Finding Remainder of $98!$ divided by $101$

We need to find the remainder when $98!$ is divided by $101$. We can use Wilson's Theorem for this problem.

Wilson's Theorem Application

Wilson's Theorem states that for any prime number $p$, we have:

$ (p-1)! \equiv -1 \pmod{p} $

In this case, $p = 101$, which is a prime number. Applying Wilson's Theorem:

$ 100! \equiv -1 \pmod{101} $

Relating $100!$ to $98!$

We can write $100!$ as:

$ 100! = 100 \times 99 \times 98! $

Substitute this back into the congruence from Wilson's Theorem:

$ (100 \times 99 \times 98!) \equiv -1 \pmod{101} $

Simplifying the Congruence

We know that $100 \equiv -1 \pmod{101}$ and $99 \equiv -2 \pmod{101}$. Substituting these values:

$ (-1 \times -2 \times 98!) \equiv -1 \pmod{101} $

$ (2 \times 98!) \equiv -1 \pmod{101} $

Since $-1 \equiv 100 \pmod{101}$, we have:

$ 2 \times 98! \equiv 100 \pmod{101} $

Solving for $98!$

To find the value of $98!$, we need to solve the congruence $2 \times 98! \equiv 100 \pmod{101}$. This involves finding the modular multiplicative inverse of $2$ modulo $101$.

We need to find a number $x$ such that $2x \equiv 1 \pmod{101}$. We can see that $2 \times 51 = 102$, and $102 \equiv 1 \pmod{101}$. So, the modular inverse of $2$ modulo $101$ is $51$.

Multiply both sides of the congruence $2 \times 98! \equiv 100 \pmod{101}$ by $51$:

$ 51 \times (2 \times 98!) \equiv 51 \times 100 \pmod{101} $

$ (51 \times 2) \times 98! \equiv 5100 \pmod{101} $

$ 1 \times 98! \equiv 5100 \pmod{101} $

Now, we find the remainder of $5100$ when divided by $101$:

$ 5100 = 50 \times 101 + 50 $

Therefore, $5100 \equiv 50 \pmod{101}$.

So, the final result is:

$ 98! \equiv 50 \pmod{101} $

The remainder when $98!$ is divided by $101$ is $50$. This value falls within the given range of $49.9$ and $50.1$.