Let us write an arbitrary natural number (for example, 2583), and then add the squares of its digits. (). Next, we do the same thing to the number obtained. Namely, . Now proceed further in the same way:
, , .
Prove that unless this procedure leads to number 1 (in which case, the number 1 will, of course, recur indefinitely), it must lead to the number 145, and the following cycle will repeat again and again:
145, 42, 20, 4, 16, 37, 58, 89.
Prove that the number is divisible by 11 for every natural k.
The number is divisible by 13, 49, 181 and 379, and is not divisible by either 5 or by 11. How can this result be confirmed?