Let $a$ be an integer such that $0 \le a \le 10$ and $a^2 \equiv a \pmod{11}$. If $a \neq 0,$ then find the value of $a$.

Answer 1

Given the problem constraints, we consider the integer \( a \) such that \( 0 \le a \le 10 \) and \( a^2 \equiv a \pmod{11} \). This congruence can be rewritten as:

\[
a^2 - a \equiv 0 \pmod{11}
\]

Factoring, we get:

\[
a(a - 1) \equiv 0 \pmod{11}
\]

Since the product \( a(a - 1) \) modulo 11 must be zero, one of the factors must be divisible by 11. Given that \( a \) and \( a - 1 \) are consecutive integers, and the only integers satisfying \( 0 \le a \le 10 \) are:

1. \( a \equiv 0 \pmod{11} \)
2. \( a - 1 \equiv 0 \pmod{11} \)

Given our range \( 0 \le a \le 10 \):

1. \( a = 0 \)
2. \( a = 1 \)

We check the values:

- For \( a = 0 \):

\[
0^2 \equiv 0 \pmod{11}
\]

This is true, but we need to find \( a \neq 0 \).

- For \( a = 1 \):

\[
1^2 \equiv 1 \pmod{11}
\]

This is also true. Consequently, the \( a \) value different from 0 fulfilling the given conditions is \( a = 1 \).

Thus, the value of \( a \) is:

\[
\boxed{1}
\]