Discussion
Modulo the prime 101, the number \((50\,!)^{2}\) is congruent to :
Answer: A
\((50!)^{2} \equiv\) -1 (mod 101).
By Wilson's theorem, \(100! \equiv -1 \pmod{101}\). Pair each factor \(k \le 50\) with \(101 - k \equiv -k\):
\[ \begin{aligned} 100! &= \prod_{k=1}^{50} k(101 - k) \\ &\equiv \prod_{k=1}^{50} (-k^{2}) = (-1)^{50}(50!)^{2} \end{aligned} \]
Since \((-1)^{50} = 1\), \((50!)^{2} \equiv 100! \equiv -1\). This also shows that -1 is a square mod 101, as it must be because \(101 \equiv 1 \pmod{4}\).
The value cannot be 0, since 101 is prime and does not divide 50!, and 1 and 2 do not satisfy Wilson's theorem here.
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