Difference between revisions of "1978 AHSME Problems/Problem 22"
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− | There can be at most one true statement on the card, eliminating <math>\textbf{(A)}, \textbf{(B)},</math> and < | + | There can be at most one true statement on the card, eliminating <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}</math>. If there are <math>0</math> true on the card, statement <math>4</math> ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is <math>\textbf{(D)}\ 3</math>, since <math>3</math> are false and only the third statement ("On this card exactly three statements are false") is correct. |
Revision as of 20:07, 29 December 2019
There can be at most one true statement on the card, eliminating and . If there are true on the card, statement ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is , since are false and only the third statement ("On this card exactly three statements are false") is correct.