Solution: The LCM of 11 and 13 is $143$. The smallest three-digit multiple of 143 is $143 \times 1 = 143$ (since $143 \times 2 = 286$ is also three-digit, but 143 is smaller). Thus, the answer is $\boxed{143}$.Question: Two marine biologists, Dr. Coral and Dr. Tide, arrive at a research station at random times between 7:00 and 8:00. If Dr. Coral arrives after Dr. Tide, what is the probability that Dr. Tide arrived before 7:45?

["Solution: The LCM of 11 and 13 is 143. For the Probability Problem: Dr. Coral and Dr. Tide arrive uniformly at random between 7:00 and 8:00, and Dr. Coral arrives after Dr. Tide. What is the probability that Dr. Tide arrived before 7:45?", "---", "Understanding the Problem\nThis classic probability scenario involves two independent, uniformly distributed arrival times within a one-hour interval. We are told Dr. Coral arrives after Dr. Tide, and we want the conditional probability that Dr. Tide arrived before 7:45.", "Let’s define:\n- $T$ = Dr. Tide’s arrival time in minutes after 7:00 (so $T \in [0, 60]$)\n- $C$ = Dr. Coral’s arrival time in minutes after 7:00 ($C \in [0, 60]$)", "We are given that $C > T$, and we want $P(T < 45 \mid C > T)$.", "---", "Step 1: Use Conditional Probability Formula\n[\nP(T < 45 \mid C > T) = \frac{P(T < 45 \ ext{ and } C > T)}{P(C > T)}\n]", "Due to symmetry in independent uniform variables, $P(C > T) = \frac{1}{2}$, since $C$ and $T$ are identically distributed and continuous—there’s no bias toward earlier or later.", "---", "Step 2: Compute the Numerator – $P(T < 45 \ ext{ and } C > T)$", "This is the area in the $T$-$C$ plane where $0 \leq T < 45$, $T < C \leq 60$.", "Graphically, this region is bounded:\n- $T$ ranges from 0 to 45\n- For each $T$, $C$ ranges from $T$ to 60", "So the area is:\n[\n\int_{T=0}^{45} \int_{C=T}^{60} dC, dT = \int_0^{45} (60 - T), dT\n]", "Calculate:\n[\n\int_0^{45} (60 - T), dT = \left[60T - \frac{T^2}{2}\right]_0^{45} = 60 \cdot 45 - \frac{45^2}{2} = 2700 - \frac{2025}{2} = 2700 - 1012.5 = 1687.5\n]", "This is the number of favorable outcomes (in scaled units). The total possible area is $60 \ imes 60 = 3600$, but since we’re working with likelihoods (proportions), we use ratios:", "So,\n[\nP(T < 45 \ ext{ and } C > T) = \frac{1687.5}{3600} = \frac{16875}{36000} = \frac{27}{64} \quad \ ext{(after simplifying)}\n]", "Wait — better to express everything proportionally:", "Since the joint distribution is uniform over $[0,60]^2$, we compute areas as fractions of total area $3600$.", "The numerator area:\n[\n\int_0^{45} (60 - T), dT = \left[60T - \frac{T^2}{2}\right]_0^{45} = 2700 - \frac{2025}{2} = \frac{5400 - 2025}{2} = \frac{3375}{2} = 1687.5\n]", "Total area = $60^2 = 3600$\nSo,\n[\nP(T < 45 \ ext{ and } C > T) = \frac{1687.5}{3600} = \frac{3375/2}{3600} = \frac{3375}{7200} = \frac{27}{64}\n]", "Wait — actually, in conditional probability with uniform variables, we can normalize without squaring:", "Alternatively, use geometric probability:\n- The total area where $C > T$ is a triangle with area $\frac{1}{2} \cdot 60 \cdot 60 = 1800$\n- The favorable region is where $T < 45$ and $C > T$. This is a trapezoid from $T=0$ to $45$, $C = T$ to $60$.\nBut split:\n- For $T \in [0,45]$: $C$ ranges from $T$ to 60 → area = $\int_0^{45} (60 - T) dt = 60 \cdot 45 - \frac{1}{2}(45)^2 = 2700 - 1012.5 = 1687.5$", "So,\n[\nP(T < 45 \mid C > T) = \frac{1687.5}{1800} = \frac{16875}{18000} = \frac{27}{32}\n]", "Wait — simplify:", "[\n\frac{1687.5}{1800} = \frac{16875}{18000} = \frac{16875 \div 375}{18000 \div 375} = \frac{45}{48} = \frac{15}{16}? \quad \ ext{No.}\n]", "Correct common denominator:", "$1687.5 / 1800 = 16875 / 18000 = 3375 / 3600 = 135 / 240 = 27 / 48 = 9/16$? Let’s compute:", "[\n1687.5 \div 1800 = \frac{1687.5}{1800} = \frac{3375}{3600} = \frac{3375 \div 75}{3600 \div 75} = \frac{45}{48} = \frac{15}{16}\n]", "Wait — $3375 \div 225 = 15$, $3600 \div 225 = 16$? $225 \cdot 15 = 3375$, $225 \cdot 16 = 3600$ — yes!", "But actually:\n[\n\frac{1687.5}{1800} = \frac{1687.5 \ imes 2}{3600 \ imes 2} = \frac{3375}{3600} = \frac{27}{32}\n]", "Check: $27 \div 32 = 0.84375$, $1687.5 / 1800 = 0.9375 = \frac{15}{16} = 0.9375$? No:\n$15/16 = 0.9375$, but $1687.5 / 1800 = 0.9375$ — yes!", "Wait:\n$1687.5 \div 1800 = ?$\n$1800 \ imes 0.9 = 1620$, $1687.5 - 1620 = 67.5$, $67.5 / 1800 = 0.0375$, so $0.9375$ — correct.\nAnd $1687.5 / 1800 = \frac{3375}{3600} = \frac{27}{32}$?\n$3375 \div 125 = 27$, $3600 \div 125 = 28.8$ — no.", "Divide numerator and denominator by 75:\n$1687.5 \div 75 = 22.5$, $1800 \div 75 = 24$ → $22.5/24 = 45/48 = 15/16$?\n$22.5 / 24 = 0.9375 = 15/16$ — yes!\nSo $1687.5 / 1800 = 15/16$", "But $15/16 = 0.9375$, and $1687.5 / 1800 = 16875 / 18000 = 3375 / 3600 = 67.5 / 80 = ?$\nBetter:\n[\n\frac{1687.5}{1800} = \frac{1687.5 \ imes 2}{3600 \ imes 2} = \frac{3375}{3600} = \frac{3375 \div 225}{3600 \div 225} = \frac{15}{16}\n]", "Yes — so:\n[\nP(T < 45 \mid C > T) = \frac{3375}{3600} \div \frac{1800}{1800} = \frac{1687.5}{1800} = \frac{15}{16}\n]", "Wait — no: conditional probability is $P(A \cap B)/P(B)$, where $P(B) = 1800/3600 = 1/2$", "So:\n[\nP(T < 45 \mid C > T) = \frac{1687.5 / 3600}{1/2} = \frac{1687.5}{1800} = \frac{15}{16}\n]", "Yes.", "So the final probability is $\boxed{\dfrac{15}{16}}$", "---", "Final Explanation\nGiven symmetry and uniform arrivals, the condition $C > T$ restricts us to the upper triangle of the unit square (in time terms normalized to 60 minutes). Within this triangle, Dr. Tide’s arrival time $T < 45$ covers 15/16 of the conditional area.", "Thus, the probability that Dr. Tide arrived before 7:45, given Dr. Coral arrived after, is $\boxed{\dfrac{15}{16}}$.", "---", "Summary:\n- Solution uses geometric probability with uniform variables\n- Condition $C > T$ gives a triangular region\n- Favorable case regions computed via integration\n- Final conditional probability: $ \frac{15}{16} $", "[\n\boxed{\dfrac{15}{16}}\n]"]