If a,b,c are integers from the set of positive integers less than 7 such that

\(\begin{align*} abc&\equiv 1\pmod 7,\\ 5c&\equiv 2\pmod 7,\\ 6b&\equiv 3+b\pmod 7, \end{align*}\)

then what is the remainder when a+b+c is divided by 7?

Guest Dec 7, 2019

#1**+1 **

We test values for b and find it is 2. We test values for c that are in the form 7x+2, and find that c=6. Then have 12a=1 mod7 so by testing values again we find that a is 3. So we do 2+3+6 = 11 and divide it by 7. Our remainder is 4.

OlympusHero Dec 7, 2019

#2**+1 **

You got it almost right !! Except that a = 3 doesn't work because 326 mod 7 = 4 instead of 1.

a=0;b=0;c=0;p=0; cycle:d=a*100+b*10+c;if(d%7==1 and d%7==2 d%7==3+b, goto loop, goto next); loop:printd," ",;p=p+1; next:c++;if(c<10, goto cycle, 0);c=0;b++;if(b<10, goto cycle, 0);b=0;c=0;a++;if(a<10, goto cycle,0);print"Total = ",p

This computer code incorporates elements of the Chinese Remainder Theorem and Modular Multiplicative Inverse. It gives the following numbers as the answer.

a = 5, b = 2, c = 6. Therefore:

526 mod 7 =1

5 * 6 mod 7=2

6 * 2 mod 7=3 + 2

Then: [a +b + c = 5 + 2 + 6] mod 7 = 6

Guest Dec 7, 2019