Puzzling Asked by María Lucía Uribe on January 19, 2021
I can't beat @Excited Raichu's a) but for b):
Correct answer by Paul Panzer on January 19, 2021
a)
Is it:
b)
Is it:
Answered by TruVortex_07 on January 19, 2021
a) Not certain this is the biggest, but I can't find anything bigger than
b) Moving 4 matches allows
which is pretty big. Not sure if it's the absolute biggest, but it's up there.
This is, of course, assuming that the digit 1 must be two sticks high. There's definitely a higher ceiling if it can only be one.
Answered by Excited Raichu on January 19, 2021
111153
Moving 2 Match sticks, the 4s can be made into 11s. And the third match stick would be moved to get 5.
Answered by user66436 on January 19, 2021
Hard to go wrong with
As obviously
Answered by Cireo on January 19, 2021
Moving 3 matches, I'd
giving
My best guess for 4 moves is to
to get
I think that's the highest so far, if each one needs two matchsticks.
Answered by Cristobol Polychronopolis on January 19, 2021
For a)
For b)
Answered by Luke Koda on January 19, 2021
Moving 4 matches to form
_ _
| | | | |_ |_
| | | | _| _|
and viewing it upside down gives S S 1111 where S 1111 is the maximum shifts function of the busy beaver game with 1111 states http://en.wikipedia.org/wiki/Busy_beaver#Maximum_shifts_function_S . S 6 is 3.515E+18267 and S 7 is already 10^10^10^10^18705353.
S S 1111 is the maximum shifts function of the busy beaver game with S 1111 states, a number uncomputably large, but still finite.
Answered by Gnubie on January 19, 2021
(b) Move 4 matches to form
_ _
| |_| _| / |_
| | _| / |_
The slash has to be squeezed in between the 3's to make 143/ε. The last character represents epsilon, an arbritarily small positive quantity used in the definition of limit in calculus, which is greater than zero to avoid division by zero.
(a) Move 3 matches to form
_ _ _
|_| | | _| / |_
| | _| _
or 473/ε. The vertical match of ε is centered over the middle bar.
Answered by Gnubie on January 19, 2021
(b) Move 4 matches to form
_ _
| |_| |_| | |
| | | _| _|
which viewed upside down is Γ Γ 661, where Γ is the gamma function http://en.wikipedia.org/wiki/Gamma_function
Γ 661 ~ 10^1576 and Γ Γ 661 ~ 10^10^1579
which is slightly larger than 7^7^473 ~ 10^10^400.
Interpreting it as Γ Γ bb1 where bb1 in hexadecimal is 2993 in decimal,
Γ 2993 ~ 10^9103 and Γ Γ 2993 is too large to compute in Wolframalpha but is larger than Γ Γ 661.
Even better are
_ _
|_| |_| | |
|| | | _| _|
Γ Γ 66¹¹ and Γ Γ bb¹¹, respectively.
Answered by Gnubie on January 19, 2021
(b) Move 4 matches to form
_ _
|_| /| /| | _|
| | _|
or 4 ↑↑ 73, where ↑ is Knuth's up-arrow notation http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation
4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4 (73 4's) is too large to compute in Wolframalpha.
_ _
|_| /| ||| | _|
| | _|
A convention allows multiple ↑'s to be specified with a superscript, so 4 ↑¹¹¹ 73 = 4 ↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑ 73 (111 ↑'s), a number too large to represent even with power towers.
Answered by Gnubie on January 19, 2021
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