The set of natural numbers $\{1,2,\dots\}$ is split into finite number of arithmetic progressions with common differences $d_1 \le d_2 \le \dots \le d_n$
Show that $$d_{n} = d_{n-1}$$
Eg: $\{4n+1\}, \{4n+3\}, \{2n\}$
Here $d_1 = 2, d_2 = d_3 = 4$.
Show that $$d_{n} = d_{n-1}$$
Eg: $\{4n+1\}, \{4n+3\}, \{2n\}$
Here $d_1 = 2, d_2 = d_3 = 4$.
Peiyush, did you come up with this problem?
ReplyDeleteHey!
DeleteThis version maybe, but variants of this are well known.
There is at least one proof (the well known case) which applies to this problem, but IIRC this problem had a simpler proof which I do not recollect.