Note the following sequences: $$1 \to 4 \to 16 \to 25 \to 22 \to 10 \to 1$$$$ 2 \to 8 \to 32 \ 11 \to 5 \to 20 \to 2$$$$3 \to 12 \to 9 \to 36 \to 27 \to 30 \to 3$$$$13 \to 13, 26 \to 26, 39 \to 39$$$$14 \to 17 \to 29 \to 38 \to 35 \to 23 \to 14$$$$19 \to 37 \to 31 \to 7 \to 28 \to 34 \to 19$$$$0 \to 0$$and of course as given in the problem: $$6 \to 24 \to 18 \to 33 \to 15 \to 21 \to 6$$ So all the numbers from 0 to 39 are represented above. Therefore once the number on the board falls below 40, it will eventually repeat. But note that the number $X = 10a + b \to a + 4b = Y$. If $X \geq 40$ then $a \geq 4$ so $3a \geq 12 > b \iff X > Y$ so the number on the board always decreases strictly. So it will eventually falls below 40. For part B, note that it $X \to Y$ then $4X \equiv Y \mod 13$ because $40a+4b \equiv a + 4b \mod 13$ So each operation multiplies the number on the board by 4 mod 13. But the initial number is divisible by 13, so the final number will also be divisible by 13. So that leaves us 3 possibilities: 13,26, or 39. Then note that the number on the board doesn't change mod 3. Because the number on the board is also divisible by 3, the final number must also be divisible by 3. So the final number is 39