The total number of prime implicants of the function $f(w, x, y, z) = \sum(0, 2, 4, 5, 6, 10)$ is _______.

Given the function $f(w, x, y, z) = \sum(0, 2, 4, 5, 6, 10)$:

Let us list the binary representation of the minterms:

• $m_0 = 0000$

• $m_2 = 0010$

• $m_4 = 0100$

• $m_5 = 0101$

• $m_6 = 0110$

• $m_{10} = 1010$

Grouping adjacent minterms:

1. Quads (groups of 4):

   • $(m_0, m_2, m_4, m_6)$ can be combined into a quad represented by $w^{\prime}z^{\prime}$ ($0--0$).

2. Pairs (groups of 2) not covered by any larger group:

   • $(m_4, m_5)$ forms a pair represented by $w^{\prime}xy^{\prime}$ ($010-$).

   • $(m_2, m_{10})$ forms a pair represented by $x^{\prime}yz^{\prime}$ ($-010$).

All other pairs (e.g., $(m_0, m_2), (m_0, m_4), (m_2, m_6), (m_4, m_6)$) are sub-cubes of the quad $(m_0, m_2, m_4, m_6)$ and therefore are not prime implicants.

Thus, the prime implicants are:

1. $w^{\prime}z^{\prime}$

2. $w^{\prime}xy^{\prime}$

3. $x^{\prime}yz^{\prime}$

The total number of prime implicants is 3.