Solution: Original area: \( 8 \times 15 = 120 \) m². New dimensions: \( 8 - 3 = 5 \) meters and \( 15 - 3 = 12 \) meters. New area: \( 5 \times 12 = 60 \) m². The decrease is \( 120 - 60 = 60 \) m². \boxed{60}Question: A geologist is studying the periodic patterns of seismic waves and notes that a specific wave pattern repeats every 24 seconds. If the current time is 2025-06-28 18:21:16, what is the smallest positive time (in seconds) after 18:21:16 when the wave pattern will repeat, such that t

["SEO-Optimized Article: When Will the Seismic Wave Pattern Repeat? Smallest Time After 18:21:16 with Remainder 6 mod 11", "Understanding the periodic behavior of seismic waves is crucial for predicting earthquake patterns and improving early warning systems. A recent study by a geologist analyzing a repeating seismic wave pattern identifies a key timing question: What is the smallest positive time—measured in seconds after 18:21:16—when the wave repeats and satisfies the mathematical condition that its duration from 18:21:16 is congruent to 6 modulo 11? Let’s explore the solution step by step.", "The wave pattern originally repeats every 24 seconds, meaning the full period is 24s. The current moment is 18:21:16, and we seek the smallest positive ( t ) such that:", "[\nt \equiv 6 \pmod{11} \quad \ ext{and} \quad t > 0\n]", "Additionally, ( t ) must correspond to a full wave cycle or a time when the pattern realigns—i.e., ( t = 24k ) for some integer ( k \geq 1 )—but with the extra constraint that ( t \equiv 6 \pmod{11} ).", "We begin by solving the congruence:", "Find the smallest positive integer ( t ) such that:\n[\nt = 24k \quad \ ext{and} \quad 24k \equiv 6 \pmod{11}\n]", "First, reduce ( 24 \mod 11 ):\n[\n24 \div 11 = 2 \ ext{ remainder } 2 \Rightarrow 24 \equiv 2 \pmod{11}\n]", "Substitute into the congruence:\n[\n2k \equiv 6 \pmod{11}\n]", "To solve for ( k ), multiply both sides by the modular inverse of 2 modulo 11. Since ( 2 \ imes 6 = 12 \equiv 1 \pmod{11} ), the inverse of 2 is 6. Thus:\n[\nk \equiv 6 \ imes 6 = 36 \equiv 36 \mod 11 = 3 \pmod{11}\n]", "So, ( k = 11m + 3 ) for integer ( m \geq 0 ). The smallest positive ( k ) is when ( m = 0 ):\n[\nk = 3 \Rightarrow t = 24k = 24 \ imes 3 = 72 \ ext{ seconds}\n]", "Now verify the congruence condition:\n[\n72 \div 11 = 6 \ ext{ remainder } 6 \Rightarrow 72 \equiv 6 \pmod{11}\n]", "This satisfies the required condition. Therefore, the smallest positive time after 18:21:16 when the seismic wave pattern repeats and the duration is congruent to 6 mod 11 is 72 seconds.", "🟢 Why This Matters\nThis precise timing helps seismologists align monitoring data with exact repeating cycles, improving detection accuracy and enabling better modeling of wave propagation across fault lines. By combining physical periodicity with number theory, researchers can predict wave returns with high confidence.", "Boxed Answer:\n[\n\boxed{72}\n]", "For future references, always check if wave repetition times align not just with periodicity, but with modular conditions—this advances both theoretical and applied seismology."]