Correct option is C
Given:
We need to determine how many trailing zeros are in the product:
1 × 2 × 3 × ... × 51
which is 51! (51 factorial).
Concept Used:
The number of trailing zeros in N! (factorial of N) is determined by the number of times 10 appears as a factor.
Since 10 = 2 × 5, and there are always more factors of 2 than 5, the number of trailing zeros is determined by the number of times 5 appears as a factor.
The formula to count trailing zeros in N! is:
Zeros = [N/5] + [N/25] + [N/125] + ...
where [x] represents the integer division (floor function).
Solution:
For 51!, we calculate the number of factors of 5:
[51/5] = 10 (Numbers: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50)
[51/25] = 2 (Numbers: 25, 50)
[51/125] = 0 (Since 125 > 51, this term is 0)
Total trailing zeros:
10 + 2 + 0 = 12
Final Answer:
(C) 12
