CMR : A ⋮ BA = 1^3 + 2^3 + ... + 98^3 + 99^3; B = 1 + 2 + 3 + ... + 99

To evaluate the expressions given in the problem, let's start by calculating \( A \) and \( B \) step by step.

### Step 1: Calculate \( A \)

The expression for \( A \) is:

\[
A = 1^3 + 2^3 + \ldots + 99^3
\]

There is a formula for the sum of the cubes of the first \( n \) natural numbers, which is:

\[
\left( \frac{n(n + 1)}{2} \right)^2
\]

For \( n = 99 \):

\[
A = \left( \frac{99(99 + 1)}{2} \right)^2 = \left( \frac{99 \cdot 100}{2} \right)^2 = (4950)^2
\]

Now, calculate \( 4950^2 \):

\[
4950^2 = 24502500
\]

Thus, we have:

\[
A = 24502500
\]

### Step 2: Calculate \( B \)

The expression for \( B \) is:

\[
B = 1 + 2 + 3 + \ldots + 99
\]

Using the formula for the sum of the first \( n \) natural numbers:

\[
S_n = \frac{n(n + 1)}{2}
\]

For \( n = 99 \):

\[
B = \frac{99(99 + 1)}{2} = \frac{99 \cdot 100}{2} = 4950
\]

### Step 3: Formulate \( A \div B \)

Now we can express \( \frac{A}{B} \):

\[
\frac{A}{B} = \frac{24502500}{4950}
\]

To compute this division, we can also simplify \( 24502500 \) by \( 4950 \):

Calculating \( \frac{24502500}{4950} \):

First, we can simplify \( 24502500 \):
\[
\frac{24502500 \div 4950}{1} = 4950
\]

Thus, we obtain:

\[
\frac{A}{B} = 4950
\]

### Conclusion

Therefore, the answer is:

\[
\boxed{4950}
\]