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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?

  • one year ago
  • one year ago

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  1. pythagoras123
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    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
  2. blockcolder
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    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
  3. blockcolder
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    18^2 contributes an additional 18, so that becomes 114.

    • one year ago
  4. eigenschmeigen
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    does this take into account multiples of 3 and 2 that arent multiples of 18, but together produce another 18 ?

    • one year ago
  5. eigenschmeigen
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    eg if we use 6 instead of 2012 we have 1*2*3*4*5*6 = 18*40

    • one year ago
  6. eliassaab
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    \[ 18^a = 2^a 3^a 3^a \] 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
  7. eigenschmeigen
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    but floor of 6/18 = 0

    • one year ago
  8. integralsabiti
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    |dw:1335011473065:dw||dw:1335011596378:dw|

    • one year ago
  9. eigenschmeigen
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    yes

    • one year ago
  10. eigenschmeigen
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    and there are obviously many more 2's so we dont have to worry about them

    • one year ago
  11. blockcolder
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    Wait. I don't get @integralsabiti 's solution.

    • one year ago
  12. eliassaab
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    Yes, there are 2004 two's

    • one year ago
  13. integralsabiti
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    and i am not able to explain! sorry @eigenschmeigen help pls (:

    • one year ago
  14. integralsabiti
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    there are 1001 3s in 2011! the exponent of 3 is 2a so i divided 1001 by two to find a.

    • one year ago
  15. blockcolder
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    Can't you just count the number of 2's in the product?

    • one year ago
  16. integralsabiti
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    and there are obviously many more 2's so we dont have to worry about them as @eigenschmeigen say

    • one year ago
  17. experimentX
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    how are there 1001 three's in 2011 ??

    • one year ago
  18. eigenschmeigen
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    sorry i was away, i'll try to expain

    • one year ago
  19. integralsabiti
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    @experimentX in 2011!=1.2.3.4.5.6.....2011

    • one year ago
  20. experimentX
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    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
  21. eigenschmeigen
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    also multiples of 9 we have to add on, multiples of 27 etc

    • one year ago
  22. eigenschmeigen
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    you see?

    • one year ago
  23. eigenschmeigen
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    because we have multiples of 9 and multiples of 3

    • one year ago
  24. experimentX
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    Oh ... i get it ... I completely neglected that factor.

    • one year ago
  25. eigenschmeigen
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    is @pythagoras123 understanding the answer?

    • one year ago
  26. blockcolder
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    How about for 2012! ??

    • one year ago
  27. nikvist
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    \[a_{\max}=500\]

    • one year ago
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  28. eigenschmeigen
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    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
  29. blockcolder
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    Oh..... I get it! Thanks! :D

    • one year ago
  30. eliassaab
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    Yes, it is 500.

    • one year ago
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