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(A) $t_n$ denotes the number of triangles with distinct integral sides chosen from $\{1, 2, 3, \ldots, n\}$.$t_{20} - t_{19}$ represents the number of

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$81$

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(A) $t_n$ denotes the number of triangles with distinct integral sides chosen from $\{1, 2, 3, \ldots, n\}$.$t_{20} - t_{19}$ represents the number of triangles with distinct sides chosen from $\{1, 2, 3, \ldots, 20\}$ such that the largest side is exactly $20$.Let the sides of the triangle be $x, y, 20$ where $x By the triangle inequality,$x + y > 20$.Since $y For a fixed $y$,the smallest side $x$ must satisfy $20 - y Thus,the number of possible values for $x$ is $y - (20 - y + 1) + 1 = 2y - 20$.Summing over $y = 11, 12, \ldots, 19$:$\sum_{y=11}^{19} (2y - 20) = 2(11+12+\ldots+19) - 20(9) = 2 	imes \frac{9}{2}(11+19) - 180 = 9(30) - 180 = 270 - 180 = 90$.Wait,the sides must be distinct. If $y=11$,$x$ can be $10$. If $y=12$,$x$ can be $9, 10, 11$. If $y=13$,$x$ can be $8, 9, 10, 11, 12$. This forms an arithmetic progression: $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81$.Thus,$t_{20} - t_{19} = 81$.

Let $t_n$ denote the number of integral-sided triangles with distinct sides chosen from $\{1, 2, 3, \ldots, n\}$. Then,$t_{20} - t_{19}$ equals

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