Finding an integer $x$ and a 3-digit prime $p$ that solves the problem $353, 46, 618, 30, 877, 235, 98, 24, 107, 188, 673$ are successive large powers of an integer $x$ modulo a 3-digit prime $p$. Find $x$ and $p$.
This question seems impossible to me, I have tried different things, but I can't get anywhere.
 A: Suppose $x^n \equiv 353 \mod p$, then $x^{n+1} \equiv 46 \mod p$. So $353x \equiv 46 \mod p$ and similarly $46x \equiv 618 \mod p$, etc. Since $343$ and $46$ are coprime, you can find $x \equiv \text{ something } \mod p$. Same with 24 and 107 (just one example); you get more equations for $x \mod p$. You also know $p > 877$ and it must be a three digit prime... you should be able to finish from here. Good luck!
Edit: There's a simpler way to conclude. By the same reasoning, $98x \equiv 24 \mod p$ and $24 x \equiv 107 \mod p$. $p>2$, so $2$ and $p$ are relatively prime, so we can divide the first congruence by 2. So $49 x \equiv 12$, and from the second congruence, $48x \equiv 214$; subtracting, $x \equiv -202 \mod p$. Substituting in the second congruence, $24(-202) \equiv 107$, that is, $p | 107 + 24*202$. Compute this number and find its prime factor decomposition. If any of the prime factors are between $878$ and $999$, it's a candidate (still need to verify that all those other powers have the right residues modulo it); if there are none, the problem is impossible. The number is $4955 = 5*991$ so it follows that $p=991$.
