$15 \cdot 22$ On the card below are four statements, and only these four:
On this card exactly one statement is false.
On this card exactly two statements are false.
On this card exactly three statements are false.
On this card exactly four statements are false.
(Assume that each statement on the card is either true or false.) Among them, the number of false statements is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(E) 4 .
(29th American High School Mathematics Examination, 1978)
[Solution] Since any pair of propositions on the card are contradictory, at most one of them is true.

Assume that none of the propositions is true, which would imply that the “fourth proposition” is true. Therefore, the statement that “there must be exactly one true proposition on the card” also holds. This contradicts the assumption.

Thus, one of the propositions must be true, in other words, “exactly three propositions are not true,” meaning the third proposition is true.
Therefore, the answer is $(D)$.