$\textbf{Problem 1.}$ [b][/b]There are less than $400$ marbles.[i][/i] If they are distributed among $3$ childrens, there is one left over if they are distributed among $7$ children, there are 2 left over. Finally if they are distributed among $5$ children, there are none left over.

What is the largest number of the marbels you have?
1. Let \( n \) be the number of marbles. According to the problem, we have the following congruences:
   \[
   n \equiv 1 \pmod{3}
   \]
   \[
   n \equiv 2 \pmod{7}
   \]
   \[
   n \equiv 0 \pmod{5}
   \]

2. We need to solve this system of congruences using the Chinese Remainder Theorem (CRT). First, we solve the congruences \( n \equiv 1 \pmod{3} \) and \( n \equiv 2 \pmod{7} \).

3. Since \( n \equiv 0 \pmod{5} \), we can write \( n = 5k \) for some integer \( k \). Substituting \( n = 5k \) into the other congruences, we get:
   \[
   5k \equiv 1 \pmod{3}
   \]
   \[
   5k \equiv 2 \pmod{7}
   \]

4. Simplify the first congruence:
   \[
   5k \equiv 1 \pmod{3} \implies 2k \equiv 1 \pmod{3}
   \]
   The multiplicative inverse of 2 modulo 3 is 2, since \( 2 \times 2 \equiv 1 \pmod{3} \). Therefore:
   \[
   k \equiv 2 \pmod{3} \implies k = 3m + 2 \text{ for some integer } m
   \]

5. Substitute \( k = 3m + 2 \) into the second congruence:
   \[
   5(3m + 2) \equiv 2 \pmod{7}
   \]
   Simplify:
   \[
   15m + 10 \equiv 2 \pmod{7} \implies m + 3 \equiv 2 \pmod{7} \implies m \equiv -1 \pmod{7} \implies m \equiv 6 \pmod{7}
   \]
   Therefore:
   \[
   m = 7p + 6 \text{ for some integer } p
   \]

6. Substitute \( m = 7p + 6 \) back into \( k = 3m + 2 \):
   \[
   k = 3(7p + 6) + 2 = 21p + 20
   \]

7. Substitute \( k = 21p + 20 \) back into \( n = 5k \):
   \[
   n = 5(21p + 20) = 105p + 100
   \]

8. To find the largest \( n \) less than 400, we solve:
   \[
   105p + 100 < 400 \implies 105p < 300 \implies p < \frac{300}{105} \approx 2.857
   \]
   The largest integer \( p \) is 2.

9. Substitute \( p = 2 \) back into \( n = 105p + 100 \):
   \[
   n = 105 \times 2 + 100 = 310
   \]

Conclusion:
\[
\boxed{310}
\]