If a, b, and c are non-negative integers less than 7 such that a+2b+3c=0(mod 7), 2a+3b+c=4(mod 7), 3a+b+2c=4 (mod 7) then determine the remainder when abc is divided by 7.
hmmm, a+b+c+(b+2c) and a+b+c+(a+2b) and a+b+c+(b+2c)
a=1;p=0; b=1;c=1;n= (a+2*b+3*c)%7;m=(2*a+3*b+c)%7;k=(3*a+b+2*c)%7;if(n==0 and m==4 and k==4, goto loop, goto next);loop:p=p+1;printp," =",a,b,c;next:a++;if(a<15, goto4,0);a=1;b++;if(b<15, goto4, 0);a=1;b=1;c++;if(c<15, goto4,0)
a=1, b=2, c=3, and abc mod 7 =(1*2*3) mod 7 =6
+118703 I did this using Matrices:
a+2b+3c=0mod72a+3b+c=4mod73a+b+2c=4mod7
[123|0(mod7)231|4(mod7)312|4(mod7)] [246|0(mod7)231|4(mod7)312|4(mod7)] [369|0(mod7)0−1−5|4(mod7)312|4(mod7)] [123|0(mod7)0−1−5|4(mod7)0−5−7|4(mod7)] [123|0(mod7)015|−4(mod7)0018|−16(mod7)] [123|0(mod7)015|−4(mod7)004|−2(mod7)]
4c=−2mod74c=5mod74c=5+7mod74c=12mod7c=3 b+5c=−4mod7b+15=−4mod7b=−19mod7b=−21+2mod7b=2 a+2b+3c=0mod7a+4+9=0mod7a=−13mod7a=−14+1mod7a=1
a=1, b=2, c=3
Under here is not supposed to display, (except as LaTex coding.)
For some reason, it keeps rendering.......
LaTex:
a+2b+3c=0\mod7\\
2a+3b+c=4\mod7\\
3a+b+2c=4\mod7\\
[123|0(mod7)231|4(mod7)312|4(mod7)]
\\~\\~\\
[246|0(mod7)231|4(mod7)312|4(mod7)]
\\~\\~\\
[369|0(mod7)0−1−5|4(mod7)312|4(mod7)]
\\~\\~\\
[123|0(mod7)0−1−5|4(mod7)0−5−7|4(mod7)]
\\~\\~\\
[123|0(mod7)015|−4(mod7)0018|−16(mod7)]
\\~\\~\\
[123|0(mod7)015|−4(mod7)004|−2(mod7)]
4c=-2\mod7\\
4c=5\;\;\mod7\\
4c=5+7\mod7\\
4c=12\;\;\mod7\\
c=3\\~\\
b+5c=-4\mod7\\
b+15=-4\mod7\\
b=-19\mod7\\
b=-21+2\mod7\\
b=2\\~\\
a+2b+3c=0\mod7\\
a+4+9=0\mod7\\
a=-13\mod7\\
a=-14+1\mod7\\
a=1
