a|(3b-c) => 可设 c=3b-xa,b|(a-c) => 可设 c=a-yb
=> 3b-xa=a-yb => (y+3)b=(x+1)a
∵a,b是不同质数,∴b|(x+1),设x+1=mb
则得y+3=ma => mc=m(a-yb)=ma-ymb=y+3-y(x+1)=3-xy
若m=0,则得x=-1,y=-3,即此时c=a+3b
又c|(2a-7b),设2a-7b=nc=na+3nb => (2-n)a=(3n+7)b
若n=0,则2a=7b => a=7,b=2,c=a+3b=7+6=13,矛盾
若n>0,则2-n=(3n+7)b/a>0 => n n>-7/3,即n=-1或-2
若n=-1 => 3a=4b =>b=3,a=4,这与a是质数矛盾
若n=-2 => 4a=b,这与b是质数矛盾
综上知m≠0,若m>0,则x+1=mb≥b≥2 =>x≥1
y+3=ma≥a≥2 =>y≥-1,3-xy=mc≥23 => xy≤-20
∴此时必有y=-1 => a=2,m=1,∴x=b-1,c=3+x=2+b
而此时2a-7b=4-7(c-2)=18-7c,=> c|(18-7c) =>c|18
这与20 xy≥26
又c|(2a-7b) => mc|(2ma-7mb),即
(3-xy)|(2y+6-7x-7) => (3-xy)|(2y-7x-1),若2y-7x-1≠0
则|2y-7x-1|≥|3-xy|=xy-3 => 2y-7x-1≤3-xy或2y-7x-1≥xy-3
=>(x+2)(y-7)≤-10 或 (x-2)(y+7)≤-12
∵此时有y-7≤-12,x+2≤-1,∴(x+2)(y-7)>0,
∴只能是(x-2)(y+7)≤-12,而x-2≤-5,y+7≤2
∵y+7≤0时,有(x-2)(y+7)≥0,∴y+7=1或2 => y=-5或-6
若y=-6,则c=xy-3=-6x-3=3(-2x-1),即c为3的倍数
这显然与c是介于20到80之间的质数相矛盾.
∴y=-5 => ma=y+3=-2 => m=-1,a=2.而c=xy-3=-5x-3,
则23≤-5x-3≤79 =>-16≤x≤-6,∵c为质数,可得
x=-8或-10或-14 => b=-x-1=7或9或13,而b为质数
∴只能取b=7和13,此时对应x=-8和-14,对应c=37和67
而(a,b,c)=(2,7,37),(2,13,67)均不满足c|(2a-7b),∴此时无解
∴m2ma-7mb=(2y-7x-1)=0
=> m(2a-7b)=0 => 2a=7b => a=7,b=2
=>x+1=2m,y+3=7m => cm=3-xy=3-(2m-1)(7m-3)
=13m-14m² => c=13-14m =>23≤13-14m≤79
=>m≥-4,分别代入m=-1,-2,-3,-4,可知
仅当m=-2时c=41为质数,进一步验证可知
(a,b,c)=(7,2,41)满足所有条件,∴a^b*c有唯一值
即a^b*c=7²·41=2009