I tried posting the following on math.stackexchange, but no answers. I can of course delete if inappropriate.
The "1089 number trick" (see e.g. here) says that if you take a three-digit number, subtract its reverse and to this answer add the reverse of the answer again, then you always get 1089.
Of course, if you spell this out the coefficients end up canceling and you're left with what you got from the carriyng, which sums to 1089.
Perhaps what we can say is really going on is that the linear combination of permutations $$(1 - \tau)+\tau(1-\tau)$$ sums to $0$ whenever $\tau$ is a transposition, implying that the computation is independent of what number we plug in.
In this way we can find other "number tricks" by using other null-relations between permutations: for instance, if $\rho\in\Sigma_3$ is the rotation $\rho(abc) = cab$, then one can use the relation $$(1-\rho)+\rho(1-\rho^2)=0$$ and obtain another "number trick" whose result is always 999 instead of 1089.
To sum up the above discussion, behind the curtain of these "number tricks" is a null-relation of permutations, which implies that the computation depends only on the carrying involved. On the other hand, we know that carrying is a cocycle and that the associated cohomology group is a certain Ext-group (see Isaksen's "A Cohomological Viewpoint on Elementary School Arithmetic", Theorem 6.6).
My question is, is it possible to interpret/formalize/generalize these "number tricks" as some sort of action of $\Sigma_n$, or $\mathbb{Z}[\Sigma_n]$ (or an augmentation subgroup thereof) on this Ext-group? And is it possible to carry out the above computations, ending up with the answer 1089 or 999, purely within some such Ext-group?
