Starting with any non-zero natural number, it is always possible to form a sequence of numbers that ends in 1, by repeatedly following the instructions below:

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Starting with any non-zero natural number, it is always possible to form a sequence of numbers that ends in 1, by repeatedly following the instructions below:
Solution

a)

$$
100 \rightarrow 99 \rightarrow 33 \rightarrow 11 \rightarrow 12 \rightarrow 4 \rightarrow 3 \rightarrow 1
$$

b) $\{27,9,3,1\},\{10,9,3,1\},\{8,9,3,1\},\{12,4,3,1\},\{6,2,3,1\}$. Analyzing the process in reverse, we must always arrive at 1, which will always be the last element of the sequence; to arrive at 1, we must always pass through 3, because if we arrive at 2, we must add 1, and if we arrive at 4, we must subtract 1, so 3 is always the second-to-last element of our sequence; to arrive at 3, we must pass through 2 or 4, as seen previously, or through 9; to arrive at 2, we must pass through 6 beforehand, to arrive at 4, we must pass through 12, and to arrive at 9, we must pass through 8, 10, or 27. Therefore, the sequences are $\{27,9,3,1\},\{10,9,3,1\}$, $\{8,9,3,1\},\{12,4,3,1\},\{6,2,3,1\}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_09dbad5281c800d666d6g-10.jpg?height=431&width=446&top_left_y=2026&top_left_x=718)
c) 21. From the previous item, we see that to arrive at 1, we must pass through 3; we also saw that to arrive at a multiple of 3, we either add one unit, subtract one unit, or perform a division by 3; but to arrive at a number that is not a multiple of 3, we can only do so through division by 3. Thus, we will start from the sequences of length 4 found in the previous item, three of which begin with a multiple of 3 and the other two with a non-multiple of 3. Each non-multiple of three is obtained by dividing by 3 from a multiple of 3 (of course), which can be obtained from 3 numbers, and in this way, these two sequences of length 4 give rise to 3 sequences of length 6 each. With each multiple of 3, we can obtain 3 numbers, only one of which is a multiple of 3, so we will have 5 sequences of length 6 for each of the three sequences. Therefore, there are $2 \cdot 3 + 3 \cdot 5 = 21$ sequences of length 6.