Question

A natural number \(n\) such that \(n!\) ends in exactly \(1000\) zeros is

• A. \(4010\)
• B. \(4000\)
• C. \(4009\)
• D. \(4004\)

Hint:

The number of trailing zeros in \(n!\) is found by counting factors of \(5\): \[ \left\lfloor \frac{n}{5}\right\rfloor+ \left\lfloor \frac{n}{25}\right\rfloor+ \left\lfloor \frac{n}{125}\right\rfloor+\cdots \]

Answer 1

The Correct Option is C

Step 1: Recall the formula for trailing zeros in \(n!\).
The number of trailing zeros in \(n!\) is determined by the number of factors of \(10\).
Since \[ 10=2\times5 \] and factors of \(2\) are more than factors of \(5\), we only count the number of factors of \(5\).
So, the number of trailing zeros in \(n!\) is \[ \left\lfloor \frac{n}{5}\right\rfloor+ \left\lfloor \frac{n}{25}\right\rfloor+ \left\lfloor \frac{n}{125}\right\rfloor+ \left\lfloor \frac{n}{625}\right\rfloor+ \cdots \]

Step 2: Check \(n=4009\).
\[ \left\lfloor \frac{4009}{5}\right\rfloor=801 \] \[ \left\lfloor \frac{4009}{25}\right\rfloor=160 \] \[ \left\lfloor \frac{4009}{125}\right\rfloor=32 \] \[ \left\lfloor \frac{4009}{625}\right\rfloor=6 \] \[ \left\lfloor \frac{4009}{3125}\right\rfloor=1 \] Now adding, \[ 801+160+32+6+1=1000 \] Thus, \(4009!\) ends in exactly \(1000\) zeros.

Step 3: Verify that it is exact.
For \(n=4010\), \[ \left\lfloor \frac{4010}{5}\right\rfloor=802 \] This would increase the number of zeros to more than \(1000\).
So, \(4010!\) does not have exactly \(1000\) zeros.
Hence, among the given options, \[ n=4009 \]

Step 4: Final conclusion.
Therefore, \[ \boxed{4009} \] which corresponds to option (3).