## pythagoras123 Group Title 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? 2 years ago 2 years ago

1. pythagoras123 Group Title

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 Group Title

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 Group Title

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

4. eigenschmeigen Group Title

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

5. eigenschmeigen Group Title

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

6. eliassaab Group Title

$18^a = 2^a 3^a 3^a$ 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 Group Title

but floor of 6/18 = 0

8. integralsabiti Group Title

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9. eigenschmeigen Group Title

yes

10. eigenschmeigen Group Title

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

11. blockcolder Group Title

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

12. eliassaab Group Title

Yes, there are 2004 two's

13. integralsabiti Group Title

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

14. integralsabiti Group Title

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

15. blockcolder Group Title

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

16. integralsabiti Group Title

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

17. experimentX Group Title

how are there 1001 three's in 2011 ??

18. eigenschmeigen Group Title

sorry i was away, i'll try to expain

19. integralsabiti Group Title

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

20. experimentX Group Title

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 Group Title

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

22. eigenschmeigen Group Title

you see?

23. eigenschmeigen Group Title

because we have multiples of 9 and multiples of 3

24. experimentX Group Title

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

25. eigenschmeigen Group Title

26. blockcolder Group Title

27. nikvist Group Title

$a_{\max}=500$

28. eigenschmeigen Group Title

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 Group Title

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

30. eliassaab Group Title

Yes, it is 500.