 Solution

The printed word “even” conflicts with the displayed numbers: all ten listed prime numbers are odd, and $2$ is the only even prime number. The scan confirms that wording.^[ref] We first construct the detector for the explicitly listed numbers, then add $2$ as requested. The literal even-prime detector is also given below.

Let the five input bits be $A,B,C,D,E$ with value $16A+8B+4C+2D+E$. Every listed number has $E=1$. For that plane, the four-variable Karnaugh map is

The entire $E=0$ plane is zero. Five valid groups in the displayed Karnaugh map give the product termsIncluding $E$, their accepted numbers are respectively $\{3,11\}$, $\{5,13\}$, $\{17,19\}$, $\{29,31\}$, and $\{3,7,19,23\}$. Their union is exactly the ten listed numbers. Hence an explicit Boolean circuit is: invert $A,B,C,D$ where needed, feed the five products into five AND gates, combine those outputs with an OR gate, and AND with $E$. Its output isAny gate with more than two inputs can be replaced by a tree of two-input gates. To include the exceptional prime number $2$, detect its exact binary representation $00010$ and add one OR branch:If “even prime” is read literally instead of following the listed numbers, the complete first detector is simplyAll three formulas classify all $32$ possible inputs, including $0$ and $1$ as nonprime.