1. The problem asks us to implement the Collatz function $f(n)$ defined as: $$ f(n) = \begin{cases} \frac{n}{2}, & \text{if } n \text{ is even} \\ 3n + 1, & \text{if } n \text{ is odd} \end{cases} $$ 2. To determine if $n$ is even or odd, we use the remainder operator $\%$ in Python: if $n \% 2 = 0$, $n$ is even; if $n \% 2 = 1$, $n$ is odd. 3. Integer division in Python is done with $//$, so when $n$ is even, $f(n) = n // 2$. 4. We test the function on $f(f(f(f(f(f(674))))))$ which should evaluate to 190. 5. Finally, we compute $f$ applied 15 times to 1071, i.e., $f^{15}(1071)$, and find the integer result. 6. Step-by-step evaluation for $f^{15}(1071)$: - $f(1071)$: 1071 is odd, so $3 \times 1071 + 1 = 3214$ - $f(3214)$: 3214 is even, so $3214 // 2 = 1607$ - $f(1607)$: odd, $3 \times 1607 + 1 = 4822$ - $f(4822)$: even, $4822 // 2 = 2411$ - $f(2411)$: odd, $3 \times 2411 + 1 = 7234$ - $f(7234)$: even, $7234 // 2 = 3617$ - $f(3617)$: odd, $3 \times 3617 + 1 = 10852$ - $f(10852)$: even, $10852 // 2 = 5426$ - $f(5426)$: even, $5426 // 2 = 2713$ - $f(2713)$: odd, $3 \times 2713 + 1 = 8140$ - $f(8140)$: even, $8140 // 2 = 4070$ - $f(4070)$: even, $4070 // 2 = 2035$ - $f(2035)$: odd, $3 \times 2035 + 1 = 6106$ - $f(6106)$: even, $6106 // 2 = 3053$ - $f(3053)$: odd, $3 \times 3053 + 1 = 9160$ So, $f^{15}(1071) = 9160$. **Final answer:** 9160