There are a family of $5$ siblings. They have a pile of at least $2$ candies and are trying to split them up
amongst themselves. If the $2$ oldest siblings share the candy equally, they will have $1$ piece of candy left over.
If the $3$ oldest siblings share the candy equally, they will also have $1$ piece of candy left over. If all $5$ siblings
share the candy equally, they will also have $1$ piece left over. What is the minimum amount of candy required
for this to be true?
1. We start by translating the problem statement into mathematical congruences. The problem states that:
   - If the 2 oldest siblings share the candy equally, they will have 1 piece left over. This can be written as:
     \[
     N \equiv 1 \pmod{2}
     \]
   - If the 3 oldest siblings share the candy equally, they will also have 1 piece left over. This can be written as:
     \[
     N \equiv 1 \pmod{3}
     \]
   - If all 5 siblings share the candy equally, they will also have 1 piece left over. This can be written as:
     \[
     N \equiv 1 \pmod{5}
     \]

2. We need to find the smallest positive integer \( N \) that satisfies all three congruences simultaneously. This can be done by finding the least common multiple (LCM) of the moduli (2, 3, and 5) and then solving for \( N \).

3. The least common multiple of 2, 3, and 5 is:
   \[
   \text{lcm}(2, 3, 5) = 30
   \]

4. Therefore, the three congruences can be combined into a single congruence:
   \[
   N \equiv 1 \pmod{30}
   \]

5. The smallest positive integer \( N \) that satisfies \( N \equiv 1 \pmod{30} \) is:
   \[
   N = 30k + 1
   \]
   where \( k \) is an integer. For the smallest solution greater than 1, we set \( k = 1 \):
   \[
   N = 30 \cdot 1 + 1 = 31
   \]

The final answer is \(\boxed{31}\).