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 $d_1 = 2, d_2 = d_3 = 4$.
