小昭 
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 回覆於: 2024/2/27 上午 02:28:13
由於可被66整除,則c為偶數
536ab98c
=53+(60+a)+(10b+9)+(80+c) (mod 99)
=202+10b+a+c (mod 99)
=4+10b+a+c (mod 99)
=4+10b+a+c (mod 33)
設a=33n-4-10b-c 其中n=1,2,3
Case One:當n=1,b=2時
則a=33(1)-4-10(1)-c
a=9-c
(c,a)=(0,9),(2,7),(4,5),(6,3),(8,1)
Case Two:當n=2,b=6時
則a=33(2)-4-10(6)-c
a=2-c
(c,a)=(0,2),(2,0)
Case Three:當n=2,b=5時
則a=33(2)-4-10(5)-c
a=12-c
(c,a)=(4,8),(6,6),(8,4)
Case Four:當n=3,b=9時
則a=33(3)-4-10(9)-c
a=5-c
(c,a)=(0,5),(2,3),(4,1)
Case Five:當n=3,b=8時
則a=33(3)-4-10(8)-c
a=15-c
(c,a)=(6,9),(8,7)
_____
所以(a,b,c)
=(9,2,0),(7,2,2),(5,2,4),(3,2,6),(1,2,8),
(2,6,0),(0,6,2),(8,5,4),(6,5,6),(4,5,8),
(5,9,0),(3,9,2),(1,9,4),(9,8,6),(7,8,8)
八位數共15個
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發表於: 2013/2/25 上午 11:18:13
