Answer :
Step-by-step explanation:
To find the remainder when \( \frac{2! \times \ldots \times 100!}{24} \) is divided, we need to consider the prime factorization of 24.
\(24 = 2^3 \times 3\)
In the factorial expression, all numbers from 2 to 100 are included. We need to find out how many factors of 2 and 3 are present in the factorial terms, because we are dividing by \(2^3 \times 3\) (which is 24).
For each multiple of 2 in the range from 2 to 100, there is at least one factor of 2. Similarly, for each multiple of 3, there is at least one factor of 3.
Counting the factors of 2:
\[\lfloor \frac{100}{2} \rfloor + \lfloor \frac{100}{4} \rfloor + \lfloor \frac{100}{8} \rfloor = 50 + 25 + 12 = 87\]
Counting the factors of 3:
\[\lfloor \frac{100}{3} \rfloor + \lfloor \frac{100}{9} \rfloor = 33 + 11 = 44\]
We have more than enough factors of 2 compared to 3, so we only need to consider the factors of 3.
Now, when 100! is divided by 24, we effectively remove all factors of 2 and 3 that make up 24. Since 3 is the limiting factor, we only need to find the highest power of 3 that divides 100!.
Using the formula for finding the highest power of a prime \( p \) that divides \( n! \) (where \( p \leq n \)), which is given by \( \lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \ldots \), we find:
\[ \lfloor \frac{100}{3} \rfloor + \lfloor \frac{100}{9} \rfloor = 33 + 11 = 44 \]
Therefore, the remainder when \( \frac{2! \times \ldots \times 100!}{24} \) is divided is the same as the remainder when \(100! \) is divided by 24.
\[ 100! \equiv 0 \ (\text{mod} \ 24) \]
So, the remainder is 0.