Question

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

• A. 240
• B. 248
• C. 256
• D. 253

Answer 1

The Correct Option is D

To find the number of trailing zeros in \(1024!\), we need to determine how many times 10 is a factor in the number. Since 10 factors arise from pairs of 2 and 5, and there are always more 2s than 5s, we only need to count the number of 5s. Use the formula:

\[\text{Number of } 5\text{s} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \left\lfloor \frac{n}{625} \right\rfloor\]

Given \(n = 1024\):

\[\left\lfloor \frac{1024}{5} \right\rfloor = 204\]

\[\left\lfloor \frac{1024}{25} \right\rfloor = 40\]

\[\left\lfloor \frac{1024}{125} \right\rfloor = 8\]

\[\left\lfloor \frac{1024}{625} \right\rfloor = 1\]

Adding these values gives the total number of 5s as factors:

\[204 + 40 + 8 + 1 = 253\]

Therefore, there are 253 trailing zeros in \(1024!\).