✔ 最佳答案
a (1~5) + b (1~8) + c (3~8) + d (2~7) = 19, 所有正整數解個數為何?
a=5, b=8, c+d=6, n={3+3, 4+2}=2a=5, b=7, c+d=7, n={3+4, 4+3, 5+2}=3a=5, b=6, c+d=8, n={3+5, 4+4, 5+3, 6+2}=4a=5, b=5, c+d=9, n={3+6, 4+5, 5+4, 6+3, 7+2}=5a=5, b=4, c+d=10, n={3+7, 4+6, 5+5, 6+4, 7+3, 8+2}=6a=5, b=3, c+d=11, n={4+7, 5+6, 6+5, 7+4, 8+3}=5a=5, b=2, c+d=12, n={5+7, 6+6, 7+5, 8+4}=4a=5, b=1, c+d=13, n={6+7, 7+6, 8+5}=3S1 = (2+3+4+5+6) + (5+4+3)= 20 + 12= 32
推廣出一般式:x = c + d = 6~10y = c + d = 11~15n = f(x) = x - 4n = g(y) = 16 - y
a=4, b=8, f(7)=3a=4, b=7, f(8)=4a=4, b=6, f(9)=5a=4, b=5, f(10)=6a=4, b=4, g(11)=5a=4, b=3, g(12)=4a=4, b=2, g(13)=3a=4, b=1, g(14)=2
S2 = (3+4+5+6) + (5+4+3+2)= 18 + 14= 32
以此類推:a=3, b=8~1S3 = f(8)+f(9)+f(10)+g(11)+g(12)+g(13)+g(14)+g(15)= (4+5+6) + (5+4+3+2+1)= 15 + 15= 30
a=2, b=8~1S4 = f(9)+f(10)+g(11)+g(12)+g(13)+g(14)+g(15)= (5+6) + (5+4+3+2+1)= 11 + 15= 26
a=1, b=8~2S5 = f(10)+g(11)+g(12)+g(13)+g(14)+g(15)= 6+5+4+3+2+1= 21
S = S1 + S2 + S3 + S4 + S5= 32 + 32 + 30 + 26 + 21= 64 + 56 + 21= 75 + 56= 131= Answer
2015-06-04 17:47:39 補充:
修正: S = 141
2015-06-08 07:15:53 補充:
(2) a=?,b=?,c=?,d=?
Ans:
a = 5
b=8, c+d=6={3+3, 4+2}
b=7, c+d=7={3+4, 4+3, 5+2}
b=6, c+d=8={3+5, 4+4, 5+3, 6+2}
b=5, c+d=9={3+6, 4+5, 5+4, 6+3, 7+2}
b=4, c+d=10={3+7, 4+6, 5+5, 6+4, 7+3, 8+2}
b=3, c+d=11={4+7, 5+6, 6+5, 7+4, 8+3}
b=2, c+d=12={5+7, 6+6, 7+5, 8+4}
b=1, c+d=13={6+7, 7+6, 8+5}
2015-06-08 07:16:58 補充:
a = 4
b=8, c+d=6={3+3, 4+2}
b=7, c+d=7={3+4, 4+3, 5+2}
b=6, c+d=8={3+5, 4+4, 5+3, 6+2}
b=5, c+d=9={3+6, 4+5, 5+4, 6+3, 7+2}
b=4, c+d=10={3+7, 4+6, 5+5, 6+4, 7+3, 8+2}
b=3, c+d=11={4+7, 5+6, 6+5, 7+4, 8+3}
b=2, c+d=12={5+7, 6+6, 7+5, 8+4}
b=1, c+d=13={6+7, 7+6, 8+5}
以下類推
2015-06-08 07:25:52 補充:
(3) 從S1算式推出了一般式,是怎麼推出來的?
Ans:
c + d = x = 6 ~ 10
f(6) = 2, f(10) = 6
=> (6,2), (10,6)
直線方程式為:
(y-2)/(x-6) = (6-2)/(10-6) = 1
=> y = x - 4
=> f(x) = x - 4
=> f(y)類推