[ce]

2

[Moveyourfeet]

haven't read the thread. My answer is 2. Done in 30 sec.
Am I a

[Moveyourfeet]

Originally Posted by .Homme.

0=1
1=0
2=0
3=0
4=-
5=-
6=1
7=0
8=2
9=1

??

This is how I did it.

[égalité]

Originally Posted by PhilePhile

oh, thanks for clarifying

There are two functions. f:Z-->Z assigns, for example, the number 1 to 0, the number 0 to 1, the number 0 to 2, etc. (where Z=the integers). Then there is a quaternary map on Z, written as a word of length 4, that simply produces the sum of the "letters" in the word. The expression "8809=6" means g((f,f,f,f)(8,8,0,9))=6, where (f,f,f,f) is the cartesian product of f with itself 4 times (it is a map from Z^4 to Z^4). The composition g((f,f,f,f)) is certainly not injective, given the brief glimpse into the behavior of the function f that was provided in the statement of the problem, and especially given the fact that decomposing an integer into a sum of 4 other integers is a highly non-unique process. Hope that makes sense gurl.

[PhilePhile]

Originally Posted by égalité

oh, thanks for clarifying

There are two functions. f:Z-->Z assigns, for example, the number 1 to 0, the number 0 to 1, the number 0 to 2, etc. (where Z=the integers). Then there is a quaternary map on Z, written as a word of length 4, that simply produces the sum of the "letters" in the word. The expression "8809=6" means g((f,f,f,f)(8,8,0,9))=6, where (f,f,f,f) is the cartesian product of f with itself 4 times (it is a map from Z^4 to Z^4). The composition g((f,f,f,f)) is certainly not injective, given the brief glimpse into the behavior of the function f that was provided in the statement of the problem, and especially given the fact that decomposing an integer into a sum of 4 other integers is a highly non-unique process. Hope that makes sense gurl.

Sorry, no . Most of us are pre-schooler here. Can you elaborate on your solution/concept .

[Sammo]

Originally Posted by égalité

oh, thanks for clarifying

There are two functions. f:Z-->Z assigns, for example, the number 1 to 0, the number 0 to 1, the number 0 to 2, etc. (where Z=the integers). Then there is a quaternary map on Z, written as a word of length 4, that simply produces the sum of the "letters" in the word. The expression "8809=6" means g((f,f,f,f)(8,8,0,9))=6, where (f,f,f,f) is the cartesian product of f with itself 4 times (it is a map from Z^4 to Z^4). The composition g((f,f,f,f)) is certainly not injective, given the brief glimpse into the behavior of the function f that was provided in the statement of the problem, and especially given the fact that decomposing an integer into a sum of 4 other integers is a highly non-unique process. Hope that makes sense gurl.

I'm studying chemical engineering and I don't understand a fuck out of this


I've always hated calculus anyway... Algebra on the other hand

[Sean.]

It's 2.

Spoiler: It's the number of enclosed spaces.

[moby]

OK, I'll play.

Basically we are computing the rank of the first homology group of each of the given topological spaces.

[égalité]

Originally Posted by moby

OK, I'll play.

Basically we are computing the rank of the first homology group of each of the given topological spaces.

Oh my I didn't notice that

[Direwolf]

Originally Posted by PhilePhile

"2" is technically not correct. The "=" should be "≈" for the "2" answer to be correct.


For example,

8096 = 5

and

9881 = 5

is not possible. It's reasonable that "9" = "6" ... but "9" + "6" = "8" ? or "0" = "9" ?

8096
8=2
0=1
9=1
6=1

9881
9=1
8=2
8=2
1=0
???

9=6=1
9+6=2

0123456789=5
0123456789=8096=9881


M=Mental
A=Abuse
T=to
H=Health

[Morning Morgan]

Originally Posted by moby

OK, I'll play.

Basically we are computing the rank of the first homology group of each of the given topological spaces.

Win.

An extension to homotopy groups would ensure that the problem can be further generalized.

[égalité]

Originally Posted by PhilePhile

Sorry, no . Most of us are pre-schooler here. Can you elaborate on your solution/concept .

Sorry I was just being unnecessarily pedantic My point is that it's perfectly legitimate for 8096 and 9881 to both "equal" 5. As moby pointed out, they are homotopy equivalent.