Here's the question you clicked on:
pythagoras123
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?
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?
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.
18^2 contributes an additional 18, so that becomes 114.
does this take into account multiples of 3 and 2 that arent multiples of 18, but together produce another 18 ?
eg if we use 6 instead of 2012 we have 1*2*3*4*5*6 = 18*40
\[ 18^a = 2^a 3^a 3^a \] 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.
but floor of 6/18 = 0
|dw:1335011473065:dw||dw:1335011596378:dw|
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
yeah ,,, but you would get multiple of 3 in every three element. I thought there would be 2011/3 three's and if you pair them up 2011/6
also multiples of 9 we have to add on, multiples of 27 etc
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! ??
so we find all the powers of three that are less than 2012: 3,9,27,81,243,729 now we do floor(2012/729) + floor(2012/243) + ... +floor(2012/3) and then divide by two because there are two threes in the PF of 18 then we floor it again and get 500
Oh..... I get it! Thanks! :D