Sum Calculation Steps

The question requires finding the sum of the squares of all two-digit numbers that are divisible by 4.

• Two-digit numbers range from 10 to 99.
• The two-digit numbers divisible by 4 are: 12, 16, 20, ..., 96.
• These numbers can be represented in the form $4m$, where $m$ is an integer.
• For the smallest number, $4m = 12$, so $m = 3$.
• For the largest number, $4m = 96$, so $m = 24$.
• Thus, the numbers are $4 \times 3, 4 \times 4, \dots, 4 \times 24$.
• The number of such terms is $24 - 3 + 1 = 22$.

Squares Sum Computation

We need to calculate the sum $S = 12^2 + 16^2 + 20^2 + \dots + 96^2$. This sum can be expressed using the representation $4m$:

$ S = \sum_{m=3}^{24} (4m)^2 $

Factor out the constant $16 = 4^2$:

$ S = \sum_{m=3}^{24} 16m^2 = 16 \sum_{m=3}^{24} m^2 $

We use the formula for the sum of the first $n$ squares: $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$.

The sum $\sum_{m=3}^{24} m^2$ is calculated as:

$ \sum_{m=3}^{24} m^2 = \left( \sum_{m=1}^{24} m^2 \right) - \left( \sum_{m=1}^{2} m^2 \right) $

Evaluating Sum of Squares

Calculate the sum of squares from 1 to 24:

$ \sum_{m=1}^{24} m^2 = \frac{24(24+1)(2 \times 24 + 1)}{6} = \frac{24 \times 25 \times 49}{6} $

$ = 4 \times 25 \times 49 = 100 \times 49 = 4900 $

Calculate the sum of squares from 1 to 2:

$ \sum_{m=1}^{2} m^2 = 1^2 + 2^2 = 1 + 4 = 5 $

Subtract to find the sum from $m=3$ to $m=24$:

$ \sum_{m=3}^{24} m^2 = 4900 - 5 = 4895 $

Final Sum Calculation

Substitute the result back into the expression for $S$:

$ S = 16 \times 4895 $

Performing the multiplication:

$ S = 78320 $

Therefore, the sum of the squares of all two-digit numbers divisible by 4 is 78320.