Find the sum of all 3 digit whole numbers that are divisible by 13?

To find the sum of all 3-digit whole numbers that are divisible by 13, we need to find the first and last number in the sequence and then use the formula for the sum of an arithmetic series.

The first 3-digit number divisible by 13 is 104 (13 * 8 = 104).
The last 3-digit number divisible by 13 is 988 (13 * 76 = 988).

To find the number of terms in the sequence, we use the formula for the nth term of an arithmetic series: an = a1 + (n - 1)d, where an is the last term, a1 is the first term, n is the number of terms, and d is the common difference. In this case, the common difference is 13.

Using the formula, we can solve for n:
988 = 104 + (n - 1) * 13
988 - 104 = 13n - 13
884 = 13n - 13
897 = 13n
n = 69

So, there are 69 terms in the sequence.

Now, we can use the formula for the sum of an arithmetic series: Sn = (n/2)(a1 + an), where Sn is the sum of the series.

Using the formula, we can find the sum:
Sn = (69/2)(104 + 988)
Sn = (34.5)(1092)
Sn = 37,620

Therefore, the sum of all 3-digit whole numbers that are divisible by 13 is 37,620.