+249 1. Find the least four-digit solution r of the congruence \(r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55} \).
2. If a,b, and c are integers from the set of positive integers less than 7 such that \(abc\equiv 1\pmod 7, 5c\equiv 2\pmod 7, 6b\equiv 3+b\pmod 7,\) then what is the remainder when a+b+c is divided by 7?
3. What is the smallest integer b>3 for which the base b number 23_b is a perfect square?
+118677 1. Find the least four-digit solution r of the congruence .
\(r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55}\)
I am not sure but I will give it a shot
\(\begin{align} r^2 + 4r + 4 &\equiv r^2 + 2r + 1 \pmod{55}\\ 2r &\equiv -3 \pmod{55}\\ 2r&=55x-3\\ 55x-2r&=3 \qquad x\in Z \qquad r \in Z \quad r \ge 1000\\ &\mbox{so if r=1000 then}\\ 55x-2000&=3\\ 55x&=2003\\ x&=37\\ \end{align}\\ \mbox{So I think that r=1000 works}\)
+118677 2. If a,b, and c are integers from the set of positive integers less than 7 such that
\(abc\equiv 1\pmod 7, 5c\equiv 2\pmod 7, 6b\equiv 3+b\pmod 7,\)
then what is the remainder when a+b+c is divided by 7?
5*6=30=2 mod7 so c=6
let ab=x
6x=1 mod 7
x=6
ab=6 (1*6) (2*3) there might be others as well.... 6 is already used I don;t know if it can be used twice.
so try 2 and 3
check 2*3*6=36=1mod7 so at least that works
If b=2 then 6*2mod7=5mod7 3+2=5 mod7 so that does work
S0 a combination that works is
a=3, b=2, c=6
a+b+c = 3+2+6 = 11 = 4mod7 The remainder is 4
