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: Count how many numbers from 1 to 1024 are multiples of 5.
Every multiple of 5 contributes at least one factor of 5. There are $\lfloor 1024/5 \rfloor = 204$ such multiples, so that gives 204 factors of 5 so far.

Step 2: Count numbers that contribute an extra factor of 5 beyond the first.
Multiples of 25 (like 25, 50, 75, ...) contribute one more factor of 5 each, beyond the one already counted in Step 1. There are $\lfloor 1024/25 \rfloor = 40$ multiples of 25 up to 1024, adding 40 more factors of 5.

Step 3: Count numbers that contribute yet another extra factor of 5.
Multiples of 125 add one more factor of 5 each beyond what's already counted: there are $\lfloor 1024/125 \rfloor = 8$ of these, adding 8 more. Multiples of 625 add one more still: there is $\lfloor 1024/625 \rfloor = 1$ such multiple (just 625 itself), adding 1 more. Since $5^5 = 3125 > 1024$, no number contributes a fifth extra factor.

Step 4: Add all the factors of 5 together.
Total factors of 5 in $1024! = 204 + 40 + 8 + 1 = 253$. Since $1024!$ has far more factors of 2 than of 5, the number of trailing zeros is limited by the factors of 5, so it equals 253.

Final Answer:
There are 253 trailing zeros in $1024!$. \[ \boxed{253} \]