Verifying that $11^{λ(m)+1} ≡ 11 \pmod m$ for $m = 41 \cdot 11$

Question

I know that $λ(m) = [40, 10] = 40$ So I need to show that $11^{41} ≡ 11 \mod 451$. From here I believe I should show that $11^{41} ≡ 11 \mod 41$ and that $11^{41} ≡ 11 \mod 11$, but I am not sure how to show the first one, and the second one (the one that is mod $11$) I know is congruent to $0 \mod 11$ so I run into a problem there.

Answer 1

Hint: The 1st one: $11^{40} = 1 \pmod{41}$ by fermat little theorem. So $11^{41} = 11 \pmod{41}$. You can continue.

Answer 2

The key idea is: $ $ if $\,p_i\,$ are distinct primes and $\,\lambda\,$ is a common multiple of all $\,p_i-1\,$ then

$\quad\lambda = (p_i\!-\!1)k_i\,\Rightarrow\, {\rm mod}\ p_i\!:\ a^{\large 1+\lambda} = a^{\large 1+(p_i-1) k_i} = a ( \color{#c00}{a^{\large p_i-1}})^{\large k}\equiv a(\color{#c00}1)^k\equiv a\ $ by $\rm\color{#c00}{little\ Fermat}$.

Thus $\,\ p_i\mid a^{1+\lambda}-a\ \Rightarrow\ \prod { p_i} = {\rm lcm}\,\{p_i\}\mid a^{1+\lambda}-a$