Question

The product 1 X 2 X 3 X 4 X ... X 2011 X 2012 = (18^a X b )where a and b are whole numbers. What is the largest value of a?

Answer 1

I think I'll make the above question clearer:
The product
\[1\times2\times3\times4\times...2011\times2012=18^{a} \times b \]
where a and b are whole numbers. What is the largest value of a?

Answer 2

There are \[\left \lfloor {2012 \over 18} \right \rfloor =111\] multiples of 18 less than 200. Taking into account 2*9 and 3*6, I'd say that a can be at most 113.

Answer 3

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

Answer 4

does this take into account multiples of 3 and 2 that arent multiples of 18, but together produce another 18 ?

Answer 5

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

Answer 6

\[
18^a = 2^a 3^a 3^a
\]
So a must be lowest power of 2 an 3 in decomposition of 2012.
If you count the power of 3 in 2012, you find 1001.

Answer 7

but floor of 6/18 = 0