Screws are sold in packs of $10$ and $12$. Harry and Sam independently go to the hardware store, and by coincidence each of them buys exactly $k$ screws. However, the number of packs of screws Harry buys is different than the number of packs Sam buys. What is the smallest possible value of $k$?
To solve this problem, we need to find the smallest number of screws, \( k \), that can be bought in two different ways using packs of 10 and 12 screws. This means we need to find the smallest \( k \) such that \( k \) can be expressed as both \( 10x \) and \( 12y \) for some integers \( x \) and \( y \), where \( x \) and \( y \) are different.

1. **Identify the Least Common Multiple (LCM):**
   The smallest number \( k \) that can be expressed as both \( 10x \) and \( 12y \) is the least common multiple (LCM) of 10 and 12. This is because the LCM is the smallest number that is a multiple of both 10 and 12.

   \[
   \text{lcm}(10, 12)
   \]

2. **Calculate the LCM:**
   To find the LCM of 10 and 12, we first find the prime factorizations of the numbers:
   \[
   10 = 2 \times 5
   \]
   \[
   12 = 2^2 \times 3
   \]

   The LCM is found by taking the highest power of each prime that appears in the factorizations:
   \[
   \text{lcm}(10, 12) = 2^2 \times 3 \times 5 = 60
   \]

3. **Verify Different Pack Combinations:**
   We need to verify that 60 screws can be bought in two different ways using packs of 10 and 12:
   \[
   60 = 10 \times 6
   \]
   \[
   60 = 12 \times 5
   \]

   Here, \( x = 6 \) and \( y = 5 \), which are different integers. Therefore, Harry and Sam can buy 60 screws in different combinations of packs.

Conclusion:
The smallest possible value of \( k \) is the least common multiple of 10 and 12, which is 60.

The final answer is \(\boxed{60}\).