pythagoras123
  • 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?
Mathematics
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SOLVED
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chestercat
  • chestercat
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pythagoras123
  • pythagoras123
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?
blockcolder
  • blockcolder
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.
blockcolder
  • blockcolder
18^2 contributes an additional 18, so that becomes 114.

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More answers

anonymous
  • anonymous
does this take into account multiples of 3 and 2 that arent multiples of 18, but together produce another 18 ?
anonymous
  • anonymous
eg if we use 6 instead of 2012 we have 1*2*3*4*5*6 = 18*40
anonymous
  • anonymous
\[ 18^a = 2^a 3^a 3^a \] 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.
anonymous
  • anonymous
but floor of 6/18 = 0
anonymous
  • anonymous
|dw:1335011473065:dw||dw:1335011596378:dw|
anonymous
  • anonymous
yes
anonymous
  • anonymous
and there are obviously many more 2's so we dont have to worry about them
blockcolder
  • blockcolder
Wait. I don't get @integralsabiti 's solution.
anonymous
  • anonymous
Yes, there are 2004 two's
anonymous
  • anonymous
and i am not able to explain! sorry @eigenschmeigen help pls (:
anonymous
  • anonymous
there are 1001 3s in 2011! the exponent of 3 is 2a so i divided 1001 by two to find a.
blockcolder
  • blockcolder
Can't you just count the number of 2's in the product?
anonymous
  • anonymous
and there are obviously many more 2's so we dont have to worry about them as @eigenschmeigen say
experimentX
  • experimentX
how are there 1001 three's in 2011 ??
anonymous
  • anonymous
sorry i was away, i'll try to expain
anonymous
  • anonymous
@experimentX in 2011!=1.2.3.4.5.6.....2011
experimentX
  • experimentX
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
anonymous
  • anonymous
also multiples of 9 we have to add on, multiples of 27 etc
anonymous
  • anonymous
you see?
anonymous
  • anonymous
because we have multiples of 9 and multiples of 3
experimentX
  • experimentX
Oh ... i get it ... I completely neglected that factor.
anonymous
  • anonymous
is @pythagoras123 understanding the answer?
blockcolder
  • blockcolder
How about for 2012! ??
nikvist
  • nikvist
\[a_{\max}=500\]
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anonymous
  • anonymous
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
blockcolder
  • blockcolder
Oh..... I get it! Thanks! :D
anonymous
  • anonymous
Yes, it is 500.

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