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#1 2008-09-17 05:31:04
- tony123
- Member
- Registered: 2007-08-03
- Posts: 230
divisible by 2008
For any
, show that
is always divisible by 2008
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#2 2008-09-17 06:26:58
- Kurre
- Member
- Registered: 2006-07-18
- Posts: 280
Re: divisible by 2008
This is easy proved by bruteforce using congruences.
2008=8*251.
5+3=8, divisible by 8.
46 and 1098 ar both divisible by 8 for n>0 (since they are even), for n=0, their sum is 1144=8*143. Thus the whole expression is divisible by 8.
now lets look at 251:
Thus the whole expression is divisible by 251, and since its divisible by 8 its also divisible by 8*251=2008. q.e.d.
(I can admit I used the computer to find the congruences for the larger numbers 
)
Last edited by Kurre (2008-09-17 06:31:45)
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#3 2008-09-17 07:20:39
- mathsyperson
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- Registered: 2005-06-22
- Posts: 4,900
Re: divisible by 2008
Nicely done!
Why did the vector cross the road?
It wanted to be normal.
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#4 2008-09-17 11:17:14
- MathsIsFun
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- Registered: 2005-01-21
- Posts: 7,713
Re: divisible by 2008
And quickly, too.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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