1978 AHSME Problems/Problem 22

There can be at most one true statement on the card, eliminating $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$ . If there are $0$ true on the card, statement $4$ ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is $\textbf{(D)}\ 3$ , since $3$ are false and only the third statement ("On this card exactly three statements are false") is correct.

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1978 AHSME Problems/Problem 22