2025 AMC 10A Problem 8

Let $T$ be the number of true statements, so $4 - T$ are false. Every statement speaks only about $T$:

• statement 1 is true exactly when $T \ge 1$;
• statement 2 is true exactly when $T \ge 2$;
• statement 3 is true exactly when $4 - T \ge 2$, that is $T \le 2$;
• statement 4 is true exactly when $4 - T \ge 1$, that is $T \le 3$.

A value of $T$ is possible only if exactly $T$ of these four conditions hold. Test all five values.

• $T = 0$: conditions 3 and 4 hold, so $2$ statements are true, not $0$.
• $T = 1$: conditions 1, 3, 4 hold, so $3$ are true, not $1$.
• $T = 2$: all four conditions hold, so $4$ are true, not $2$.
• $T = 3$: conditions 1, 2, 4 hold and condition 3 fails, so $3$ are true. Consistent.
• $T = 4$: conditions 1, 2, 4 hold, so $3$ are true, not $4$.

Only $T = 3$ survives: statements 1, 2 and 4 are true and statement 3 is false. Reading it back confirms the arrangement, since "at least two of these statements are false" is indeed false when exactly one is false. The number of false statements is $4 - 3 = 1$.

The answer is $\boxed{\textbf{(B)}\ 1}$.

Self-referential statements have no truth value until the whole assignment is fixed, so the only sound method is to guess the global quantity, here the count $T$, and check whether the guess reproduces itself. Because all four statements mention nothing but how many are true or false, the puzzle has a single unknown and just five candidates, each testable in a line. The same fixed-point check handles any list of "at least $k$ of these are true or false" statements.

• Finding that three statements are true and marking 3, choice (D), when the question asks how many are false.
• Reading statements 2 and 3 as a flat contradiction and concluding that two statements must be false, choice (C).
• Judging each statement against the other three only; every statement is also one of "these statements" and must be counted in its own tally.