Let $$f(n) = \left[\frac{1}{3} + \frac{3n}{100}\right]n$$, where $$[n]$$ denotes the greatest integer less than or equal to n. Then $$\sum_{n=1}^{56} f(n)$$ is equal to:

The function is defined as $$ f(n) = \left[ \frac{1}{3} + \frac{3n}{100} \right] n $$, where $$ [x] $$ denotes the greatest integer less than or equal to $$ x $$. We need to compute $$ \sum_{n=1}^{56} f(n) $$.

First, let $$ a_n = \frac{1}{3} + \frac{3n}{100} $$. Then $$ f(n) = n \cdot [a_n] $$. We determine the value of $$ [a_n] $$ for $$ n $$ from 1 to 56.

Calculate $$ a_n $$ for the boundaries:

• At $$ n = 1 $$: $$ a_1 = \frac{1}{3} + \frac{3 \cdot 1}{100} = \frac{1}{3} + \frac{3}{100} $$. Using a common denominator of 300, $$ \frac{1}{3} = \frac{100}{300} $$ and $$ \frac{3}{100} = \frac{9}{300} $$, so $$ a_1 = \frac{100}{300} + \frac{9}{300} = \frac{109}{300} \approx 0.3633 $$. Thus, $$ [a_1] = 0 $$.
• At $$ n = 56 $$: $$ a_{56} = \frac{1}{3} + \frac{3 \cdot 56}{100} = \frac{1}{3} + \frac{168}{100} = \frac{1}{3} + \frac{42}{25} $$. Using a common denominator of 75, $$ \frac{1}{3} = \frac{25}{75} $$ and $$ \frac{42}{25} = \frac{126}{75} $$, so $$ a_{56} = \frac{25}{75} + \frac{126}{75} = \frac{151}{75} \approx 2.0133 $$. Thus, $$ [a_{56}] = 2 $$.

Now, find when $$ [a_n] $$ changes. Solve for when $$ a_n \lt 1 $$:

$$$ \frac{1}{3} + \frac{3n}{100} \lt 1 \implies \frac{3n}{100} \lt 1 - \frac{1}{3} = \frac{2}{3} \implies 3n \lt \frac{2}{3} \cdot 100 = \frac{200}{3} \implies 9n \lt 200 \implies n \lt \frac{200}{9} \approx 22.222. $$$

Since $$ n $$ is an integer, $$ n \leq 22 $$. Check $$ n = 22 $$: $$ a_{22} = \frac{1}{3} + \frac{3 \cdot 22}{100} = \frac{1}{3} + \frac{66}{100} = \frac{1}{3} + \frac{33}{50} $$. Using a common denominator of 150, $$ \frac{1}{3} = \frac{50}{150} $$ and $$ \frac{33}{50} = \frac{99}{150} $$, so $$ a_{22} = \frac{50}{150} + \frac{99}{150} = \frac{149}{150} \approx 0.9933 \lt 1 $$. Thus, $$ [a_{22}] = 0 $$. For $$ n = 23 $$: $$ a_{23} = \frac{1}{3} + \frac{3 \cdot 23}{100} = \frac{1}{3} + \frac{69}{100} $$. Using a common denominator of 300, $$ \frac{1}{3} = \frac{100}{300} $$ and $$ \frac{69}{100} = \frac{207}{300} $$, so $$ a_{23} = \frac{100}{300} + \frac{207}{300} = \frac{307}{300} \approx 1.0233 \gt 1 $$. Thus, $$ [a_{23}] = 1 $$. Therefore, for $$ n = 1 $$ to $$ 22 $$, $$ [a_n] = 0 $$.

Next, solve for when $$ a_n \lt 2 $$:

$$$ \frac{1}{3} + \frac{3n}{100} \lt 2 \implies \frac{3n}{100} \lt 2 - \frac{1}{3} = \frac{5}{3} \implies 3n \lt \frac{5}{3} \cdot 100 = \frac{500}{3} \implies 9n \lt 500 \implies n \lt \frac{500}{9} \approx 55.555. $$$

Since $$ n $$ is an integer, $$ n \leq 55 $$. Check $$ n = 55 $$: $$ a_{55} = \frac{1}{3} + \frac{3 \cdot 55}{100} = \frac{1}{3} + \frac{165}{100} = \frac{1}{3} + \frac{33}{20} $$. Using a common denominator of 60, $$ \frac{1}{3} = \frac{20}{60} $$ and $$ \frac{33}{20} = \frac{99}{60} $$, so $$ a_{55} = \frac{20}{60} + \frac{99}{60} = \frac{119}{60} \approx 1.9833 \lt 2 $$. Thus, $$ [a_{55}] = 1 $$. For $$ n = 56 $$, as before, $$ [a_{56}] = 2 $$. Therefore, for $$ n = 23 $$ to $$ 55 $$, $$ [a_n] = 1 $$, and for $$ n = 56 $$, $$ [a_n] = 2 $$.

Now, express $$ f(n) $$:

• For $$ n = 1 $$ to $$ 22 $$: $$ f(n) = n \cdot 0 = 0 $$.
• For $$ n = 23 $$ to $$ 55 $$: $$ f(n) = n \cdot 1 = n $$.
• For $$ n = 56 $$: $$ f(n) = 56 \cdot 2 = 112 $$.

The sum is:

$$$ \sum_{n=1}^{56} f(n) = \sum_{n=1}^{22} 0 + \sum_{n=23}^{55} n + 112 = 0 + \sum_{n=23}^{55} n + 112. $$$

Compute $$ \sum_{n=23}^{55} n $$. The sum of the first $$ m $$ integers is $$ \frac{m(m+1)}{2} $$, so:

$$$ \sum_{n=23}^{55} n = \sum_{n=1}^{55} n - \sum_{n=1}^{22} n = \frac{55 \cdot 56}{2} - \frac{22 \cdot 23}{2} = \frac{3080}{2} - \frac{506}{2} = 1540 - 253 = 1287. $$$

Alternatively, the number of terms from 23 to 55 is $$ 55 - 23 + 1 = 33 $$, and the sum is $$ \frac{\text{number of terms} \cdot (\text{first term} + \text{last term})}{2} = \frac{33 \cdot (23 + 55)}{2} = \frac{33 \cdot 78}{2} = 33 \cdot 39 = 1287 $$.

Add the term for $$ n = 56 $$:

$$$ \sum_{n=1}^{56} f(n) = 1287 + 112 = 1399. $$$

Hence, the correct answer is Option C.