## pythagoras123 3 years 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?

1. 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?

2. 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.

3. blockcolder

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

4. eigenschmeigen

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

5. eigenschmeigen

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

6. eliassaab

$18^a = 2^a 3^a 3^a$ 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.

7. eigenschmeigen

but floor of 6/18 = 0

8. integralsabiti

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9. eigenschmeigen

yes

10. eigenschmeigen

and there are obviously many more 2's so we dont have to worry about them

11. blockcolder

Wait. I don't get @integralsabiti 's solution.

12. eliassaab

Yes, there are 2004 two's

13. integralsabiti

and i am not able to explain! sorry @eigenschmeigen help pls (:

14. integralsabiti

there are 1001 3s in 2011! the exponent of 3 is 2a so i divided 1001 by two to find a.

15. blockcolder

Can't you just count the number of 2's in the product?

16. integralsabiti

and there are obviously many more 2's so we dont have to worry about them as @eigenschmeigen say

17. experimentX

how are there 1001 three's in 2011 ??

18. eigenschmeigen

sorry i was away, i'll try to expain

19. integralsabiti

@experimentX in 2011!=1.2.3.4.5.6.....2011

20. 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

21. eigenschmeigen

also multiples of 9 we have to add on, multiples of 27 etc

22. eigenschmeigen

you see?

23. eigenschmeigen

because we have multiples of 9 and multiples of 3

24. experimentX

Oh ... i get it ... I completely neglected that factor.

25. eigenschmeigen

26. blockcolder

27. nikvist

$a_{\max}=500$

28. eigenschmeigen

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

29. blockcolder

Oh..... I get it! Thanks! :D

30. eliassaab

Yes, it is 500.