Show that the remainder when 2^1990 (2 to the power of 1990) is divided by 1990 equals 1024.

(In reply to Step by Step by Charlie)

That process never exceeds ordinary non-extended precision, and the successive powers of 1024 mod 1990 are
1836, 1504, 1826, 1214, 1376, 104, 1026, 1894, 1196, 854, 886, 1814, 866, 1234,
1956, 1004, 1256, 604, 1596, 514, 976, 444, 936, 1274, 1126, 814, 1716, 14,
406, 1824, 1156, 1684, 1076, 1354, 1456, 434, 646, 824, 16, 464, 1516, 184,
1356, 1514, 126, 1664, 496, 454, 1226, 1724, 246, 1164, 1916, 1834, 1446, 144,
196, 1704, 1656, 264, 1686, 1134, 1046, 484, 106, 1084, 1586, 224, 526, 1324,
586, 1074, 1296, 1764, 1406, 974, 386, 1244, 256, 1454, 376, 954, 1796, 344,
26, 754, 1966, 1294, 1706, 1714, 1946, 714, 806, 1484, 1246, 314, 1146, 1394,
626, 244, 1106, 234, 816, 1774, 1696, 1424, 1496, 1594, 456, 1284, 1416, 1264,
836, 364, 606, 1654, 206, 4, 116, 1374, 46, 1334, 876, 1524, 416, 124, 1606,
804, 1426, 1554, 1286, 1474, 956, 1854, 36, 1044, 426, 414, 66, 1914, 1776,
1754, 1116, 524, 1266, 894, 56, 1624, 1326, 644, 766, 324, 1436, 1844, 1736,
594, 1306, 64, 1856, 94, 736, 1444, 86, 504, 686, 1984, 1816, 924, 926, 984,
676, 1694, 1366, 1804, 576, 784, 846, 654, 1056, 774, 556, 204, 1936, 424, 356,
374, 896, 114, 1316, 354, 316, 1204, 1086, 1644, 1906, 1544, 996, 1024, this last being 1024^199 mod 1990 or 2^1990 mod 1990.

Posted by Charlie on 2003-03-02 09:35:51