and 3, 6, 12. In each of them, each number is obtained from the previous one according to the same rule.

a) Find this rule.

b) Find all natural numbers that transform into themselves (according to this rule).

c) Prove that the number $2^{1991}$ will become a single-digit number after several transformations.
a) The law can be guessed, noticing, for example, that while the number is a single digit, it doubles, and then - apparently not. And the fact that 10 turns into 2 suggests that it is not the number itself that is doubled, but the sum of its digits. Thus, the sought law has been discovered: "Double the sum of the digits."

Of course, this is not a proof in the strict mathematical sense of the word. For example, this way one can "prove" that the number sixty is divisible by all numbers. Indeed, 60 is divisible by 1, by 2, by 3, by 4, by 5, by 6... However, for solving the problem, it is only necessary to find a "sufficiently simple" rule, following which one can obtain

such a sequence. And the ability to see, to feel the pattern (which was required in this problem) is no less important for a mathematician than the ability to reason strictly! If you find any other (but also "sufficiently simple") law that gives the sequences 2, 4, 8, 16, 14, 10, 2 and 3, 6, 12, please write to us (and on the Olympiad, such a solution would also be accepted!).

b) See the solution to problem 5 for grades $5-6$.

c) Notice that if a number is at least three digits, then the sum of its digits is less than the number itself. This means the number will decrease until it becomes a two-digit or a single-digit number. The only remaining danger is to get stuck in a "fixed point" - 18. But this is impossible in our case, since the initial number was not divisible by 9 (prove rigorously the following property of our law for generating one number from another: if a certain number is divisible by 9, then the number from which it was obtained is also divisible by 9).

## Answer

a) Double the sum of the digits; b) 18.