若正整数a,b,c满足(a-1)(a+bc-b-c)=115,求a+b+c的最大值
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既然是正整数,那么,(a-1) 与 (a+bc-b-c) 的积是 115,那么,它们肯定是 115 的因数。
因为 115 = 5×23
那么就有:
a - 1 = 5
a + bc - b - c = 23
则
a = 6
bc - b - c = 17
bc - c -b + 1 = 18
c(b-1) - (b-1) = 18
(c-1)(b-1) = 18 = 1×18 = 2×9 = 3×6
同样的道理,有:
c-1 = 1, b-1 = 18 或 c-1=18, b-1 = 1。即 b = 19, c=2,或 b=2, c=19
或
c-1 = 2, b-1 = 9 或 c - 1 = 9, b-1 = 2。即 b = 10, c = 3,或 b = 3, c = 10
或
c-1 = 3, b-1 = 6 或 c-1 = 6, b-1 = 3。即 b=7, c=4,或 b = 4, c = 7
综上所述,a、b 、c 的可能组合是:
(5, 19, 2)、(5, 2, 19)、(5, 10, 3)、(5, 3, 10)、(5, 7, 4)、(5, 4, 7)
那么,a + b + c 的最大值就等于:
=5 + 19 + 2
= 26
希望能够帮到你!
因为 115 = 5×23
那么就有:
a - 1 = 5
a + bc - b - c = 23
则
a = 6
bc - b - c = 17
bc - c -b + 1 = 18
c(b-1) - (b-1) = 18
(c-1)(b-1) = 18 = 1×18 = 2×9 = 3×6
同样的道理,有:
c-1 = 1, b-1 = 18 或 c-1=18, b-1 = 1。即 b = 19, c=2,或 b=2, c=19
或
c-1 = 2, b-1 = 9 或 c - 1 = 9, b-1 = 2。即 b = 10, c = 3,或 b = 3, c = 10
或
c-1 = 3, b-1 = 6 或 c-1 = 6, b-1 = 3。即 b=7, c=4,或 b = 4, c = 7
综上所述,a、b 、c 的可能组合是:
(5, 19, 2)、(5, 2, 19)、(5, 10, 3)、(5, 3, 10)、(5, 7, 4)、(5, 4, 7)
那么,a + b + c 的最大值就等于:
=5 + 19 + 2
= 26
希望能够帮到你!
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