Moto Hell - The Motorola Modding Community

General => Off topic => Topic started by: Exploited on October 08, 2007, 11:49:08 am



Title: 2007
Post by: Exploited on October 08, 2007, 11:49:08 am
You must solve that problem until the end of the year, because it is related to 2007 :)

The sum of five numbers is 2007: A + B + C + D + E = 2007

The sum of three of the numbers is 1000: A + B + C = 1000

Can you proof that the multiply of the numbers can be divided to 4 without residual? (eg you can split A.B.C.D.E on 4 equal parts with no remainder)?


Title: Re: 2007
Post by: motomaniacs on October 08, 2007, 02:47:27 pm
damn very hard to solve....


Title: Re: 2007
Post by: Joka X on October 08, 2007, 10:43:44 pm
i don't think i can solve it.......i'm bad at math....


Title: Re: 2007
Post by: error398 on October 08, 2007, 11:52:17 pm
thinking..


Title: Re: 2007
Post by: 3sixthy on October 09, 2007, 12:10:49 am
mmm thinkin' 2 lol


Title: Re: 2007
Post by: motomaniacs on October 10, 2007, 08:31:27 pm
after take a lot of time thinking
the answer is I GAVE UP :P


Title: Re: 2007
Post by: keaglez on October 11, 2007, 01:35:25 am
Im not bad in math at school, but this is........too................... :(


Title: Re: 2007
Post by: Exploited on October 11, 2007, 10:21:19 am
Solution:

First let we say what A, B and C cannot be - they cannot be all odd (beause the sum is even) and they cannot be two even + one odd (again for the same reason).

The variants left are: all even or two odd and one even

1. If all of them are even - it is obvious that their multiplication can be divided on 4:
A*B*C*D*E/4 = (A/2)*(B/2)*C*D*E => because both A and B are even - they can be divided on 2 without remainder => 1. is  Prooved

2. Let we check the variant if A and B are odd and C is even. Let's see what is happening with D and E:
A+B+C+D+E = 2007 => D+E = 2007-(A+B+C) => D+E = 1007, which means that one of (D,E) is odd and one is even. => from the five numbers we have one even from (A,B,C) and one even from (D,E) => we have total 3 odd and 2 even - again like one we have (A*B*C/2)*(D*E/2) =>... prooved!

1. + 2. = PROOVED :)


Title: Re: 2007
Post by: Jithin on October 11, 2007, 08:31:47 pm
Thanks for that answer Phil... I need to improve :)


Title: Re: 2007
Post by: huatz84 on October 21, 2007, 06:18:13 am
Wow,It's so hard for me...I'm bad in math.I always got zero at school..:D


Title: Re: 2007
Post by: Exploited on October 21, 2007, 08:14:29 am
zero?


Title: Re: 2007
Post by: Jithin on October 21, 2007, 08:15:46 am
zero?
ROFL ;D


Title: Re: 2007
Post by: huatz84 on October 21, 2007, 09:31:02 am
Zero means the lowest score like the shape of egg..:D


Title: Re: 2007
Post by: keaglez on October 22, 2007, 10:55:48 pm
Ow...how can you finnished your school then? ???

Haha... Just kiddin... ;D