Let $a$, $b$, and $c$ be integers such that $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$. What is the remainder when $a^2 + b^2 + c^2$ is divided by 7?

Answer 1

Given that \(a \equiv 1 \pmod{7}\), \(b \equiv 2 \pmod{7}\), and \(c \equiv 6 \pmod{7}\), we seek the remainder of \(a^2 + b^2 + c^2\) when divided by 7.

First, we calculate each square modulo 7.

For \(a \equiv 1 \pmod{7}\):
\[
a^2 \equiv 1^2 \equiv 1 \pmod{7}
\]

For \(b \equiv 2 \pmod{7}\):
\[
b^2 \equiv 2^2 \equiv 4 \pmod{7}
\]

For \(c \equiv 6 \pmod{7}\):
\[
c^2 \equiv 6^2 \equiv 36 \pmod{7}
\]
Calculating \(36 \mod 7\):
\[
36 \div 7 = 5 \quad \text{remainder} \, 1 \quad \Rightarrow \quad 36 \equiv 1 \pmod{7}
\]

Summing the results:
\[
a^2 + b^2 + c^2 \equiv 1 + 4 + 1 \pmod{7}
\]
\[
6 \pmod{7}
\]

Thus, the remainder when \(a^2 + b^2 + c^2\) is divided by 7 is
\[
\boxed{6}
\]