Question

How many zeros would be there in \(1024!\)

• A. 240
• B. 248
• C. 256
• D. 253

Hint:

Count the trailing zeros using the highest power of 5 dividing 1024 factorial, since factors of 2 are always more plentiful.

Answer 1

The Correct Option is D

Step 1: Recall why factorials end in zeros.
A trailing zero comes from a factor of 10, which is \(2 \times 5\). In any factorial there are always many more factors of 2 than factors of 5, so the number of trailing zeros in \(n!\) equals the highest power of 5 that divides \(n!\).

Step 2: Use the standard formula for the highest power of 5.
Highest power of 5 in \(1024!\) \(= \left\lfloor \frac{1024}{5} \right\rfloor + \left\lfloor \frac{1024}{25} \right\rfloor + \left\lfloor \frac{1024}{125} \right\rfloor + \left\lfloor \frac{1024}{625} \right\rfloor\), where \(\lfloor \cdot \rfloor\) means take only the integer part and drop the decimal part.

Step 3: Work out each term.
\(\left\lfloor \frac{1024}{5} \right\rfloor = 204\), \(\left\lfloor \frac{1024}{25} \right\rfloor = 40\), \(\left\lfloor \frac{1024}{125} \right\rfloor = 8\), \(\left\lfloor \frac{1024}{625} \right\rfloor = 1\). The next power, \(5^5 = 3125\), is bigger than 1024, so we stop here.

Step 4: Add up the terms.
Total \(= 204 + 40 + 8 + 1 = 253\).

Final Answer:
\(1024!\) has 253 trailing zeros. \[ \boxed{253} \]