Question:

Step 1: Define $g(x) = xf(x) - 1$. Since $f(i) = \frac{1}{i}$ for $i = 1, 2, 3, 4$, it follows that $g(i) = i \cdot f(i) - 1 = i \cdot \frac{1}{i} - 1 = 1 - 1 = 0$ for $i = 1, 2, 3, 4$. Thus, $g(x)$ has roots at $x = 1, 2, 3, 4$. Step 2: Determine the form of $g(x)$. Since $f(x)$ is a monic polynomial of degree 4, $f(x) = x^4 + c_3x^3 + c_2x^2 + c_1x + c_0$. Then $g(x) = xf(x) - 1 = x(x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) - 1 = x^5 + c_3x^4 + c_2x^3 + c_1x^2 + c_0x - 1$. Therefore, $g(x)$ is a monic polynomial of degree 5. Since $g(x)$ has roots $1, 2, 3, 4$, we can write $g(x)$ in the form: $$g(x) = (x-1)(x-2)(x-3)(x-4)(x-a)$$ for some constant $a$, because $g(x)$ is monic of degree 5. Step 3: Find the value of $a$. Evaluate $g(x)$ at $x=0$: $g(0) = 0 \cdot f(0) - 1 = -1$. Substitute $x=0$ into the factored form of $g(x)$: $$g(0) = (-1)(-2)(-3)(-4)(-a) = 24(-a) = -24a$$ Equating the two expressions for $g(0)$: $$-24a = -1 \Rightarrow a = \frac{1}{24}$$ Thus, the polynomial $g(x)$ is: $$xf(x) - 1 = (x-1)(x-2)(x-3)(x-4)\left(x - \frac{1}{24}\right)$$ Step 4: Calculate $f(5)$. Substitute $x=5$ into the equation for $xf(x)-1$: $$5f(5) - 1 = (5-1)(5-2)(5-3)(5-4)\left(5 - \frac{1}{24}\right)$$ $$5f(5) - 1 = (4)(3)(2)(1)\left(\frac{120-1}{24}\right)$$ $$5f(5) - 1 = 24 \cdot \frac{119}{24}$$ $$5f(5) - 1 = 119$$ $$5f(5) = 120$$ $$f(5) = 24$$ Step 5: Determine the number of zeroes at the end of $f(5)!$. We need to find the number of trailing zeroes in $24!$. This is determined by the number of factors of 5 in its prime factorization. Using Legendre's formula: $$\left\lfloor\frac{24}{5}\right\rfloor + \left\lfloor\frac{24}{25}\right\rfloor = 4 + 0 = 4$$ The number of zeroes at the end of $f(5)!$ is 4. Step 6: Determine the number of divisors of $f(5)$. $f(5) = 24$. The prime factorization of 24 is $2^3 \cdot 3^1$. The number of divisors is given by the product of one more than each exponent in the prime factorization: $$(3+1)(1+1) = 4 \cdot 2 = 8$$ The number of divisors of $f(5)$ is 8. Step 7: Determine the sum of even divisors of $f(5)$. The divisors of $24 = 2^3 \cdot 3^1$ are of the form $2^a \cdot 3^b$ where $0 \le a \le 3$ and $0 \le b \le 1$. Even divisors must have $a \ge 1$. The sum of even divisors is: $$(2^1+2^2+2^3)(3^0+3^1) = (2+4+8)(1+3) = (14)(4) = 56$$ The sum of even divisors of $f(5)$ is 56. Step 8: Determine the sum of odd divisors of $f(5)$. Odd divisors of $24 = 2^3 \cdot 3^1$ are those divisors where the power of 2 is $2^0$. These are the divisors of $3^1$. The odd divisors are $3^0 = 1$ and $3^1 = 3$. The sum of odd divisors is $1+3=4$.