Is that $a^{d}\equiv 1\pmod n$? If it is, then it follows from the following result: The exponent of a finite abelian group occurs as the order of an element of the group. This, in turn, is clear from the structure theorem for finite abelian groups: they are finite direct products of cyclic groups.

Let \[n=\prod_{i}p_{i}^{k_{i}}\] , then ring \[Z/nZ=\sum_{i}(Z/p_{i}^{k_{i}}Z.\] Used $(Z/p^{k}Z)^{*}$ (p>2) are ciclic followed your result.

If two elements $u_{1}$ and $u_{2}$ have orders $n_{1}$ and $n_{2}$, then let $(n_{1}, n_{2})=m$, and let $n_{i}=mk_{i}$ for $i=1, 2$. Then $(k_{1}, k_{2})=1 \implies \exists a, b\in \mathbb{Z}, ak_{1}+bk_{2}=1$, with $(a, k_{2})=(b, k_{1})=1$. Then consider $u_{1}^{b}u_{2}^{a}=u_{3}$, and some work with orders yields that $n_{1}, n_{2}$ divide the order of $u_{3}$.

$n=2^{\alpha_{1}}p_{2}^{\alpha_{2}}p_{3}^{\alpha_{3}}...p_{m}^{\alpha_{m}}$ We will prove that $LCM(\phi(p_{2}^{\alpha_{2}}), \phi(p_{3}^{\alpha_{3}}), ..., \phi(p_{m}^{\alpha_{m}}))|d$. Let $A$ be the set of residues $\mod{n}$ which are primitive roots $\mod{p_{i}^{\alpha_{i}}}$ for $i=2,3,...m$. By the Chinese Remainder Theorem $|A|>0$ and the claim is obvious. Now, from the set $A$ we can choose such $g$ that $ord_{2}(g)$ is largest and then it's clear that $d=ord_{2}(g) \cdot LCM(\phi(p_{2}^{\alpha_{2}}), \phi(p_{3}^{\alpha_{3}}), ..., \phi(p_{m}^{\alpha_{m}}))|$. Also $ord_{n}(g)=d$

sorry, I don't know what is the meaning of $ord_{n}b$?

$ord_{n}(b)=a$ iff $a$ is the smallest positive integer such that $b^{a}\equiv 1 \mod{n}$ (if such exists).

Thank you very much ,TomciO !

Hm...isn't this a very well known fact of the carmichael function $\lambda$? (since it is obvious that $d=\lambda(n)$ - or it is even by definition, depending how one defines $\lambda$). A strange problem to propose for a competition

Yes, it is a bit strange... It's obvious... It's not even necessary to invoke Carmichael function...