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18^2 contributes an additional 18, so that becomes 114.

eg if we use 6 instead of 2012 we have 1*2*3*4*5*6 = 18*40

but floor of 6/18 = 0

|dw:1335011473065:dw||dw:1335011596378:dw|

yes

and there are obviously many more 2's so we dont have to worry about them

Wait. I don't get @integralsabiti 's solution.

Yes, there are 2004 two's

and i am not able to explain! sorry @eigenschmeigen help pls (:

there are 1001 3s in 2011! the exponent of 3 is 2a so i divided 1001 by two to find a.

Can't you just count the number of 2's in the product?

and there are obviously many more 2's so we dont have to worry about them
as @eigenschmeigen say

how are there 1001 three's in 2011 ??

sorry i was away, i'll try to expain

@experimentX in 2011!=1.2.3.4.5.6.....2011

also multiples of 9 we have to add on, multiples of 27 etc

you see?

because we have multiples of 9 and multiples of 3

Oh ... i get it ... I completely neglected that factor.

is @pythagoras123 understanding the answer?

How about for 2012! ??

\[a_{\max}=500\]

Oh..... I get it! Thanks! :D

Yes, it is 500.