I must be missing something. Why is this significant?
Earlier this year a solution for k=33 was found [1], so 42 was the next unknown value.
[1] Brooker, A., "CRACKING THE PROBLEM WITH 33", https://people.maths.bris.ac.uk/~maarb/papers/cubesv1.pdf
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I must be missing something. Why is this significant?
Earlier this year a solution for k=33 was found [1], so 42 was the next unknown value.
[1] Brooker, A., "CRACKING THE PROBLEM WITH 33", https://people.maths.bris.ac.uk/~maarb/papers/cubesv1.pdf
It's also the answer to life, the universe and everything. Which is nice.
In Python >>> (-80538738812075974)**3 + 80435758145817515**3 + 12602123297335631**3 42
dc -e '_80538738812075974 3 ^ 80435758145817515 3 ^ 12602123297335631 3 ^ + + p'It's also the answer to life, the universe and everything. Which is nice.
Why doesn't this evaluate in Excel? =(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3 Returns 1.09785E+36
Btw:
julia> (-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3
42
42 is also the answer to the universe. ;-)
Earlier quoted context omitted.
If you can break up a reasonably-sized table (perhaps 120 or so) of numbers into their respective sums of cubes, breaking many types of encryption is arbitrary.
Guessing you meant "trivial". Do you have some info about how this helps break encryption?
Why doesn't this evaluate in Excel? =(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3 Returns 1.09785E+36
same as Google. https://www.google.com/search?sxsrf=ACYBGNQRb1b08L-cU-JY3GHr...
https://www.bing.com/search?q=%3D%28-80538738812075974%29%5E...
Earlier quoted context omitted.
If you can break up a reasonably-sized table (perhaps 120 or so) of numbers into their respective sums of cubes, breaking many types of encryption is arbitrary.
Guessing you meant "trivial". Do you have some info about how this helps break encryption?