To find the number of zeros at the end of \( 100! \), we need to determine how many times \( 10 \) divides into \( 100! \). Since \( 10 = 2 \times 5 \), every pair of factors of 2 and 5 contributes a factor of 10, resulting in one zero at the end of the number.
The number of factors of 5 in \( 100! \) will determine the number of zeros at the end, because there are always more factors of 2 than factors of 5 in any factorial. To find the number of factors of 5, we use the following formula: \[ \left\lfloor \frac{100}{5} \right\rfloor + \left\lfloor \frac{100}{25} \right\rfloor \] The first term counts the multiples of 5, and the second term counts the multiples of 25 (which contribute an extra factor of 5). Let's calculate it: \[ \left\lfloor \frac{100}{5} \right\rfloor = 20 \] \[ \left\lfloor \frac{100}{25} \right\rfloor = 4 \] Thus, the total number of factors of 5 in \( 100! \) is: \[ 20 + 4 = 24 \] Therefore, the number of zeros at the end of \( 100! \) is \( 24 \).
\[ \boxed{24} \]
Directions: In Question Numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option from the following:
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): For any two prime numbers $p$ and $q$, their HCF is 1 and LCM is $p + q$.
Reason (R): For any two natural numbers, HCF × LCM = product of numbers.