Solution: The sum of squares formula gives $S = rac{15 imes 16 imes 31}{6} = 1240$. Compute $1240 \mod 13$: $13 imes 95 = 1235$, so $1240 - 1235 = 5$. The remainder is $oxed{5}$.Question: A science educator is designing a lesson on probability using a digital simulation that generates random 6-digit numbers using only the digits 1, 2, and 3. What is the probability that a randomly generated number has exactly two consecutive digits that are the same?

["Learning Probability Through Digit Patterns: A Simulation-Based Approach with Consecutive Matching", "In modern STEM education, introducing probability concepts through engaging simulations helps students visualize abstract ideas. A compelling hands-on activity involves generating six-digit numbers using only the digits 1, 2, and 3, then analyzing the likelihood of specific patterns—such as exactly two consecutive identical digits. This exploration not only reinforces combinatorics but also strengthens logical reasoning in probability.", "### The Problem: Counting Valid 6-Digit Numbers with Exactly Two Consecutive Matching Digits", "Each digit in the number is independently chosen from {1, 2, 3}. We are to compute the probability that a randomly generated 6-digit number (no leading zeros, but here all digits are ≥1, so no issues) contains exactly two consecutive digits that are the same, and no more than that elsewhere in the number.", "Let’s define:\n- A consecutive match as two adjacent digits that are equal: e.g., "112...", "333...", but not "111..." (which contains three in a row).\n- We want sequences where exactly one pair of consecutive equal digits appears, and no three or more identical adjacent digits.", "For example, valid examples include:\n- 112123 (one pair: 11)\n- 221321 (valid)\n- Invalid: 112233 (three pairs but also "22" and "33", and possibly more than one “exactly two” claims? But if both “11” and “22” are present, that’s two separate pairs—if not overlapping and total cumulative count is “exactly two” such pairs, clarity matters).", "But to match the educational intent—clean counting—let’s clarify:\nWe target sequences with exactly one instance of two consecutive identical digits, and no digit repeated three times consecutively in the entire number.", "This is a combinatorial probability problem requiring careful enumeration.", "---", "### Step 1: Total number of 6-digit numbers with digits in {1,2,3}", "Each of 6 positions has 3 choices:", "$$\n\ ext{Total} = 3^6 = 729\n$$", "---", "### Step 2: Count favorable outcomes — sequences with exactly one pair of consecutive identical digits, no triple repeats", "This is complex due to overlapping cases and boundary conditions. We use a constructive counting method.", "We model the string of 6 digits $ d_1d_2d_3d_4d_5d_6 $, where digits ∈ {1,2,3}, such that:\n- Exactly one index $ i \in {1,2,3,4,5} $ satisfies $ d_i = d_{i+1} $\n- For all $ i = 1 $ to $ 5 $, $ d_i <br/>\ne d_{i+1} $ whenever $ i $ is not the only match (to avoid multiple matches or triple repeats)\n- No three consecutive digits are equal.", "We consider the structure of such sequences: a run of length 2 (“AA”) embedded in a background of runs of length 1 (“A”) and isolated digits, with no run longer than 2, and exactly one run of length 2.", "Let’s define a “run” as a maximal consecutive sequence of the same digit.", "We want exactly one run of length 2, and all other runs of length 1 — and no run longer than 2.", "Possible run-length patterns for a 6-digit number with exactly one pair (i.e., one run of length 2, rest length 1, total run count = number of runs ≥ 5):", "Let $ r $ = total number of runs.\nSince we have one run of length 2, and the rest (5 runs) of length 1, then $ r = 6 $.\n(6 runs: five single 1-digit blocks and one double 2-digit block → total digits: $ 1+1+1+1+1+2 = 7 $ → too many!)", "Wait: 5 runs of length 1 = 5 digits, plus one run of length 2 = 2 digits → total 7 digits. But we have only 6 digits.", "So impossibility?", "Wait: 5 runs of length 1 = 5, plus 1 run of length 2 → total length 7 → not possible.", "Thus, no such sequence exists with exactly one run of length 2 and 6 total digits, because 6 = sum of run lengths ⇒ if one run is length 2, remaining 4 runs must sum to 4, so each length 1 → 5 runs, total length 2 + 5 = 7 → contradiction.", "Therefore, the only way to have exactly one pair of consecutive equal digits with no triple repetition is to have two runs: one of length 2, five runs of length 1 → total positions: 2 + 5 = 7 → still too long.", "Wait — contradiction again.", "Any sequence with exactly one instance of two consecutive equal digits must have run structure with one run of length 2 and five runs of length 1 → total length = 2 + 5×1 = 7 → exceeds 6.", "Hence, impossible.", "But what if the pair is not isolated? Suppose we have runs: e.g., [AA][B][C][D][E] → runs: AA (len 2), then singles. Total length: 2+1+1+1+1 = 6 → 5 runs, one of length 2, others length 1 → total length 6.", "Number of runs: 5.", "We need exactly one pair of consecutive equals — i.e., exactly one run of length ≥2, and no run longer than 2 (to avoid triple repeats). So maximum run length is 2.", "Thus, only possible configuration is: six digits, one run of length 2, and four rings of length 1 → total length: 2 + 4 = 6.", "But 6 ÷ 1 run of len 2 + 4 of len 1 = 6 digits → yes!", "Wait: number of runs = 5 (one of length 2, four of length 1).\nSequence: e.g., AABCD? No — runs: AA, B, C, D, E → five runs: AA (2), B(1), C(1), D(1), E(1). Total 2+1+1+1+1=6. But number of runs is 5, each run 1 or 2 → total length 6. Only one run has length 2 → satisfies “exactly one pair of consecutive equal digits”.", "Can we have a run of length 3? Then we’d have multiple consecutive equals, but more than one “pair” — e.g., "111" contains "11" and "11" overlapping — but we count adjacent equal pairs, so "111" has two such pairs: positions 1–2 and 2–3. So one run of length 3 → two instances of consecutive identical digits → violates “exactly two” if interpreted as two such pairs, but our goal is exactly one occurrence of consecutive equality — i.e., only one $ d_i = d_{i+1} $.", "But "111" has two such transitions — not allowed.", "So to have exactly one $ i $ such that $ d_i = d_{i+1} $, and no digit repeated thrice, the only possible pattern is: one pair "AA" embedded in a sequence with five runs total — meaning four single digits, one double — total 6 digits.", "Number of runs: 5.", "One run of length 2, four of length 1 → total digits: 2 + 4 = 6 → valid.", "So only valid shape: five runs → one of length 2, four of length 1 → total length 6.", "Now, we count how many such sequences exist, with digits in {1,2,3}, no run longer than 2, and exactly one run of length 2.", "---", "### Step 3: Count valid configurations", "We proceed by:", "1. Choosing positions for the run of two identical digits.\n2. Assigning the digit (1,2, or 3).\n3. Assigning the remaining 4 digits (in single runs) such that:\n - No two adjacent digits form another equal pair.\n - No digit appears three times in a row (already avoided).\n - The adjacent digits to the run do not create another pair.", "We must ensure:\n- The run of two is not adjacent to the same digit on one side (would make triple if adjacent), or on both sides (only if internal).\n- But since it’s a single block, we only need:\n - The digit to the left of the pair (if exists) ≠ A\n - The digit to the right of the pair (if exists) ≠ A\n - Also, the first digit of the run (A) must not equal digit immediately before it — already ensured by constraints.\n - And the last digit of the run (A) must not equal digit after — also ensured.", "But to avoid overlapping pairs, we must ensure that:\n- If the double run is surrounded, the neighboring digits are different from A.\n- Also, if a single-digit block is adjacent to the double, its digit ≠ A.", "So we model the run structure as a sequence of 5 runs: one of length 2 (call this R2), four of length 1 (R1).", "We need to count the number of ways to:\n- Assign run types (which digit, and which position is double)\n- Assign digits to runs\n- Ensure no adjacent equal digits across run boundaries", "Let’s denote the runs as $ R_1, R_2, R_3, R_4, R_5 $, order left to right.", "Exactly one $ R_i $ has length 2, the other four have length 1.", "There are $ \binom{5}{1} = 5 $ ways to choose which run is the double.", "For each such choice, we assign a digit to it (3 choices: 1,2,3).", "Now, the remaining 4 runs are single digits, each ∈ {1,2,3}, but:\n- Each adjacent pair of runs must not have equal digits (especially avoid creating new runs of length 2)\n- So: if two adjacent runs are both length 1 and share the same digit, they form a new run of length 2 → invalid.\n- Also, a single digit next to a double block must differ from the double digit.", "So the assignment must ensure that:\n- Between runs, if both are single, digits differ.\n- A single run adjacent to a double run must have digit ≠ A (A = digit in double run).", "We must model this depending on whether the double run is at the start, end, or middle.", "Due to symmetry, we can compute for each position and account for constraints.", "But to simplify, let’s use a known technique: inclusion with structural enumeration.", "However, due to complexity, we instead compute total number of such sequences via constructive counting.", "Let’s fix the position of the double run, say at position $ i $ (1st, 2nd, 3rd, 4th, or 5th run).", "Let $ L = 2 $ (length of double), so the run occupies 2 adjacent digits.", "Total length = 6.", "The 5 runs take up positions: the sum of run lengths is 6.", "We now consider each possible location of the double run.", "---", "Case 1: Double run is at position 1 (first two digits)\nRuns: R1 (double), R2, R3, R4, R5 (all singles)\nLet digits: A,A,B,C,D,E\nConstraints:\n- B ≠ A (to avoid triple before)\n- C ≠ A (to avoid adjacent pair at end of R1 and start of R2)\n- D ≠ B\n- E ≠ C", "And digits ∈ {1,2,3}", "Step-by-step:", "- Choose A: 3 choices\n- B: ≠ A → 2 choices\n- C: ≠ A → 2 choices (but can = B? Yes, as long as ≠ A)\n- D: ≠ B → 2 choices\n- E: ≠ C → 2 choices", "So: 3 × 2 × 2 × 2 × 2 = 3 × 16 = 48", "But wait — is R1 used for double digit, and R2–R5 for single digits, all different types? Yes — only one double run.", "So 48 sequences for this configuration.", "Case 2: Double run at middle — positions 2 and 3 → runs R1, R2 (double), R3, R4, R5\nSequence: B,A,A,C,D,E\nConstraints:\n- A ≠ B (left side)\n- C ≠ A (to avoid B=A,A,C → but only adjacent equality matters) — adjacent pairs: B-A (ok), A-A (valid), A-C → so need C ≠ A? Not directly — but A-C forms a pair only if C=A? No: A (run) and A (run) →"]