Find the minimal $t$, $t\in\mathbb{N}$ such that $113^t \equiv 1\bmod(455)$

Find the minimal $$t$$, $$t\in\mathbb{N}$$ such that $$113^t \equiv 1\bmod(455)$$.

I received this problem in introductory number theory so I have tried to use the basic properties of modular congruences and other division-based properties. So I have tried identifying the following, Using the property that $$a\equiv b\bmod(n)$$ and $$m\mid n$$ then $$a\equiv b\bmod(m)$$ and then, \begin{align} 113^t \equiv1\bmod(5) \\ 113^t \equiv1\bmod(7) \\ 113^t \equiv1\bmod(13) \\ \end{align} Then we get a system of linear congruences and then maybe if we let $$113^t$$ be $$x$$ then we can get $$x$$ and find a minimal $$t$$ satisfying that $$x$$. But does not seem to really work. Anyway, you could use Eulers theorem and the Eulers totient function and get $$t=288$$ which I don't know if is correct since I think $$288$$ is actually the maximal maybe(not sure)? But I want to solve this with more elementary methods without such concepts.

• Actually, $113^{12}\equiv 1 \bmod 455$, see this duplicate, or many others. Oct 26, 2023 at 11:47
• Note that when you are working modulo $5$, $113$ is the same as $3$; modulo $7$, $113=1$, and modulo $13$, $113=9=-4$. Oct 26, 2023 at 11:50

If $$113^{288}\equiv1\pmod{455}$$ then $$113^{144}\equiv \pm 1\pmod{455}$$. You can verify that $$113^{144}\equiv 1\pmod{455}$$ so $$113^{12^2}\equiv 1\pmod{455}$$ and $$113^{12}\equiv 1\pmod{455}$$. Since $$113^{6}\equiv 274\pmod{455}$$. You find now that the minimum exponent is $$12$$.