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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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pythagoras123Best ResponseYou've already chosen the best response.0
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?
 one year ago

blockcolderBest ResponseYou've already chosen the best response.1
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.
 one year ago

blockcolderBest ResponseYou've already chosen the best response.1
18^2 contributes an additional 18, so that becomes 114.
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
does this take into account multiples of 3 and 2 that arent multiples of 18, but together produce another 18 ?
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
eg if we use 6 instead of 2012 we have 1*2*3*4*5*6 = 18*40
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
\[ 18^a = 2^a 3^a 3^a \] 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.
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
but floor of 6/18 = 0
 one year ago

integralsabitiBest ResponseYou've already chosen the best response.1
dw:1335011473065:dwdw:1335011596378:dw
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
and there are obviously many more 2's so we dont have to worry about them
 one year ago

blockcolderBest ResponseYou've already chosen the best response.1
Wait. I don't get @integralsabiti 's solution.
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
Yes, there are 2004 two's
 one year ago

integralsabitiBest ResponseYou've already chosen the best response.1
and i am not able to explain! sorry @eigenschmeigen help pls (:
 one year ago

integralsabitiBest ResponseYou've already chosen the best response.1
there are 1001 3s in 2011! the exponent of 3 is 2a so i divided 1001 by two to find a.
 one year ago

blockcolderBest ResponseYou've already chosen the best response.1
Can't you just count the number of 2's in the product?
 one year ago

integralsabitiBest ResponseYou've already chosen the best response.1
and there are obviously many more 2's so we dont have to worry about them as @eigenschmeigen say
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
how are there 1001 three's in 2011 ??
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
sorry i was away, i'll try to expain
 one year ago

integralsabitiBest ResponseYou've already chosen the best response.1
@experimentX in 2011!=1.2.3.4.5.6.....2011
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
also multiples of 9 we have to add on, multiples of 27 etc
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
because we have multiples of 9 and multiples of 3
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Oh ... i get it ... I completely neglected that factor.
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
is @pythagoras123 understanding the answer?
 one year ago

blockcolderBest ResponseYou've already chosen the best response.1
How about for 2012! ??
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
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
 one year ago

blockcolderBest ResponseYou've already chosen the best response.1
Oh..... I get it! Thanks! :D
 one year ago
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