Let $$f(x) = 2x^n + \lambda$$, $$\lambda \in \mathbb{R}$$, $$n \in \mathbb{N}$$, and $$f(4) = 133$$, $$f(5) = 255$$. Then the sum of all the positive integer divisors of $$(f(3) - f(2))$$ is

We need to find the sum of all positive integer divisors of $$(f(3) - f(2))$$.

Given $$f(x) = 2x^n + \lambda$$:

$$f(4) = 2 \cdot 4^n + \lambda = 133 \quad \text{...(i)}$$

$$f(5) = 2 \cdot 5^n + \lambda = 255 \quad \text{...(ii)}$$

Subtracting (i) from (ii):

$$2(5^n - 4^n) = 122 \implies 5^n - 4^n = 61$$

Testing values of $$n$$:

• $$n = 1$$: $$5 - 4 = 1 \neq 61$$
• $$n = 2$$: $$25 - 16 = 9 \neq 61$$
• $$n = 3$$: $$125 - 64 = 61$$ ✓

So $$n = 3$$. From (i): $$\lambda = 133 - 2(64) = 133 - 128 = 5$$.

$$f(3) = 2(27) + 5 = 59$$

$$f(2) = 2(8) + 5 = 21$$

$$f(3) - f(2) = 59 - 21 = 38$$

$$38 = 2 \times 19$$

Divisors: $$1, 2, 19, 38$$

Sum of divisors $$= 1 + 2 + 19 + 38 = 60$$

The correct answer is Option B: $$60$$.