How would I solve a problem like for 1005! is divisible by 10 to the nth power, what is the highest value of n? I don't know how to do it without figuring out the entire 1005!.

If the question is what is the largest single factor in 1005! then 10^3.
If the spirit is what is the largest n possible it may help to try somethings with a smaller number like 5!.
5! =120 divisible by 10^1. The 2*5=10.
Similarliy 10! has 2x5 and 10 as factors within it so 10^2.

I'll probably miss some factors for 1005!
Note that you can only use each factor 1 time

1,10,100,1000,2*5,20*50,200*500,4*25,40*250,8*125
Summing up the exponents n is at least 23.

A little bit more thinking since my last post. 10^n is now at least 10^28.
By using the same style of thinking but using the larger factors
1,10,100,1000
2,5 20,50, 200,500

40,25 (instead of 4 and 25)
400,250 (instead of 40 and 250)
800,125 (insetad of 8 or 80 and 125)
as in my previous answer.

3rd time a charm? Math is not my strongest.
I see I omitted many combinations of 2 and 5
such as 12*15, 22*25 etc..
There are 101 combinations
Also 20,30... (2x10,3*10) 100 of those.
I looked at the combination of 200,500 ..what of 600,400? 6*100,4*100
finally 1000 by itself.
N=1*101 +1*100+2*10+3*1=222.

It's hard to think in this size of exponent

Double counts in my last answer...& I can't add.

100 is found both in the 10's and 100s so it should not be counted in 10s as 10^1 and again as 10^2. but 10^1 + 10^1

N=101 (pairs of 2&5) +100 (tens) +10 (hundreds) +1(Thousands) =212 total

I think I'll quit thinking about this