Let n be a natural number. The number of consecutive zeros at the end of the expansion of n! is <strong>EXACTLY</strong> 2. How many values of n are possible?
- A. 3
- B. 4
- C. 5 ✓
- D. More than 5
Correct Answer: C. 5
Explanation
The number of trailing zeros in n! is determined by \lfloor \frac{n}{5} \rfloor + \lfloor \frac{n}{25} \rfloor + \dots. For exactly 2 trailing zeros, \lfloor \frac{n}{5} \rfloor = 2. This is satisfied by integers n = 10, 11, 12, 13, 14. Thus, 5 values of n are possible.
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