9514 1404 393
Answer:
N = 250
Explanation:
The way your question is written, the answer is a number of 2582 digits, much too large to show here. We assume you actually want N such that 10^N is the largest power of 10 that will divide 1005!.
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That factor is limited by the number of factors of 2 and 5 in 1005!. Of those, there are far more factors of 2, so the actual limit is the number of factors of 5 in 1005!.
Every 5th number is divisible by 5, so there are 1005/5 = 201 numbers divisible by 5 in 1005!.
Every 5th one of those is divisible by another factor of 5, so an additional 201/5 ≈ 40.
Every 5th one of those is divisible by another factor of 5, so an additional 40/5 = 8.
Finally, every 5th one of those has yet another factor of 5, so an additional 8/5 ≈ 1.
Then the total number of factors of 5 in 1005! is 201 +40 +8 + 1 = 250.
That is, the number 1005! will end in 250 zeros. It will be divisible by 10^250.
N = 250