Question

How many continious zeros are there at the end of the expansion of 69!? (N! is the product of all the positive integers from N to 1)

• A. 15
• B. 16
• C. 17
• D. 18

Answer 1

The Correct Option is A

Finding the Number of Trailing Zeros in \( N! \):

The number of trailing zeros in \( N! \) is given by the formula:

\[ \sum_{i=1}^{\infty} \left\lfloor \frac{N}{5^i} \right\rfloor \] 

Calculation for \( 69! \):

• First term: \[ \left\lfloor \frac{69}{5} \right\rfloor = 13 \]
• Second term: \[ \left\lfloor \frac{69}{25} \right\rfloor = 2 \]
• Third term: \[ \left\lfloor \frac{69}{125} \right\rfloor = 0 \]

Total Trailing Zeros:

\[ 13 + 2 + 0 = 15 \]

Final Answer:

Thus, the correct answer is 15 (Option A).