The sum of all those terms, of the arithmetic progression 3, 8, 13, ..., 373, which are not divisible by 3, is equal to _______.

Find the sum of terms in AP 3, 8, 13, ..., 373 that are NOT divisible by 3. We first determine the total number of terms: since $$a = 3, d = 5, a_n = 373$$ and $$373 = 3 + (n-1)(5)$$, it follows that $$n = 75$$. The total sum of all 75 terms is $$ S = \frac{75}{2}(3 + 373) = \frac{75 \times 376}{2} = 14100 $$. To identify the terms divisible by 3, observe the sequence begins 3, 8, 13, 18, 23, 28, 33, ... and a general term is $$3 + 5(k-1) = 5k - 2$$, which is divisible by 3 when $$5k - 2 \equiv 0 \pmod{3}$$, i.e., $$2k \equiv 2 \pmod{3}$$ so $$k \equiv 1 \pmod{3}$$. Thus the indices are $$k = 1, 4, 7, ..., 73$$, giving a sub-AP with first term 3, common difference 15, and last term 363. The number of terms in this sub-AP is $$\frac{363 - 3}{15} + 1 = 25$$. The sum of these 25 terms is $$ S_3 = \frac{25}{2}(3 + 363) = \frac{25 \times 366}{2} = 4575 $$. Subtracting this from the total sum gives $$14100 - 4575 = 9525$$.

The answer is 9525.