第十四屆培正數學邀請賽第20題求解

Obviously, the last 3 digits is 444 only (000 is not allowed).Let the 'good' number be (100a + 10b + c), so,(100a + 10b + c)² = 1000x + 444
==> 10000a² + 2000ab + 100b² + 200ac + 20bc + c² = 1000x + 444
==> 1000(10a² + 2ab) + 100(b² + 2ac) + 10(2bc) + c² = 1000x + 444
==> 100(b² + 2ac) + 10(2bc) + c² = 1000y + 444
∴ c = 2 or 8

Case c = 2 :
100(b² + 4a) + 10(4b) + 4 = 1000y + 444
==> 10(b² + 4a) + 4b = 100y + 44
∴ b = 1 or 6
When (b, c) = (1, 2),
10(1 + 4a) + 4 = 100y + 44
==> 1 + 4a = 10y + 4
==> 4a = 10y + 3 ⋯⋯ no suitable value for a.
When (b, c) = (6, 2)
10(36 + 4a) + 24 = 100y + 44
==> 6 + 4a + 2 = 10p + 14
==> 4a = 10p + 6
==> a = 4, 9, 14, 19, ...
∴ (a, b, c) = (4, 6, 2), (9, 6, 2), (14, 6, 2), (19, 6, 2), ... ⋯⋯⋯⋯ [i]

Case c = 8 :
100(b² + 16a) + 10(16b) + 64 = 1000y + 444
==> 10(b² + 6a) + 16b = 100q + 38
∴ b = 3 or 8
When (b, c) = (3, 8),
10(9 + 6a) + 48 = 100q + 38
==> 9 + 6a = 10r + 9
==> 6a = 10r
==> a = 0, 5, 10, 15, ...
∴ (a, b, c) = (0, 3, 8), (5, 3, 8), (10, 3, 8), (15, 3, 8), ... ⋯⋯⋯⋯[(ii]
When (b, c) = (8, 8)
10(64 + 6a) + 128 = 100q + 38
==> 4 + 6a + 12 = 10r + 23
==> 6a = 10r + 7 ⋯⋯ no suitable value for a.

Arrange all the 'good' numbers (in [i], [ii]) in ascending order :
38, 462, 538, 962, 1038, 1462, 1538, 1962, ...
∴ the second 'good' number is 462.

可否參考一下這個？

http://www.qbyte.org/puzzles/p089s.html

462^2 = 213444