note on it...
Also post its link in this note as a comment...
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Here, I've created a note for this.
tan 45
5/4=1
Solution : int a=5,b=4,c; c=a/b; (hence c=1)
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Ya....and even c=5%4
5%4=1
5 modulo 4
5 = 101 (binary notation) 4 = 100 So. 5 xor 4 = 001
Cot45= 1
5% 4
⌊ln(5.4)⌋
Log54 to the base 54
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5 and 4 can be used only once..
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(5 xor 4) in binary
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5/4 remainder is 1 5mod4 is 1
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5%4=1...Yes,thats one of the ways that I knew.
totient function of 4 x .5
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here 0.5 is not possible...
-(4-5)=1 The mirror effect
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No minus sign.........
5%4 =1 , 5⊕4 = 1, ⌊45⌋ =1, ⌈54⌉ =1 ,
5%4 = 1 :P .. The % is the modulus sign.. It gives the remainder of 2 non- floating numbers ! As the remainder for floating numbers is zero always!
*5 mod 4 =1; *(5+4) mod 4=1;
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5 (mod 4) is possible but (5+4)(mod 4) is not possible...only one time 5...
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oh....kk sry
(factorial(5)/factorial(4))/5
5-4 floor(5/4) ceiling(5/4) 54/54
1= 5!/(5×4!)
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5 and 4 only once...
5#/4!
abs(5/4)
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Why not 5 mod 4 p^2
(Actually p^2 is Piyush Patnaik)
floor(ln(5)/ln(4)) and ceil(ln(4)/ln(5)) I hope ln(x), the natural logarithm, is acceptable here.
Mod(5,4)
(4/5)X(5/4)
(5*4)^0 =1
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0 not allowed and
5 and 4 not allowed twice.
Dim x as integer; x=5/4
5%4=1 what say guys ?
We can use 5 XNOR 4 == 1 101 XNOR 100 == 1 What you say?
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5 xnor 4 = 101⨁100=001=110
But 5 xor 1 = 001.
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Sorry my fault, It's not a valid solution. Solution should be like, we can not only use 5 and 4? ryte?
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(cos 5) ^2 + (sin 5) ^2 = 1
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2 cannot be used.
can i use concantenation?
54(5∣4)=1
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You can use concatenations
But two times 5 and 4 not allowed.
Using Exclusive OR gate. 5=101(b) 4=100(b) In Ex-OR logic output will high if input level is different. Here after computation output is 001(b) its decimal is 1
Write a comment or ask a question... 5!/(5*4!)=1
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No two times 5
5%4 = 1
5 \times 4 / 5\times 4
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5 and 4 can be used only once...
5 modules 4 = 1
5/4
5^{0} \times 4^{0}
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Good one but you used 0
can you please explain the notation....??
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It is 5 raised to 0 divided by 4 raised to 0
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5×4=20.Then 20÷20=1
!0
The 3 solution i no is Lint of 4/5 Zint of 5/4 Tan45
(4-5)^{2}
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'-' sign should not be used and extra numbers cant be used.
Mine is easier 5/4 x 4/5
((5×4)/5)/4 or ((4×5)/4)/5. They work
(5^2+4^2+4)/45
( 5 / 4 ) * ( 4 / 5 ) or ( 5 / 5 ) / (4 / 4) something like this :D
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Nope...you cant use 5 and 4 two times.
Tan(45)=1
y=54xthen[dx2d2y]!=1
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You cannot use 2...
4 mod(5)=1
5%4 = 1 ie., 5 modulus of 4 = 1
5mod4,⌊45⌋,⌈54⌉,round45,round544!Γ(5),cot(45deg)
5%4=1or equivalently 5mod4=1
4!γ =1
uuhgjjjjk
(5/4)*(4/5)
sin(4+5)=⌈0.3955...⌉=1
5^4
54/54 =1,45/45=1,5%4=1,(5*5)%4!
Taking (5+4)- (4+4) = 1
[5/4]
a^0 = 1
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Nope....0 not to be used.
f\left( x \right)=x\ f(5)-f(4)=1
How about .5 repeated plus .4 repeated? Does that break the rule for multiple usages of the digits if the horizontal line notation is used?
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yep..breaks the rule.
(5×4)/(10×2)
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10 and 2 cant be used.
54/54
(54)/(54)
5 x 4 -19
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nvm... Can't have a - sign
(54)/(54)
5^0 times 4^0 = 1
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Wrong answer 0 can't be used
nope...0 cant be used.
I'LL BLOW ALL OF YOUR MINDS!!!! |4-5|=1