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

16 A + 8 B + 4 C + 2 D + E

. Every listed number has

E = 1

. For that plane, the four-variable Karnaugh map is

$A B \ C D$	$00$	$01$	$11$	$10$
$00$	$0$	$1$	$1$	$1$
$01$	$0$	$1$	$0$	$1$
$11$	$0$	$0$	$1$	$1$
$10$	$1$	$1$	$1$	$0$

The entire

E = 0

plane is zero. Five valid groups in the displayed Karnaugh map give the product terms

\overline{A} \overline{C} D, \overline{A} C \overline{D}, A \overline{B} \overline{C}, A BC, \overline{B} D .

Including

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 is

P_{odd} = E (\overline{A} \overline{C} D \lor \overline{A} C \overline{D} \lor A \overline{B} \overline{C} \lor A BC \lor \overline{B} D) .

Any gate with more than two inputs can be replaced by a tree of two-input gates. To include the exceptional prime number

2

00010

and add one OR branch:

P_{all} = P_{odd} \lor \overline{A} \overline{B} \overline{C} D \overline{E} .

If “even prime” is read literally instead of following the listed numbers, the complete first detector is simply

P_{even} = \overline{A} \overline{B} \overline{C} D \overline{E} .

All three formulas classify all

32

possible inputs, including

0

and

1

as nonprime.

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