金声玉亮2.0模型评测报告

问题

Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes $1$ token or $4$ tokens from the stack. The player who removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ such that there is a strategy that guarantees that Bob wins, regardless of Alice’s moves. Please think step by step, and then display the answer at the very end of your response. The answer is an integer consisting of exactly 3 digits (including leading zeros), ranging from 000 to 999, inclusive. For example, the answer might be 068 or 972. If you cannot determine the correct answer, take your best guess. Remember to have the three digits as the last part of the response.

答案

模型	模型答案	正确答案
金声玉亮2.0	To determine the number of positive integers \( n \) less than or equal to 2024 such that Bob has a winning strategy in the game, we need to analyze the game's dynamics. ### Step-by-Step Analysis 1. Game Rules Recap: - Alice and Bob take turns removing 1 or 4 tokens from a stack of \( n \) tokens. - Alice goes first. - The player who removes the last token wins. 2. Winning and Losing Positions: - We need to identify the positions where Bob can guarantee a win, regardless of Alice's moves. - A key insight is to look for positions where the player to move next is forced into a losing position. 3. Pattern Recognition: - Let's start by examining small values of \( n \): - \( n = 1 \): Alice wins immediately by taking 1 token. - \( n = 2 \): Alice wins by taking 1 token, leaving 1 for Bob. - \( n = 3 \): Alice wins by taking 1 token, leaving 2 for Bob. - \( n = 4 \): Alice wins by taking 4 tokens. - \( n = 5 \): Alice can take 1 token, leaving 4 for Bob, who then wins. - \( n = 6 \): Alice can take 1 token, leaving 5 for Bob, who then wins. - \( n = 7 \): Alice can take 1 token, leaving 6 for Bob, who then wins. - \( n = 8 \): Alice can take 4 tokens, leaving 4 for Bob, who then wins. - \( n = 9 \): Alice can take 1 token, leaving 8 for Bob, who then wins. 4. General Pattern: - From the above, it seems that positions \( n = 1, 2, 3, 4 \) are winning positions for Alice. - Positions \( n = 5, 6, 7, 8 \) are winning positions for Bob. - This suggests a pattern where every 4th position starting from 5 is a winning position for Bob. 5. Verification and Extension: - To confirm, we need to check if this pattern holds for larger \( n \): - For \( n = 9 \), Alice can take 1 token, leaving 8 for Bob, who wins. - For \( n = 10 \), Alice can take 1 token, leaving 9 for Bob, who wins. - For \( n = 11 \), Alice can take 1 token, leaving 10 for Bob, who wins. - For \( n = 12 \), Alice can take 4 tokens, leaving 8 for Bob, who wins. - This confirms that every 4th position starting from 5 is a winning position for Bob. 6. Counting the Winning Positions for Bob: - We need to count how many such positions exist up to 2024. - The sequence of such positions is \( 5, 9, 13, \ldots \). - This is an arithmetic sequence with the first term \( a = 5 \) and common difference \( d = 4 \). 7. Finding the Number of Terms: - The \( k \)-th term of the sequence is given by \( a_k = 5 + (k-1) \cdot 4 = 4k + 1 \). - We need \( 4k + 1 \leq 2024 \): \[ 4k + 1 \leq 2024 \implies 4k \leq 2023 \implies k \leq \frac{2023}{4} \approx 505.75 \] - Since \( k \) must be an integer, the largest possible \( k \) is 505. 8. Conclusion: - There are 505 positions where Bob can guarantee a win. Thus, the number of positive integers \( n \) less than or equal to 2024 such that Bob has a winning strategy is: \[ \boxed{505} \]	809
deepseek-r1	The players take turns removing 1 or 4 tokens, with the goal of taking the last token. Bob can win if \( n \) is a losing position (i.e., no matter Alice's move, Bob can force a win). By analyzing positions, losing positions occur at \( n \equiv 2, 5, 7, \text{ or } 10 \pmod{10} \). These repeat every 10 numbers. For \( n \leq 2024 \): - Each block of 10 numbers contains 4 losing positions. - There are \( \left\lfloor \frac{2024}{10} \right\rfloor = 202 \) full blocks, contributing \( 202 \times 4 = 808 \) losing positions. - The remaining \( 2024 - 2020 = 4 \) numbers (2021–2024) correspond to positions 1–4 in the cycle. Only 2022 (position 2) is a losing position. Total losing positions: \( 808 + 1 = 809 \). Answer: 809	809
gpto1	To solve this problem, we'll identify all the positions (number of tokens \( n \)) where Bob, as the second player, can guarantee a win regardless of Alice's moves. In two-player impartial games like this, we often categorize positions into P-positions (Previous player can force a win) and N-positions (Next player can force a win). Step 1: Define the Base Case - When \( n = 0 \), the player whose turn it is loses (since there are no tokens to remove). So, \( n = 0 \) is a P-position. Step 2: Recurrence Relations For \( n > 0 \), define: - \( f(n) = 1 \) if either \( f(n-1) = 0 \) or \( f(n-4) = 0 \), meaning the player can move to a P-position and thus the current position is an N-position. - \( f(n) = 0 \) otherwise. Step 3: Compute \( f(n) \) for \( n \geq 1 \) By applying the recurrence and starting from \( n = 1 \), we compute \( f(n) \): - For \( n = 1 \): Since \( f(1-1) = f(0) = 0 \), \( f(1) = 1 \). - For \( n = 2 \): \( f(2-1) = f(1) = 1 \); \( f(2-4) \) is invalid. Both options lead to N-positions, so \( f(2) = 0 \). - Continue this process up to \( n = 2024 \). Step 4: Identify P-positions Upon computing, we observe a pattern: - The P-positions occur at \( n \) where \( n \mod 5 \equiv 0 \) or \( n \mod 5 \equiv 2 \). Step 5: Count the Number of P-positions We need to determine how many positive integers \( n \leq 2024 \) satisfy \( n \mod 5 \equiv 0 \) or \( n \mod 5 \equiv 2 \): - Numbers where \( n \mod 5 = 0 \): - First number: \( n = 5 \) - Last number: \( n = 2020 \) - Number of terms: \( \frac{2020 - 5}{5} + 1 = 404 \). - Numbers where \( n \mod 5 = 2 \): - First number: \( n = 2 \) - Last number: \( n = 2022 \) - Number of terms: \( \frac{2022 - 2}{5} + 1 = 405 \). Step 6: Total Number of P-positions Total P-positions (where Bob can win): - \( 404 (n \mod 5 = 0) + 405 (n \mod 5 = 2) = 809 \). Thus, there are 809 positive integers \( n \leq 2024 \) where Bob has a winning strategy. Answer: 809	809
gpt-4o-2024-08-06	To solve this problem, we need to determine the winning and losing positions for Alice and Bob. A position is a losing position if the player whose turn it is cannot force a win with optimal play, and a winning position if the player can force a win with optimal play. Let's analyze the game: 1. Base Cases: - If \( n = 0 \), the player to move loses because there are no tokens to take. So, 0 is a losing position. - If \( n = 1 \), Alice can take 1 token and win. So, 1 is a winning position. - If \( n = 2 \), Alice can take 1 token, leaving 1 for Bob, who will then win. So, 2 is a losing position. - If \( n = 3 \), Alice can take 1 token, leaving 2 for Bob, who will then lose. So, 3 is a winning position. - If \( n = 4 \), Alice can take all 4 tokens and win. So, 4 is a winning position. - If \( n = 5 \), Alice can take 4 tokens, leaving 1 for Bob, who will then win. So, 5 is a losing position. 2. General Strategy: - A position \( n \) is a losing position if every move leaves the opponent in a winning position. - A position \( n \) is a winning position if there is at least one move that leaves the opponent in a losing position. 3. Pattern Recognition: - From the base cases, we see that 0 and 2 are losing positions. - Let's continue to find the pattern for small values of \( n \). 4. Continuing the Pattern: - \( n = 6 \): Alice can take 1 token, leaving 5 (losing for Bob). So, 6 is winning. - \( n = 7 \): Alice can take 1 token, leaving 6 (winning for Bob). So, 7 is losing. - \( n = 8 \): Alice can take 4 tokens, leaving 4 (winning for Bob). So, 8 is losing. - \( n = 9 \): Alice can take 4 tokens, leaving 5 (losing for Bob). So, 9 is winning. - \( n = 10 \): Alice can take 4 tokens, leaving 6 (winning for Bob). So, 10 is losing. - \( n = 11 \): Alice can take 4 tokens, leaving 7 (losing for Bob). So, 11 is winning. - \( n = 12 \): Alice can take 4 tokens, leaving 8 (losing for Bob). So, 12 is winning. - \( n = 13 \): Alice can take 4 tokens, leaving 9 (winning for Bob). So, 13 is losing. - \( n = 14 \): Alice can take 4 tokens, leaving 10 (losing for Bob). So, 14 is winning. - \( n = 15 \): Alice can take 4 tokens, leaving 11 (winning for Bob). So, 15 is losing. - \( n = 16 \): Alice can take 4 tokens, leaving 12 (winning for Bob). So, 16 is losing. - \( n = 17 \): Alice can take 4 tokens, leaving 13 (losing for Bob). So, 17 is winning. 5. Pattern Identification: - The losing positions are 0, 2, 5, 7, 8, 10, 13, 15, 16, ... - Observing the differences: 2, 3, 2, 1, 2, 3, 2, 1, 2, ... - The pattern repeats every 5 numbers: 2, 3, 2, 1, 2. 6. Finding Losing Positions up to 2024: - The sequence of losing positions is periodic with period 5. - The sequence starts at 0 and follows the pattern: 0, 2, 5, 7, 8, 10, 13, 15, 16, 18, 21, 23, 24, 26, 29, 31, 32, 34, 37, 39, 40, 42, 45, 47, 48, 50, 53, 55, 56, 58, 61, 63, 64, 66, 69, 71, 72, 74, 77, 79, 80, 82, 85, 87, 88, 90, 93, 95, 96, 98, 101, 103, 104, 106, 109, 111, 112, 114, 117, 119, 120, 122, 125, 127, 128, 130, 133, 135, 136, 138, 141, 143, 144, 146, 149, 151, 152, 154, 157, 159, 160, 162, 165, 167, 168, 170, 173, 175, 176, 178, 181, 183, 184, 186, 189, 191, 192, 194, 197, 199, 200, 202, 205, 207, 208, 210, 213, 215, 216, 218, 221, 223, 224, 226, 229, 231, 232, 234, 237, 239, 240, 242, 245, 247, 248, 250, 253, 255, 256, 258, 261, 263, 264, 266, 269, 271, 272, 274, 277, 279, 280, 282, 285, 287, 288, 290, 293, 295, 296, 298, 301, 303, 304, 306, 309, 311, 312, 314, 317, 319, 320, 322, 325, 327, 328, 330, 333, 335, 336, 338, 341, 343, 344, 346, 349, 351, 352, 354, 357, 359, 360, 362, 365, 367, 368, 370, 373, 375, 376, 378, 381, 383, 384, 386, 389, 391, 392, 394, 397, 399, 400, 402, 405, 407, 408, 410, 413, 415, 416, 418, 421, 423, 424, 426, 429, 431, 432, 434, 437, 439, 440, 442, 445, 447, 448, 450, 453, 455, 456, 458, 461, 463, 464, 466, 469, 471, 472, 474, 477, 479, 480, 482, 485, 487, 488, 490, 493, 495, 496, 498, 501, 503, 504, 506, 509, 511, 512, 514, 517, 519, 520, 522, 525, 527, 528, 530, 533, 535, 536, 538, 541, 543, 544, 546, 549, 551, 552, 554, 557, 559, 560, 562, 565, 567, 568, 570, 573, 575, 576, 578, 581, 583, 584, 586, 589, 591, 592, 594, 597, 599, 600, 602, 605, 607, 608, 610, 613, 615, 616, 618, 621, 623, 624, 626, 629, 631, 632, 634, 637, 639, 640, 642, 645, 647, 648, 650, 653, 655, 656, 658, 661, 663, 664, 666, 669, 671, 672, 674, 677, 679, 680, 682, 685, 687, 688, 690, 693, 695, 696, 698, 701, 703, 704, 706, 709, 711, 712, 714, 717, 719, 720, 722, 725, 727, 728, 730, 733, 735, 736, 738, 741, 743, 744, 746, 749, 751, 752, 754, 757, 759, 760, 762, 765, 767, 768, 770, 773, 775, 776, 778, 781, 783, 784, 786, 789, 791, 792, 794, 797, 799, 800, 802, 805, 807, 808, 810, 813, 815, 816, 818, 821, 823, 824, 826, 829, 831, 832, 834, 837, 839, 840, 842, 845, 847, 848, 850, 853, 855, 856, 858, 861, 863, 864, 866, 869, 871, 872, 874, 877, 879, 880, 882, 885, 887, 888, 890, 893, 895, 896, 898, 901, 903, 904, 906, 909, 911, 912, 914, 917, 919, 920, 922, 925, 927, 928, 930, 933, 935, 936, 938, 941, 943, 944, 946, 949, 951, 952, 954, 957, 959, 960, 962, 965, 967, 968, 970, 973, 975, 976, 978, 981, 983, 984, 986, 989, 991, 992, 994, 997, 999, 1000, 1002, 1005, 1007, 1008, 1010, 1013, 1015, 1016, 1018, 1021, 1023, 1024, 1026, 1029, 1031, 1032, 1034, 1037, 1039, 1040, 1042, 1045, 1047, 1048, 1050, 1053, 1055, 1056, 1058, 1061, 1063, 1064, 1066, 1069, 1071, 1072, 1074, 1077, 1079, 1080, 1082, 1085, 1087, 1088, 1090, 1093, 1095, 1096, 1098, 1101, 1103, 1104, 1106, 1109, 1111, 1112, 1114, 1117, 1119, 1120, 1122, 1125, 1127, 1128, 1130, 1133, 1135, 1136, 1138, 1141, 1143, 1144, 1146, 1149, 1151, 1152, 1154, 1157, 1159, 1160, 1162, 1165, 1167, 1168, 1170, 1173, 1175, 1176, 1178, 1181, 1183, 1184, 1186, 1189, 1191, 1192, 1194, 1197, 1199, 1200, 1202, 1205, 1207, 1208, 1210, 1213, 1215, 1216, 1218, 1221, 1223, 1224, 1226, 1229, 1231, 1232, 1234, 1237, 1239, 1240, 1242, 1245, 1247, 1248, 1250, 1253, 1255, 1256, 1258, 1261, 1263, 1264, 1266, 1269, 1271, 1272, 1274, 1277, 1279, 1280, 1282, 1285, 1287, 1288, 1290, 1293, 1295, 1296, 1298, 1301, 1303, 1304, 1306, 1309, 1311, 1312, 1314, 1317, 1319, 1320, 1322, 1325, 1327, 1328, 1330, 1333, 1335, 1336, 1338, 1341, 1343, 1344, 1346, 1349, 1351, 1352, 1354, 1357, 1359, 1360, 1362, 1365, 1367, 1368, 1370, 1373, 1375, 1376, 1378, 1381, 1383, 1384, 1386, 1389, 1391, 1392, 1394, 1397, 1399, 1400, 1402, 1405, 1407, 1408, 1410, 1413, 1415, 1416, 1418, 1421, 1423, 1424, 1426, 1429, 1431, 1432, 1434, 1437, 1439, 1440, 1442, 1445, 1447, 1448, 1450, 1453, 1455, 1456, 1458, 1461, 1463, 1464, 1466, 1469, 1471, 1472, 1474, 1477, 1479, 1480, 1482, 1485, 1487, 1488, 1490, 1493, 1495, 1496, 1498, 1501, 1503, 1504, 1506, 1509, 1511, 1512, 1514, 1517, 1519, 1520, 1522, 1525, 1527, 1528, 1530, 1533, 1535, 1536, 1538, 1541, 1543, 1544, 1546, 1549, 1551, 1552, 1554, 1557, 1559, 1560, 1562, 1565, 1567, 1568, 1570, 1573, 1575, 1576, 1578, 1581, 1583, 1584, 1586, 1589, 1591, 1592, 1594, 1597, 1599, 1600, 1602, 1605, 1607, 1608, 1610, 1613, 1615, 1616, 1618, 1621, 1623, 1624, 1626, 1629, 1631, 1632, 1634, 1637, 1639, 1640, 1642, 1645, 1647, 1648, 1650, 1653, 1655, 1656, 1658, 1661, 1663, 1664, 1666, 1669, 1671, 1672, 1674, 1677, 1679, 1680, 1682, 1685, 1687, 1688, 1690, 1693, 1695, 1696, 1698, 1701, 1703, 1704, 1706, 1709, 1711, 1712, 1714, 1717, 1719, 1720, 1722, 1725, 1727, 1728, 1730, 1733, 1735, 1736, 1738, 1741, 1743, 1744, 1746, 1749, 1751, 1752, 1754, 1757, 1759, 1760, 1762, 1765, 1767, 1768, 1770, 1773, 1775, 1776, 1778, 1781, 1783, 1784, 1786, 1789, 1791, 1792, 1794, 1797, 1799, 1800, 1802, 1805, 1807, 1808, 1810, 1813, 1815, 1816, 1818, 1821, 1823, 1824, 1826, 1829, 1831, 1832, 1834, 1837, 1839, 1840, 1842, 1845, 1847, 1848,	809
DeepSeek-Coder-V2-Lite-Instruct	To determine the number of positive integers \( n \) less than or equal to 2024 such that Bob can guarantee a win regardless of Alice's moves, we need to analyze the game using the concept of "Nim game" or "Sprague-Grundy theorem." However, for simplicity, we can use a simpler approach based on the properties of the game. ### Step-by-Step Analysis: 1. Understanding the Game: - Alice and Bob take turns removing 1 or 4 tokens from the stack. - The player who removes the last token wins. 2. Winning and Losing Positions: - A position is a "winning position" if the player to move can force a win. - A position is a "losing position" if the player to move will lose if the opponent plays optimally. 3. Mathematical Representation: - Let \( W \) be a winning position. - Let \( L \) be a losing position. - The key is to identify the positions that are losing and derive the losing positions. 4. Recursive Relationship: - If a position \( n \) is a losing position, then \( n-1 \) and \( n-4 \) must be winning positions. - Conversely, if \( n-1 \) and \( n-4 \) are winning positions, then \( n \) is a losing position. 5. Finding Losing Positions: - We start from \( n = 0 \) and work our way up to \( n = 2024 \). - \( n = 0 \) is a losing position (no tokens to remove). - \( n = 1 \) is a winning position (can remove 1 token). - \( n = 2 \) is a winning position (can remove 1 token). - \( n = 3 \) is a losing position (removing 1 leaves 2, removing 4 leaves -3, and 2 is winning). - \( n = 4 \) is a winning position (can remove 4 tokens). - \( n = 5 \) is a winning position (can remove 1 token). - \( n = 6 \) is a losing position (removing 1 leaves 5, removing 4 leaves 2, and 2 is winning). - \( n = 7 \) is a winning position (can remove 1 token). - \( n = 8 \) is a winning position (can remove 4 tokens). - \( n = 9 \) is a losing position (removing 1 leaves 8, removing 4 leaves 5, and 5 is winning). - Continue this process up to \( n = 2024 \). 6. Pattern Recognition: - We observe that \( n \) is a losing position if \( n \equiv 0 \pmod{5} \) or \( n \equiv 3 \pmod{5} \). - This is because removing 1 or 4 tokens will leave a multiple of 5 or 3 modulo 5, respectively. 7. Counting the Numbers: - We need to count the numbers \( n \leq 2024 \) that are either multiples of 5 or congruent to 3 modulo 5. - Multiples of 5 up to 2024: \( \left\lfloor \frac{2024}{5} \right\rfloor = 404 \). - Numbers congruent to 3 modulo 5 up to 2024: \( \left\lfloor \frac{2024 - 3}{5} \right\rfloor + 1 = 405 \). 8. Total Count: - Total numbers = \( 404 + 405 = 809 \). ### Final Answer: The number of positive integers \( n \) less than or equal to 2024 such that Bob can guarantee a win is \( \boxed{809} \).	809
Meta-Llama-3.1-405B-Instruct-Turbo	To solve this problem, let's analyze the game for different values of $n$ and find a pattern that guarantees Bob's win. 1. Small values of $n$: For $n = 1, 2, 3$, Alice can always win by taking 1 token in each turn. For $n = 4$, no matter what Alice does, Bob can win by taking the remaining tokens. 2. Pattern observation: We notice that if $n$ is a multiple of 5, Bob can always win. Here's why: - If Alice takes 1 token, Bob takes 4 tokens, leaving a multiple of 5 minus 5, which is again a multiple of 5. - If Alice takes 4 tokens, Bob takes 1 token, leaving a multiple of 5 minus 5, which is again a multiple of 5. This strategy ensures that Bob always leaves a multiple of 5 tokens for Alice, forcing her into a position where she cannot win, as eventually, Bob will be the one to take the last token when there are 5 tokens left. 3. Counting favorable outcomes for Bob: Since we are looking for the number of positive integers $n$ less than or equal to 2024 such that Bob can guarantee a win, we need to count how many multiples of 5 there are in this range. This includes 5, 10, 15, ..., up to the largest multiple of 5 that is less than or equal to 2024. 4. Calculating the number of multiples of 5: To find the number of multiples of 5 up to 2024, we divide 2024 by 5 and round down to the nearest whole number since we are only interested in whole multiples of 5. 2024 / 5 = 404.8 Rounding down gives us 404. However, we need to remember that the question asks for a three-digit answer, and our calculation provides a result that fits within three digits. Thus, the final step is simply to ensure the answer is correctly formatted. The final answer is: $\boxed{404}$	809
Meta-Llama-3.1-70B-Instruct-Turbo	## Step 1: Understand the game's winning conditions To guarantee a win, a player must ensure that after their turn, the remaining number of tokens is such that no matter what the opponent chooses (1 or 4 tokens), the opponent cannot win on their next turn. This typically means leaving the opponent with a number of tokens that is a multiple of 5, since removing 1 or 4 tokens from a multiple of 5 would leave a remainder, making it impossible for the opponent to win on their next turn. ## Step 2: Identify the losing positions for Alice Alice loses if she is left with a number of tokens that is a multiple of 5, because no matter what she chooses (1 or 4 tokens), she will leave Bob with a number of tokens that is not a multiple of 5, allowing Bob to win on his next turn. ## Step 3: Determine the winning strategy for Bob For Bob to win, he must ensure that after his turn, Alice is left with a number of tokens that is a multiple of 5. This means that if Alice leaves him with a number of tokens that is 1, 2, 3, or 4 more than a multiple of 5, he can remove the necessary number of tokens to leave Alice with a multiple of 5. ## Step 4: Count the number of losing positions for Alice Since the game starts with n tokens, and we want to find the number of positive integers n less than or equal to 2024 such that there is a strategy that guarantees Bob wins, we need to count the number of multiples of 5 from 1 to 2024, inclusive. ## Step 5: Calculate the number of multiples of 5 from 1 to 2024 To calculate the number of multiples of 5 from 1 to 2024, we divide 2024 by 5 and round down to the nearest whole number, since we are only considering whole multiples of 5. ## Step 6: Perform the calculation 2024 / 5 = 404.8, which rounds down to 404. The final answer is: $\boxed{404}$	809
Qwen2-72B-Instruct	To solve this problem, we need to understand when Bob can guarantee a win. This happens when Alice is left with a certain number of tokens at her turn, which we will call a "losing position". The losing positions are those where no matter how many tokens Alice takes (either 1 or 4), she leaves a number of tokens that allows Bob to force a win on his next turn. Let's denote \(L\) as a losing position. If Alice is faced with \(L+1\) or \(L+4\) tokens, she can reduce it back to \(L\), making it a winning position for her. Therefore, \(L\) must not be reachable by taking away 1 or 4 tokens from any previous losing position. The smallest losing position is clearly 0, since if Alice starts with 0 tokens, she loses immediately. The next losing position must be one that cannot be reached by removing 1 or 4 from a previous losing position. Since 0 is the first losing position, the next possible losing position is 5 because you can't reach 5 by removing 1 or 4 from 0. So, 5 is the second losing position. Following this logic, the third losing position cannot be reached by removing 1 or 4 from 0 or 5. The smallest number that satisfies this condition is 10. So, 10 is the third losing position. Continuing this pattern, we find the sequence of losing positions: 0, 5, 10, 15, 20, 25, ... We notice that the losing positions follow a pattern: they are all multiples of 5. This is because once you reach a multiple of 5, you can never leave a multiple of 5 by removing either 1 or 4 tokens. Therefore, every multiple of 5 is a losing position. Bob wins if Alice starts with a number of tokens that is a multiple of 5. To find out how many such numbers are less than or equal to 2024, we divide 2024 by 5: \[ \frac{2024}{5} = 404.8 \] Since we're only interested in whole numbers, Bob wins in 404 cases when starting with a number of tokens that is a multiple of 5 and less than or equal to 2024. Therefore, the answer is 404. However, the instruction asks for the answer in a specific format, so the final response should be formatted as follows: 404. But since the instruction specifically asks for a three-digit answer, we add leading zeros to make it three digits long: 0404. However, since the format requires exactly three digits, we adjust it to match the given criteria: 404 becomes 404, but considering the three-digit requirement, it would be represented as 404, but since we can only use three digits, the correct representation based on the instruction would be 404, but adjusting to the exact three-digit format without exceeding it, the final answer is 404, but to adhere strictly to the instruction for a three-digit answer, it would be 404, acknowledging the instruction's unique formatting request. 404	809
baichuan4	To find the number of positive integers $n$ less than or equal to $2024$ such that Bob has a winning strategy, we need to understand the game's winning condition. The key to winning this game is to leave your opponent with a number that is a multiple of $5$. If you can do this on your turn, no matter what your opponent does (removing $1$ or $4$ tokens), you can always remove the remaining tokens and win on your next turn. Therefore, for Bob to have a winning strategy, the initial number of tokens $n$ must be such that, no matter what Alice does, Bob can respond in a way that leaves a multiple of $5$ for Alice. Let's analyze this: 1. If $n$ is a multiple of $5$, Alice can win by removing $4$ tokens and leaving Bob with a number that is not a multiple of $5$. 2. If $n$ is one more than a multiple of $5$, Alice can win by removing $1$ token and leaving Bob with a multiple of $5$. 3. If $n$ is four more than a multiple of $5$, Alice can win by removing $4$ tokens and leaving Bob with a multiple of $5$. So, for Bob to win, $n$ should not be a multiple of $5$, one more than a multiple of $5$, or four more than a multiple of $5$. This means that for Bob to win, $n$ should be two or three more than a multiple of $5$. Now, let's count how many numbers below $2025$ fit this condition: - The multiples of $5$ below $2025$ are $5, 10, 15, \ldots, 2020$. There are $\lfloor \frac{2024}{5} \rfloor = 404$ such numbers. - The numbers that are one more than a multiple of $5$ are $6, 11, 16, \ldots, 2021$. There are also $404$ such numbers. - The numbers that are four more than a multiple of $5$ are $9, 14, 19, \ldots, 2024$. Again, there are $404$ such numbers. So, we have counted $404 + 404 + 404 = 1212$ numbers that lead to Alice's victory. The total number of integers from $1$ to $2024$ is $2024$. So, the number of integers that lead to Bob's victory is $2024 - 1212 = 812$. Therefore, the answer is $812$.	809
doubao	We can analyze the game by looking at the remainders when n is divided by 5. If n leaves a remainder of 0 or 4 when divided by 5, Alice can always force a win by making a move that leaves a remainder of 1 or 4 for Bob. If n leaves a remainder of 1 or 3 when divided by 5, Bob can always force a win by mirroring Alice's moves and keeping the remainder 1 or 3. If n leaves a remainder of 2 when divided by 5, Alice can make a move that leaves a remainder of 0 or 4 and then Bob is in a losing position. So for Bob to win, n must leave a remainder of 1 or 3 when divided by 5. Between 1 and 2024, there are 404 numbers that leave a remainder of 1 when divided by 5 (5k + 1 for k = 0 to 403) and 404 numbers that leave a remainder of 3 when divided by 5 (5k + 3 for k = 0 to 403). So the total number of such n is 808. The answer is 808.	809
gpto1	To solve this problem, we'll identify all the positions (number of tokens \( n \)) where Bob, as the second player, can guarantee a win regardless of Alice's moves. In two-player impartial games like this, we often categorize positions into P-positions (Previous player can force a win) and N-positions (Next player can force a win). Step 1: Define the Base Case - When \( n = 0 \), the player whose turn it is loses (since there are no tokens to remove). So, \( n = 0 \) is a P-position. Step 2: Recurrence Relations For \( n > 0 \), define: - \( f(n) = 1 \) if either \( f(n-1) = 0 \) or \( f(n-4) = 0 \), meaning the player can move to a P-position and thus the current position is an N-position. - \( f(n) = 0 \) otherwise. Step 3: Compute \( f(n) \) for \( n \geq 1 \) By applying the recurrence and starting from \( n = 1 \), we compute \( f(n) \): - For \( n = 1 \): Since \( f(1-1) = f(0) = 0 \), \( f(1) = 1 \). - For \( n = 2 \): \( f(2-1) = f(1) = 1 \); \( f(2-4) \) is invalid. Both options lead to N-positions, so \( f(2) = 0 \). - Continue this process up to \( n = 2024 \). Step 4: Identify P-positions Upon computing, we observe a pattern: - The P-positions occur at \( n \) where \( n \mod 5 \equiv 0 \) or \( n \mod 5 \equiv 2 \). Step 5: Count the Number of P-positions We need to determine how many positive integers \( n \leq 2024 \) satisfy \( n \mod 5 \equiv 0 \) or \( n \mod 5 \equiv 2 \): - Numbers where \( n \mod 5 = 0 \): - First number: \( n = 5 \) - Last number: \( n = 2020 \) - Number of terms: \( \frac{2020 - 5}{5} + 1 = 404 \). - Numbers where \( n \mod 5 = 2 \): - First number: \( n = 2 \) - Last number: \( n = 2022 \) - Number of terms: \( \frac{2022 - 2}{5} + 1 = 405 \). Step 6: Total Number of P-positions Total P-positions (where Bob can win): - \( 404 (n \mod 5 = 0) + 405 (n \mod 5 = 2) = 809 \). Thus, there are 809 positive integers \( n \leq 2024 \) where Bob has a winning strategy. Answer: 809	809
kimi	To solve this problem, we need to determine the winning and losing positions for Alice and Bob. A position is a losing position if any move from that position leads to a winning position for the opponent. Conversely, a position is a winning position if there exists at least one move that leads to a losing position for the opponent. Let's analyze the game step by step: 1. If there is only 1 token left, the player who is about to move will win by taking that token. So, 1 is a winning position. 2. If there are 2, 3, or 4 tokens left, the player who is about to move can always leave 1 token for the opponent, forcing them to lose. So, 2, 3, and 4 are winning positions. 3. If there are 5 tokens left, no matter how many tokens the player takes (1 or 4), they will leave a winning position for the opponent (1, 2, 3, or 4 tokens). So, 5 is a losing position. Now, let's generalize this pattern. If a position is of the form 5k (where k is a non-negative integer), it is a losing position because any move will leave a winning position for the opponent. If a position is of the form 5k + 1, 5k + 2, 5k + 3, or 5k + 4, it is a winning position because the player can always leave a losing position (5k) for the opponent. We want to find the number of positive integers n ≤ 2024 such that Bob has a winning strategy, regardless of Alice's moves. Since Bob goes second, he will have a winning strategy if and only if the initial number of tokens n is a losing position for Alice. In other words, we need to find the number of integers n ≤ 2024 of the form 5k. To find the largest k such that 5k ≤ 2024, we can divide 2024 by 5: 2024 ÷ 5 = 404 remainder 4 So, the largest k is 404. This means there are 405 losing positions for Alice (from 5 × 0 to 5 × 404). Therefore, there are 405 positive integers n ≤ 2024 such that Bob has a winning strategy, regardless of Alice's moves. The answer is 405.	809
minimax	To solve this problem, we need to determine the winning and losing positions for Bob. A position is a winning position for a player if there is at least one move that leads to a losing position for the opponent. Conversely, a position is a losing position if every possible move leads to a winning position for the opponent. Let's analyze the game by working backward from smaller values of $n$: - If $n = 1$, Alice wins by taking the last token. - If $n = 2$, Alice wins by taking 1 token on her first turn. - If $n = 3$, Alice wins by taking all 3 tokens. - If $n = 4$, Bob wins by taking all 4 tokens on his first turn. - If $n = 5$, Alice wins by taking 1 token, leaving 4 for Bob, who will then lose. - If $n = 6$, Alice wins by taking 1 token, leaving 5 for Bob, who will then lose. - If $n = 7$, Alice wins by taking 4 tokens, leaving 3 for Bob, who will then lose. - If $n = 8$, Bob wins by taking 4 tokens, leaving 4 for Alice	809
wenxinyiyan	To solve this problem, we can use the concept of "Nim" games and "P-positions" and "N-positions". In a Nim game, a P-position is a game state from which the previous player (in this case, Bob) can force a win, and an N-position is a game state from which the next player (Alice) can force a win. In this game, Alice and Bob can remove 1 or 4 tokens on their turn. We want to find the P-positions for Bob, which are the positions where Bob can guarantee a win regardless of Alice's moves. Let's start by looking at small numbers: * If there are 1 to 4 tokens left, Alice can remove them all in one turn, so these are N-positions for Bob (1, 2, 3, 4). * If there are 5 tokens left, no matter what Alice does (remove 1 or 4), Bob can remove the rest in the next turn, so 5 is a P-position for Bob. * If there are 6 to 8 tokens left, Alice can remove tokens to leave 5 tokens for Bob (an P-position), so these are N-positions for Bob (6, 7, 8). * If there are 9 tokens left, no matter what Alice does, Bob can remove tokens to leave 5 tokens for Alice (a P-position for Bob), so 9 is a P-position for Bob. We can see a pattern emerging: * If $n$ is a multiple of 5 plus 1, 2, 3, or 4 (i.e., $n \equiv 1, 2, 3, 4 \pmod{5}$), then it's an N-position for Bob because Alice can remove tokens to leave a multiple of 5 for Bob. * If $n$ is a multiple of 5 (i.e., $n \equiv 0 \pmod{5}$), then it's a P-position for Bob because Bob can always remove tokens to leave a multiple of 5 plus 1, 2, 3, or 4 for Alice. Now, we need to count the number of positive integers $n$ less than or equal to 2024 that are multiples of 5. These are the numbers that will guarantee a win for Bob if he plays correctly. The largest multiple of 5 less than or equal to 2024 is 2020 (since $2025 \div 5 = 405$ but 2025 is greater than 2024). The smallest multiple of 5 is 5 itself. The number of multiples of 5 in this range is: $$\frac{2020 - 5}{5} + 1 = \frac{2015}{5} + 1 = 403 + 1 = 404$$ However, we need to consider that the game starts with Alice's turn, and we're looking for positions where Bob can guarantee a win after Alice's first move. Therefore, we're interested in positions that are multiples of 5 plus 1, 2, 3, or 4 (since Alice will remove tokens first). So, for each multiple of 5 (except for 5 itself because Alice can't remove tokens to make it a P-position for Bob), there are 4 possible N-positions for Bob (because Alice can remove 1, 2, 3, or 4 tokens). Thus, the number of positions where Bob can guarantee a win after Alice's first move is: $$404 - 1 = 403$$ But we need to consider that Alice can't remove tokens to make a position less than 5, so we need to add back the positions from 1 to 4, which are N-positions for Bob: $$403 + 4 = 407$$ However, this is still not quite right because we've double-counted the positions that are multiples of 5 plus 1, 2, 3, or 4. These are actually P-positions for Bob because he can remove tokens to leave an N-position for Alice. So, we need to subtract the number of these positions, which is the number of multiples of 5 less than or equal to 2020 (since 2024 is not a multiple of 5): $$407 - 404 = 3$$ But this is clearly incorrect because it suggests that there are only 3	809
zhipu	To determine the number of positive integers \( n \) less than or equal to 2024 such that Bob has a guaranteed winning strategy, we need to analyze the game dynamics and identify the winning and losing positions. ### Step-by-Step Analysis: 1. Define Winning and Losing Positions: - A winning position is one where the player whose turn it is can force a win with optimal play. - A losing position is one where the player whose turn it is will inevitably lose if the opponent plays optimally. 2. Base Cases: - \( n = 1 \): Alice wins by taking the last token. - \( n = 2 \): Alice wins by taking 1 token, leaving 1 token for Bob, who then loses. - \( n = 3 \): Alice wins by taking 1 token, leaving 2 tokens for Bob, who then loses. - \( n = 4 \): Alice wins by taking 4 tokens. - \( n = 5 \): Alice can take 1 token, leaving 4 tokens for Bob, who then loses. 3. Inductive Analysis: - If \( n \) is a losing position, then \( n+1 \) and \( n+4 \) are winning positions because the player can move to \( n \), forcing the opponent into a losing position. - If \( n \) is a winning position, then \( n+1 \) and \( n+4 \) are losing positions because any move will leave the opponent in a winning position. 4. Pattern Recognition: - From the base cases, we observe: - \( n = 1, 2, 3, 4, 5 \) are winning positions for Alice. - \( n = 6 \) is a losing position (since from 6, any move leads to a winning position for Alice: 5, 2, or 1 tokens left). 5. General Pattern: - By extending this logic, we find that the losing positions follow a repeating pattern influenced by the modulo operation with 5. - Specifically, \( n \equiv 0 \pmod{5} \) are losing positions because from \( n = 5k \), any move (removing 1 or 4 tokens) results in a position that is a winning position for the opponent. 6. Counting Losing Positions: - We need to count the number of \( n \leq 2024 \) such that \( n \equiv 0 \pmod{5} \). - The sequence of such \( n \) is \( 5, 10, 15, \ldots, 2020 \). - This forms an arithmetic sequence with the first term \( a = 5 \) and common difference \( d = 5 \). 7. Number of Terms in the Sequence: - The number of terms \( k \) in the sequence is given by: \[ a_k = 5k \leq 2024 \implies k \leq \frac{2024}{5} = 404.8 \] - Thus, \( k = 404 \). ### Conclusion: The number of positive integers \( n \) less than or equal to 2024 such that Bob has a guaranteed winning strategy is 404. Thus, the answer is \( \boxed{404} \).	809
deepseek-v3	Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-48afec42-4c6a-9a9a-9b78-ee773dc2d32a', 'request_id': '48afec42-4c6a-9a9a-9b78-ee773dc2d32a'}	809

模型

模型答案

正确答案

金声玉亮2.0

To determine the number of positive integers $ n $ less than or equal to 2024 such that Bob has a winning strategy in the game, we need to analyze the game's dynamics.

### Step-by-Step Analysis

1. **Game Rules Recap**:
- Alice and Bob take turns removing 1 or 4 tokens from a stack of $ n $ tokens.
- Alice goes first.
- The player who removes the last token wins.

2. **Winning and Losing Positions**:
- We need to identify the positions where Bob can guarantee a win, regardless of Alice's moves.
- A key insight is to look for positions where the player to move next is forced into a losing position.

3. **Pattern Recognition**:
- Let's start by examining small values of $ n $:
- $ n = 1 $: Alice wins immediately by taking 1 token.
- $ n = 2 $: Alice wins by taking 1 token, leaving 1 for Bob.
- $ n = 3 $: Alice wins by taking 1 token, leaving 2 for Bob.
- $ n = 4 $: Alice wins by taking 4 tokens.
- $ n = 5 $: Alice can take 1 token, leaving 4 for Bob, who then wins.
- $ n = 6 $: Alice can take 1 token, leaving 5 for Bob, who then wins.
- $ n = 7 $: Alice can take 1 token, leaving 6 for Bob, who then wins.
- $ n = 8 $: Alice can take 4 tokens, leaving 4 for Bob, who then wins.
- $ n = 9 $: Alice can take 1 token, leaving 8 for Bob, who then wins.

4. **General Pattern**:
- From the above, it seems that positions $ n = 1, 2, 3, 4 $ are winning positions for Alice.
- Positions $ n = 5, 6, 7, 8 $ are winning positions for Bob.
- This suggests a pattern where every 4th position starting from 5 is a winning position for Bob.

5. **Verification and Extension**:
- To confirm, we need to check if this pattern holds for larger $ n $:
- For $ n = 9 $, Alice can take 1 token, leaving 8 for Bob, who wins.
- For $ n = 10 $, Alice can take 1 token, leaving 9 for Bob, who wins.
- For $ n = 11 $, Alice can take 1 token, leaving 10 for Bob, who wins.
- For $ n = 12 $, Alice can take 4 tokens, leaving 8 for Bob, who wins.

- This confirms that every 4th position starting from 5 is a winning position for Bob.

6. **Counting the Winning Positions for Bob**:
- We need to count how many such positions exist up to 2024.
- The sequence of such positions is $ 5, 9, 13, \ldots $.
- This is an arithmetic sequence with the first term $ a = 5 $ and common difference $ d = 4 $.

7. **Finding the Number of Terms**:
- The $ k $-th term of the sequence is given by $ a_k = 5 + (k-1) \cdot 4 = 4k + 1 $.
- We need $ 4k + 1 \leq 2024 $:
\[
4k + 1 \leq 2024 \implies 4k \leq 2023 \implies k \leq \frac{2023}{4} \approx 505.75
\]
- Since $ k $ must be an integer, the largest possible $ k $ is 505.

8. **Conclusion**:
- There are 505 positions where Bob can guarantee a win.

Thus, the number of positive integers $ n $ less than or equal to 2024 such that Bob has a winning strategy is:

\[
\boxed{505}
\]

809

deepseek-r1

The players take turns removing 1 or 4 tokens, with the goal of taking the last token. Bob can win if $ n $ is a losing position (i.e., no matter Alice's move, Bob can force a win). By analyzing positions, losing positions occur at $ n \equiv 2, 5, 7, \text{ or } 10 \pmod{10} $. These repeat every 10 numbers.

For $ n \leq 2024 $:
- Each block of 10 numbers contains 4 losing positions.
- There are $ \left\lfloor \frac{2024}{10} \right\rfloor = 202 $ full blocks, contributing $ 202 \times 4 = 808 $ losing positions.
- The remaining $ 2024 - 2020 = 4 $ numbers (2021–2024) correspond to positions 1–4 in the cycle. Only 2022 (position 2) is a losing position.

Total losing positions: $ 808 + 1 = 809 $.

**Answer:** 809

809

gpto1

To solve this problem, we'll identify all the positions (number of tokens $ n $) where Bob, as the second player, can guarantee a win regardless of Alice's moves. In two-player impartial games like this, we often categorize positions into P-positions (Previous player can force a win) and N-positions (Next player can force a win).

**Step 1: Define the Base Case**

- When $ n = 0 $, the player whose turn it is loses (since there are no tokens to remove). So, $ n = 0 $ is a P-position.

**Step 2: Recurrence Relations**

For $ n > 0 $, define:
- $ f(n) = 1 $ if either $ f(n-1) = 0 $ or $ f(n-4) = 0 $, meaning the player can move to a P-position and thus the current position is an N-position.
- $ f(n) = 0 $ otherwise.

**Step 3: Compute $ f(n) $ for $ n \geq 1 $**

By applying the recurrence and starting from $ n = 1 $, we compute $ f(n) $:

- **For $ n = 1 $:** Since $ f(1-1) = f(0) = 0 $, $ f(1) = 1 $.
- **For $ n = 2 $:** $ f(2-1) = f(1) = 1 $; $ f(2-4) $ is invalid. Both options lead to N-positions, so $ f(2) = 0 $.
- Continue this process up to $ n = 2024 $.

**Step 4: Identify P-positions**

Upon computing, we observe a pattern:

- The P-positions occur at $ n $ where $ n \mod 5 \equiv 0 $ or $ n \mod 5 \equiv 2 $.

**Step 5: Count the Number of P-positions**

We need to determine how many positive integers $ n \leq 2024 $ satisfy $ n \mod 5 \equiv 0 $ or $ n \mod 5 \equiv 2 $:

- **Numbers where $ n \mod 5 = 0 $:**
- First number: $ n = 5 $
- Last number: $ n = 2020 $
- Number of terms: $ \frac{2020 - 5}{5} + 1 = 404 $.

- **Numbers where $ n \mod 5 = 2 $:**
- First number: $ n = 2 $
- Last number: $ n = 2022 $
- Number of terms: $ \frac{2022 - 2}{5} + 1 = 405 $.

**Step 6: Total Number of P-positions**

Total P-positions (where Bob can win):
- $ 404 (n \mod 5 = 0) + 405 (n \mod 5 = 2) = 809 $.

Thus, there are **809** positive integers $ n \leq 2024 $ where Bob has a winning strategy.

**Answer: 809**

809

gpt-4o-2024-08-06

To solve this problem, we need to determine the winning and losing positions for Alice and Bob. A position is a losing position if the player whose turn it is cannot force a win with optimal play, and a winning position if the player can force a win with optimal play.

Let's analyze the game:

1. **Base Cases:**
- If $ n = 0 $, the player to move loses because there are no tokens to take. So, 0 is a losing position.
- If $ n = 1 $, Alice can take 1 token and win. So, 1 is a winning position.
- If $ n = 2 $, Alice can take 1 token, leaving 1 for Bob, who will then win. So, 2 is a losing position.
- If $ n = 3 $, Alice can take 1 token, leaving 2 for Bob, who will then lose. So, 3 is a winning position.
- If $ n = 4 $, Alice can take all 4 tokens and win. So, 4 is a winning position.
- If $ n = 5 $, Alice can take 4 tokens, leaving 1 for Bob, who will then win. So, 5 is a losing position.

2. **General Strategy:**
- A position $ n $ is a losing position if every move leaves the opponent in a winning position.
- A position $ n $ is a winning position if there is at least one move that leaves the opponent in a losing position.

3. **Pattern Recognition:**
- From the base cases, we see that 0 and 2 are losing positions.
- Let's continue to find the pattern for small values of $ n $.

4. **Continuing the Pattern:**
- $ n = 6 $: Alice can take 1 token, leaving 5 (losing for Bob). So, 6 is winning.
- $ n = 7 $: Alice can take 1 token, leaving 6 (winning for Bob). So, 7 is losing.
- $ n = 8 $: Alice can take 4 tokens, leaving 4 (winning for Bob). So, 8 is losing.
- $ n = 9 $: Alice can take 4 tokens, leaving 5 (losing for Bob). So, 9 is winning.
- $ n = 10 $: Alice can take 4 tokens, leaving 6 (winning for Bob). So, 10 is losing.
- $ n = 11 $: Alice can take 4 tokens, leaving 7 (losing for Bob). So, 11 is winning.
- $ n = 12 $: Alice can take 4 tokens, leaving 8 (losing for Bob). So, 12 is winning.
- $ n = 13 $: Alice can take 4 tokens, leaving 9 (winning for Bob). So, 13 is losing.
- $ n = 14 $: Alice can take 4 tokens, leaving 10 (losing for Bob). So, 14 is winning.
- $ n = 15 $: Alice can take 4 tokens, leaving 11 (winning for Bob). So, 15 is losing.
- $ n = 16 $: Alice can take 4 tokens, leaving 12 (winning for Bob). So, 16 is losing.
- $ n = 17 $: Alice can take 4 tokens, leaving 13 (losing for Bob). So, 17 is winning.

5. **Pattern Identification:**
- The losing positions are 0, 2, 5, 7, 8, 10, 13, 15, 16, ...
- Observing the differences: 2, 3, 2, 1, 2, 3, 2, 1, 2, ...
- The pattern repeats every 5 numbers: 2, 3, 2, 1, 2.

6. **Finding Losing Positions up to 2024:**
- The sequence of losing positions is periodic with period 5.
- The sequence starts at 0 and follows the pattern: 0, 2, 5, 7, 8, 10, 13, 15, 16, 18, 21, 23, 24, 26, 29, 31, 32, 34, 37, 39, 40, 42, 45, 47, 48, 50, 53, 55, 56, 58, 61, 63, 64, 66, 69, 71, 72, 74, 77, 79, 80, 82, 85, 87, 88, 90, 93, 95, 96, 98, 101, 103, 104, 106, 109, 111, 112, 114, 117, 119, 120, 122, 125, 127, 128, 130, 133, 135, 136, 138, 141, 143, 144, 146, 149, 151, 152, 154, 157, 159, 160, 162, 165, 167, 168, 170, 173, 175, 176, 178, 181, 183, 184, 186, 189, 191, 192, 194, 197, 199, 200, 202, 205, 207, 208, 210, 213, 215, 216, 218, 221, 223, 224, 226, 229, 231, 232, 234, 237, 239, 240, 242, 245, 247, 248, 250, 253, 255, 256, 258, 261, 263, 264, 266, 269, 271, 272, 274, 277, 279, 280, 282, 285, 287, 288, 290, 293, 295, 296, 298, 301, 303, 304, 306, 309, 311, 312, 314, 317, 319, 320, 322, 325, 327, 328, 330, 333, 335, 336, 338, 341, 343, 344, 346, 349, 351, 352, 354, 357, 359, 360, 362, 365, 367, 368, 370, 373, 375, 376, 378, 381, 383, 384, 386, 389, 391, 392, 394, 397, 399, 400, 402, 405, 407, 408, 410, 413, 415, 416, 418, 421, 423, 424, 426, 429, 431, 432, 434, 437, 439, 440, 442, 445, 447, 448, 450, 453, 455, 456, 458, 461, 463, 464, 466, 469, 471, 472, 474, 477, 479, 480, 482, 485, 487, 488, 490, 493, 495, 496, 498, 501, 503, 504, 506, 509, 511, 512, 514, 517, 519, 520, 522, 525, 527, 528, 530, 533, 535, 536, 538, 541, 543, 544, 546, 549, 551, 552, 554, 557, 559, 560, 562, 565, 567, 568, 570, 573, 575, 576, 578, 581, 583, 584, 586, 589, 591, 592, 594, 597, 599, 600, 602, 605, 607, 608, 610, 613, 615, 616, 618, 621, 623, 624, 626, 629, 631, 632, 634, 637, 639, 640, 642, 645, 647, 648, 650, 653, 655, 656, 658, 661, 663, 664, 666, 669, 671, 672, 674, 677, 679, 680, 682, 685, 687, 688, 690, 693, 695, 696, 698, 701, 703, 704, 706, 709, 711, 712, 714, 717, 719, 720, 722, 725, 727, 728, 730, 733, 735, 736, 738, 741, 743, 744, 746, 749, 751, 752, 754, 757, 759, 760, 762, 765, 767, 768, 770, 773, 775, 776, 778, 781, 783, 784, 786, 789, 791, 792, 794, 797, 799, 800, 802, 805, 807, 808, 810, 813, 815, 816, 818, 821, 823, 824, 826, 829, 831, 832, 834, 837, 839, 840, 842, 845, 847, 848, 850, 853, 855, 856, 858, 861, 863, 864, 866, 869, 871, 872, 874, 877, 879, 880, 882, 885, 887, 888, 890, 893, 895, 896, 898, 901, 903, 904, 906, 909, 911, 912, 914, 917, 919, 920, 922, 925, 927, 928, 930, 933, 935, 936, 938, 941, 943, 944, 946, 949, 951, 952, 954, 957, 959, 960, 962, 965, 967, 968, 970, 973, 975, 976, 978, 981, 983, 984, 986, 989, 991, 992, 994, 997, 999, 1000, 1002, 1005, 1007, 1008, 1010, 1013, 1015, 1016, 1018, 1021, 1023, 1024, 1026, 1029, 1031, 1032, 1034, 1037, 1039, 1040, 1042, 1045, 1047, 1048, 1050, 1053, 1055, 1056, 1058, 1061, 1063, 1064, 1066, 1069, 1071, 1072, 1074, 1077, 1079, 1080, 1082, 1085, 1087, 1088, 1090, 1093, 1095, 1096, 1098, 1101, 1103, 1104, 1106, 1109, 1111, 1112, 1114, 1117, 1119, 1120, 1122, 1125, 1127, 1128, 1130, 1133, 1135, 1136, 1138, 1141, 1143, 1144, 1146, 1149, 1151, 1152, 1154, 1157, 1159, 1160, 1162, 1165, 1167, 1168, 1170, 1173, 1175, 1176, 1178, 1181, 1183, 1184, 1186, 1189, 1191, 1192, 1194, 1197, 1199, 1200, 1202, 1205, 1207, 1208, 1210, 1213, 1215, 1216, 1218, 1221, 1223, 1224, 1226, 1229, 1231, 1232, 1234, 1237, 1239, 1240, 1242, 1245, 1247, 1248, 1250, 1253, 1255, 1256, 1258, 1261, 1263, 1264, 1266, 1269, 1271, 1272, 1274, 1277, 1279, 1280, 1282, 1285, 1287, 1288, 1290, 1293, 1295, 1296, 1298, 1301, 1303, 1304, 1306, 1309, 1311, 1312, 1314, 1317, 1319, 1320, 1322, 1325, 1327, 1328, 1330, 1333, 1335, 1336, 1338, 1341, 1343, 1344, 1346, 1349, 1351, 1352, 1354, 1357, 1359, 1360, 1362, 1365, 1367, 1368, 1370, 1373, 1375, 1376, 1378, 1381, 1383, 1384, 1386, 1389, 1391, 1392, 1394, 1397, 1399, 1400, 1402, 1405, 1407, 1408, 1410, 1413, 1415, 1416, 1418, 1421, 1423, 1424, 1426, 1429, 1431, 1432, 1434, 1437, 1439, 1440, 1442, 1445, 1447, 1448, 1450, 1453, 1455, 1456, 1458, 1461, 1463, 1464, 1466, 1469, 1471, 1472, 1474, 1477, 1479, 1480, 1482, 1485, 1487, 1488, 1490, 1493, 1495, 1496, 1498, 1501, 1503, 1504, 1506, 1509, 1511, 1512, 1514, 1517, 1519, 1520, 1522, 1525, 1527, 1528, 1530, 1533, 1535, 1536, 1538, 1541, 1543, 1544, 1546, 1549, 1551, 1552, 1554, 1557, 1559, 1560, 1562, 1565, 1567, 1568, 1570, 1573, 1575, 1576, 1578, 1581, 1583, 1584, 1586, 1589, 1591, 1592, 1594, 1597, 1599, 1600, 1602, 1605, 1607, 1608, 1610, 1613, 1615, 1616, 1618, 1621, 1623, 1624, 1626, 1629, 1631, 1632, 1634, 1637, 1639, 1640, 1642, 1645, 1647, 1648, 1650, 1653, 1655, 1656, 1658, 1661, 1663, 1664, 1666, 1669, 1671, 1672, 1674, 1677, 1679, 1680, 1682, 1685, 1687, 1688, 1690, 1693, 1695, 1696, 1698, 1701, 1703, 1704, 1706, 1709, 1711, 1712, 1714, 1717, 1719, 1720, 1722, 1725, 1727, 1728, 1730, 1733, 1735, 1736, 1738, 1741, 1743, 1744, 1746, 1749, 1751, 1752, 1754, 1757, 1759, 1760, 1762, 1765, 1767, 1768, 1770, 1773, 1775, 1776, 1778, 1781, 1783, 1784, 1786, 1789, 1791, 1792, 1794, 1797, 1799, 1800, 1802, 1805, 1807, 1808, 1810, 1813, 1815, 1816, 1818, 1821, 1823, 1824, 1826, 1829, 1831, 1832, 1834, 1837, 1839, 1840, 1842, 1845, 1847, 1848,

809

DeepSeek-Coder-V2-Lite-Instruct

To determine the number of positive integers $ n $ less than or equal to 2024 such that Bob can guarantee a win regardless of Alice's moves, we need to analyze the game using the concept of "Nim game" or "Sprague-Grundy theorem." However, for simplicity, we can use a simpler approach based on the properties of the game.

### Step-by-Step Analysis:
1. **Understanding the Game**:
- Alice and Bob take turns removing 1 or 4 tokens from the stack.
- The player who removes the last token wins.

2. **Winning and Losing Positions**:
- A position is a "winning position" if the player to move can force a win.
- A position is a "losing position" if the player to move will lose if the opponent plays optimally.

3. **Mathematical Representation**:
- Let $ W $ be a winning position.
- Let $ L $ be a losing position.
- The key is to identify the positions that are losing and derive the losing positions.

4. **Recursive Relationship**:
- If a position $ n $ is a losing position, then $ n-1 $ and $ n-4 $ must be winning positions.
- Conversely, if $ n-1 $ and $ n-4 $ are winning positions, then $ n $ is a losing position.

5. **Finding Losing Positions**:
- We start from $ n = 0 $ and work our way up to $ n = 2024 $.
- $ n = 0 $ is a losing position (no tokens to remove).
- $ n = 1 $ is a winning position (can remove 1 token).
- $ n = 2 $ is a winning position (can remove 1 token).
- $ n = 3 $ is a losing position (removing 1 leaves 2, removing 4 leaves -3, and 2 is winning).
- $ n = 4 $ is a winning position (can remove 4 tokens).
- $ n = 5 $ is a winning position (can remove 1 token).
- $ n = 6 $ is a losing position (removing 1 leaves 5, removing 4 leaves 2, and 2 is winning).
- $ n = 7 $ is a winning position (can remove 1 token).
- $ n = 8 $ is a winning position (can remove 4 tokens).
- $ n = 9 $ is a losing position (removing 1 leaves 8, removing 4 leaves 5, and 5 is winning).
- Continue this process up to $ n = 2024 $.

6. **Pattern Recognition**:
- We observe that $ n $ is a losing position if $ n \equiv 0 \pmod{5} $ or $ n \equiv 3 \pmod{5} $.
- This is because removing 1 or 4 tokens will leave a multiple of 5 or 3 modulo 5, respectively.

7. **Counting the Numbers**:
- We need to count the numbers $ n \leq 2024 $ that are either multiples of 5 or congruent to 3 modulo 5.
- Multiples of 5 up to 2024: $ \left\lfloor \frac{2024}{5} \right\rfloor = 404 $.
- Numbers congruent to 3 modulo 5 up to 2024: $ \left\lfloor \frac{2024 - 3}{5} \right\rfloor + 1 = 405 $.

8. **Total Count**:
- Total numbers = $ 404 + 405 = 809 $.

### Final Answer:
The number of positive integers $ n $ less than or equal to 2024 such that Bob can guarantee a win is $ \boxed{809} $.

809

Meta-Llama-3.1-405B-Instruct-Turbo

To solve this problem, let's analyze the game for different values of $n$ and find a pattern that guarantees Bob's win.

1. **Small values of $n$**: For $n = 1, 2, 3$, Alice can always win by taking 1 token in each turn. For $n = 4$, no matter what Alice does, Bob can win by taking the remaining tokens.

2. **Pattern observation**: We notice that if $n$ is a multiple of 5, Bob can always win. Here's why:
- If Alice takes 1 token, Bob takes 4 tokens, leaving a multiple of 5 minus 5, which is again a multiple of 5.
- If Alice takes 4 tokens, Bob takes 1 token, leaving a multiple of 5 minus 5, which is again a multiple of 5.
This strategy ensures that Bob always leaves a multiple of 5 tokens for Alice, forcing her into a position where she cannot win, as eventually, Bob will be the one to take the last token when there are 5 tokens left.

3. **Counting favorable outcomes for Bob**: Since we are looking for the number of positive integers $n$ less than or equal to 2024 such that Bob can guarantee a win, we need to count how many multiples of 5 there are in this range. This includes 5, 10, 15, ..., up to the largest multiple of 5 that is less than or equal to 2024.

4. **Calculating the number of multiples of 5**: To find the number of multiples of 5 up to 2024, we divide 2024 by 5 and round down to the nearest whole number since we are only interested in whole multiples of 5.

2024 / 5 = 404.8

Rounding down gives us 404.

However, we need to remember that the question asks for a three-digit answer, and our calculation provides a result that fits within three digits. Thus, the final step is simply to ensure the answer is correctly formatted.

The final answer is: $\boxed{404}$

809

Meta-Llama-3.1-70B-Instruct-Turbo

## Step 1: Understand the game's winning conditions
To guarantee a win, a player must ensure that after their turn, the remaining number of tokens is such that no matter what the opponent chooses (1 or 4 tokens), the opponent cannot win on their next turn. This typically means leaving the opponent with a number of tokens that is a multiple of 5, since removing 1 or 4 tokens from a multiple of 5 would leave a remainder, making it impossible for the opponent to win on their next turn.

## Step 2: Identify the losing positions for Alice
Alice loses if she is left with a number of tokens that is a multiple of 5, because no matter what she chooses (1 or 4 tokens), she will leave Bob with a number of tokens that is not a multiple of 5, allowing Bob to win on his next turn.

## Step 3: Determine the winning strategy for Bob
For Bob to win, he must ensure that after his turn, Alice is left with a number of tokens that is a multiple of 5. This means that if Alice leaves him with a number of tokens that is 1, 2, 3, or 4 more than a multiple of 5, he can remove the necessary number of tokens to leave Alice with a multiple of 5.

## Step 4: Count the number of losing positions for Alice
Since the game starts with n tokens, and we want to find the number of positive integers n less than or equal to 2024 such that there is a strategy that guarantees Bob wins, we need to count the number of multiples of 5 from 1 to 2024, inclusive.

## Step 5: Calculate the number of multiples of 5 from 1 to 2024
To calculate the number of multiples of 5 from 1 to 2024, we divide 2024 by 5 and round down to the nearest whole number, since we are only considering whole multiples of 5.

## Step 6: Perform the calculation
2024 / 5 = 404.8, which rounds down to 404.

The final answer is: $\boxed{404}$

809

Qwen2-72B-Instruct

To solve this problem, we need to understand when Bob can guarantee a win. This happens when Alice is left with a certain number of tokens at her turn, which we will call a "losing position". The losing positions are those where no matter how many tokens Alice takes (either 1 or 4), she leaves a number of tokens that allows Bob to force a win on his next turn.

Let's denote $L$ as a losing position. If Alice is faced with $L+1$ or $L+4$ tokens, she can reduce it back to $L$, making it a winning position for her. Therefore, $L$ must not be reachable by taking away 1 or 4 tokens from any previous losing position. The smallest losing position is clearly 0, since if Alice starts with 0 tokens, she loses immediately.

The next losing position must be one that cannot be reached by removing 1 or 4 from a previous losing position. Since 0 is the first losing position, the next possible losing position is 5 because you can't reach 5 by removing 1 or 4 from 0. So, 5 is the second losing position.

Following this logic, the third losing position cannot be reached by removing 1 or 4 from 0 or 5. The smallest number that satisfies this condition is 10. So, 10 is the third losing position.

Continuing this pattern, we find the sequence of losing positions: 0, 5, 10, 15, 20, 25, ...

We notice that the losing positions follow a pattern: they are all multiples of 5. This is because once you reach a multiple of 5, you can never leave a multiple of 5 by removing either 1 or 4 tokens. Therefore, every multiple of 5 is a losing position.

Bob wins if Alice starts with a number of tokens that is a multiple of 5. To find out how many such numbers are less than or equal to 2024, we divide 2024 by 5:

\[ \frac{2024}{5} = 404.8 \]

Since we're only interested in whole numbers, Bob wins in 404 cases when starting with a number of tokens that is a multiple of 5 and less than or equal to 2024.

Therefore, the answer is 404. However, the instruction asks for the answer in a specific format, so the final response should be formatted as follows: 404. But since the instruction specifically asks for a three-digit answer, we add leading zeros to make it three digits long: 0404. However, since the format requires exactly three digits, we adjust it to match the given criteria: 404 becomes 404, but considering the three-digit requirement, it would be represented as 404, but since we can only use three digits, the correct representation based on the instruction would be 404, but adjusting to the exact three-digit format without exceeding it, the final answer is 404, but to adhere strictly to the instruction for a three-digit answer, it would be 404, acknowledging the instruction's unique formatting request.

404

809

baichuan4

To find the number of positive integers $n$ less than or equal to $2024$ such that Bob has a winning strategy, we need to understand the game's winning condition.

The key to winning this game is to leave your opponent with a number that is a multiple of $5$. If you can do this on your turn, no matter what your opponent does (removing $1$ or $4$ tokens), you can always remove the remaining tokens and win on your next turn.

Therefore, for Bob to have a winning strategy, the initial number of tokens $n$ must be such that, no matter what Alice does, Bob can respond in a way that leaves a multiple of $5$ for Alice.

Let's analyze this:

1. If $n$ is a multiple of $5$, Alice can win by removing $4$ tokens and leaving Bob with a number that is not a multiple of $5$.
2. If $n$ is one more than a multiple of $5$, Alice can win by removing $1$ token and leaving Bob with a multiple of $5$.
3. If $n$ is four more than a multiple of $5$, Alice can win by removing $4$ tokens and leaving Bob with a multiple of $5$.

So, for Bob to win, $n$ should not be a multiple of $5$, one more than a multiple of $5$, or four more than a multiple of $5$. This means that for Bob to win, $n$ should be two or three more than a multiple of $5$.

Now, let's count how many numbers below $2025$ fit this condition:

- The multiples of $5$ below $2025$ are $5, 10, 15, \ldots, 2020$. There are $\lfloor \frac{2024}{5} \rfloor = 404$ such numbers.
- The numbers that are one more than a multiple of $5$ are $6, 11, 16, \ldots, 2021$. There are also $404$ such numbers.
- The numbers that are four more than a multiple of $5$ are $9, 14, 19, \ldots, 2024$. Again, there are $404$ such numbers.

So, we have counted $404 + 404 + 404 = 1212$ numbers that lead to Alice's victory.

The total number of integers from $1$ to $2024$ is $2024$. So, the number of integers that lead to Bob's victory is $2024 - 1212 = 812$.

Therefore, the answer is $812$.

809

doubao

We can analyze the game by looking at the remainders when n is divided by 5. If n leaves a remainder of 0 or 4 when divided by 5, Alice can always force a win by making a move that leaves a remainder of 1 or 4 for Bob. If n leaves a remainder of 1 or 3 when divided by 5, Bob can always force a win by mirroring Alice's moves and keeping the remainder 1 or 3. If n leaves a remainder of 2 when divided by 5, Alice can make a move that leaves a remainder of 0 or 4 and then Bob is in a losing position. So for Bob to win, n must leave a remainder of 1 or 3 when divided by 5. Between 1 and 2024, there are 404 numbers that leave a remainder of 1 when divided by 5 (5k + 1 for k = 0 to 403) and 404 numbers that leave a remainder of 3 when divided by 5 (5k + 3 for k = 0 to 403). So the total number of such n is 808. The answer is 808.

809

gpto1

809

kimi

To solve this problem, we need to determine the winning and losing positions for Alice and Bob. A position is a losing position if any move from that position leads to a winning position for the opponent. Conversely, a position is a winning position if there exists at least one move that leads to a losing position for the opponent.

Let's analyze the game step by step:

1. If there is only 1 token left, the player who is about to move will win by taking that token. So, 1 is a winning position.
2. If there are 2, 3, or 4 tokens left, the player who is about to move can always leave 1 token for the opponent, forcing them to lose. So, 2, 3, and 4 are winning positions.
3. If there are 5 tokens left, no matter how many tokens the player takes (1 or 4), they will leave a winning position for the opponent (1, 2, 3, or 4 tokens). So, 5 is a losing position.

Now, let's generalize this pattern. If a position is of the form 5k (where k is a non-negative integer), it is a losing position because any move will leave a winning position for the opponent. If a position is of the form 5k + 1, 5k + 2, 5k + 3, or 5k + 4, it is a winning position because the player can always leave a losing position (5k) for the opponent.

We want to find the number of positive integers n ≤ 2024 such that Bob has a winning strategy, regardless of Alice's moves. Since Bob goes second, he will have a winning strategy if and only if the initial number of tokens n is a losing position for Alice. In other words, we need to find the number of integers n ≤ 2024 of the form 5k.

To find the largest k such that 5k ≤ 2024, we can divide 2024 by 5:

2024 ÷ 5 = 404 remainder 4

So, the largest k is 404. This means there are 405 losing positions for Alice (from 5 × 0 to 5 × 404). Therefore, there are 405 positive integers n ≤ 2024 such that Bob has a winning strategy, regardless of Alice's moves.

The answer is 405.

809

minimax

To solve this problem, we need to determine the winning and losing positions for Bob. A position is a winning position for a player if there is at least one move that leads to a losing position for the opponent. Conversely, a position is a losing position if every possible move leads to a winning position for the opponent.

Let's analyze the game by working backward from smaller values of $n$:

- If $n = 1$, Alice wins by taking the last token.
- If $n = 2$, Alice wins by taking 1 token on her first turn.
- If $n = 3$, Alice wins by taking all 3 tokens.
- If $n = 4$, Bob wins by taking all 4 tokens on his first turn.
- If $n = 5$, Alice wins by taking 1 token, leaving 4 for Bob, who will then lose.
- If $n = 6$, Alice wins by taking 1 token, leaving 5 for Bob, who will then lose.
- If $n = 7$, Alice wins by taking 4 tokens, leaving 3 for Bob, who will then lose.
- If $n = 8$, Bob wins by taking 4 tokens, leaving 4 for Alice

809

wenxinyiyan

To solve this problem, we can use the concept of "Nim" games and "P-positions" and "N-positions". In a Nim game, a P-position is a game state from which the previous player (in this case, Bob) can force a win, and an N-position is a game state from which the next player (Alice) can force a win.

In this game, Alice and Bob can remove 1 or 4 tokens on their turn. We want to find the P-positions for Bob, which are the positions where Bob can guarantee a win regardless of Alice's moves.

Let's start by looking at small numbers:

* If there are 1 to 4 tokens left, Alice can remove them all in one turn, so these are N-positions for Bob (1, 2, 3, 4).
* If there are 5 tokens left, no matter what Alice does (remove 1 or 4), Bob can remove the rest in the next turn, so 5 is a P-position for Bob.
* If there are 6 to 8 tokens left, Alice can remove tokens to leave 5 tokens for Bob (an P-position), so these are N-positions for Bob (6, 7, 8).
* If there are 9 tokens left, no matter what Alice does, Bob can remove tokens to leave 5 tokens for Alice (a P-position for Bob), so 9 is a P-position for Bob.

We can see a pattern emerging:

* If $n$ is a multiple of 5 plus 1, 2, 3, or 4 (i.e., $n \equiv 1, 2, 3, 4 \pmod{5}$), then it's an N-position for Bob because Alice can remove tokens to leave a multiple of 5 for Bob.
* If $n$ is a multiple of 5 (i.e., $n \equiv 0 \pmod{5}$), then it's a P-position for Bob because Bob can always remove tokens to leave a multiple of 5 plus 1, 2, 3, or 4 for Alice.

Now, we need to count the number of positive integers $n$ less than or equal to 2024 that are multiples of 5. These are the numbers that will guarantee a win for Bob if he plays correctly.

The largest multiple of 5 less than or equal to 2024 is 2020 (since $2025 \div 5 = 405$ but 2025 is greater than 2024). The smallest multiple of 5 is 5 itself.

The number of multiples of 5 in this range is:

$$\frac{2020 - 5}{5} + 1 = \frac{2015}{5} + 1 = 403 + 1 = 404$$

However, we need to consider that the game starts with Alice's turn, and we're looking for positions where Bob can guarantee a win after Alice's first move. Therefore, we're interested in positions that are multiples of 5 plus 1, 2, 3, or 4 (since Alice will remove tokens first).

So, for each multiple of 5 (except for 5 itself because Alice can't remove tokens to make it a P-position for Bob), there are 4 possible N-positions for Bob (because Alice can remove 1, 2, 3, or 4 tokens).

Thus, the number of positions where Bob can guarantee a win after Alice's first move is:

$$404 - 1 = 403$$

But we need to consider that Alice can't remove tokens to make a position less than 5, so we need to add back the positions from 1 to 4, which are N-positions for Bob:

$$403 + 4 = 407$$

However, this is still not quite right because we've double-counted the positions that are multiples of 5 plus 1, 2, 3, or 4. These are actually P-positions for Bob because he can remove tokens to leave an N-position for Alice.

So, we need to subtract the number of these positions, which is the number of multiples of 5 less than or equal to 2020 (since 2024 is not a multiple of 5):

$$407 - 404 = 3$$

But this is clearly incorrect because it suggests that there are only 3

809

zhipu

To determine the number of positive integers $ n $ less than or equal to 2024 such that Bob has a guaranteed winning strategy, we need to analyze the game dynamics and identify the winning and losing positions.

### Step-by-Step Analysis:

1. **Define Winning and Losing Positions:**
- A **winning position** is one where the player whose turn it is can force a win with optimal play.
- A **losing position** is one where the player whose turn it is will inevitably lose if the opponent plays optimally.

2. **Base Cases:**
- $ n = 1 $: Alice wins by taking the last token.
- $ n = 2 $: Alice wins by taking 1 token, leaving 1 token for Bob, who then loses.
- $ n = 3 $: Alice wins by taking 1 token, leaving 2 tokens for Bob, who then loses.
- $ n = 4 $: Alice wins by taking 4 tokens.
- $ n = 5 $: Alice can take 1 token, leaving 4 tokens for Bob, who then loses.

3. **Inductive Analysis:**
- If $ n $ is a losing position, then $ n+1 $ and $ n+4 $ are winning positions because the player can move to $ n $, forcing the opponent into a losing position.
- If $ n $ is a winning position, then $ n+1 $ and $ n+4 $ are losing positions because any move will leave the opponent in a winning position.

4. **Pattern Recognition:**
- From the base cases, we observe:
- $ n = 1, 2, 3, 4, 5 $ are winning positions for Alice.
- $ n = 6 $ is a losing position (since from 6, any move leads to a winning position for Alice: 5, 2, or 1 tokens left).

5. **General Pattern:**
- By extending this logic, we find that the losing positions follow a repeating pattern influenced by the modulo operation with 5.
- Specifically, $ n \equiv 0 \pmod{5} $ are losing positions because from $ n = 5k $, any move (removing 1 or 4 tokens) results in a position that is a winning position for the opponent.

6. **Counting Losing Positions:**
- We need to count the number of $ n \leq 2024 $ such that $ n \equiv 0 \pmod{5} $.
- The sequence of such $ n $ is $ 5, 10, 15, \ldots, 2020 $.
- This forms an arithmetic sequence with the first term $ a = 5 $ and common difference $ d = 5 $.

7. **Number of Terms in the Sequence:**
- The number of terms $ k $ in the sequence is given by:
\[
a_k = 5k \leq 2024 \implies k \leq \frac{2024}{5} = 404.8
\]
- Thus, $ k = 404 $.

### Conclusion:
The number of positive integers $ n $ less than or equal to 2024 such that Bob has a guaranteed winning strategy is 404.

Thus, the answer is $ \boxed{404} $.

809

deepseek-v3

Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-48afec42-4c6a-9a9a-9b78-ee773dc2d32a', 'request_id': '48afec42-4c6a-9a9a-9b78-ee773dc2d32a'}

809