- #1
Bipolarity
- 776
- 2
I'm trying to prove something in modular arithmetic that I came upon across my studies in comp sci. Consider a set of natural numbers [itex] {n_{1},n_{2},n_{3},...n_{k}} [/itex]
Consider two more natural numbers [itex]m[/itex] and [itex]p[/itex] such that
[tex] (\sum^{k}_{i=1}n_{i} ) \ mod \ m = p [/tex]
Now prove that
[itex]((((n_{1} \ mod \ m + n_{2}) \ mod \ m + n_{3}) \ mod \ m + n_{4}) \ mod \ m + ... + n_{k}) \ mod \ m = p [/itex]
All help would be appreciated.
BiP
Consider two more natural numbers [itex]m[/itex] and [itex]p[/itex] such that
[tex] (\sum^{k}_{i=1}n_{i} ) \ mod \ m = p [/tex]
Now prove that
[itex]((((n_{1} \ mod \ m + n_{2}) \ mod \ m + n_{3}) \ mod \ m + n_{4}) \ mod \ m + ... + n_{k}) \ mod \ m = p [/itex]
All help would be appreciated.
BiP