Invariance Math Example 4

Solution

1. 1
   Start: $n=6$, $6 \pmod{3}=0$. Subtract 1: $n=5$, $5\pmod{3}=2$. Parity of $n\pmod{3}$ changed — not an invariant.
2. 2
   Try: parity of $n$. Start even ($n=6$). Subtract 1 gives odd; divide by 2 gives 3 (odd). From odd, subtract 1 gives even; can't divide.
3. 3
   Invariant found: the value $n$ is always a positive integer, and it strictly decreases toward 1 with each move (both operations reduce $n$). The positivity of $n$ is preserved.