someone plz help me with this math problem-
Find three consecutive odd whole numbers such that sum of squares is a four digit number whose digits are all same.
Math help!!!
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- Posts:56
- Joined:Fri Feb 18, 2011 11:30 pm
Re: Math help!!!
the possible 4 digits numbers are 1111,3333,5555,7777,9999.(sum of 3 odd squares is an odd number)
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- Posts:56
- Joined:Fri Feb 18, 2011 11:30 pm
Re: Math help!!!
i find these three numbers.it's $(41)^2+(43^2)+(45)^2=5555$.
i find it in this method $(2n-1)^2+(2n+1)^2+(2n+3)^2=12(n^2+n+1)-1$. it means the $4$ digit number$+1$ is divisible by $12$ then i try $1112,3334,5556...$ etc.
i find it in this method $(2n-1)^2+(2n+1)^2+(2n+3)^2=12(n^2+n+1)-1$. it means the $4$ digit number$+1$ is divisible by $12$ then i try $1112,3334,5556...$ etc.
Re: Math help!!!
that's a BDMO prob 2009
http://www.matholympiad.org.bd/forum/vi ... f=41&t=419
http://www.matholympiad.org.bd/forum/vi ... f=41&t=419
Try not to become a man of success but rather to become a man of value.-Albert Einstein
Re: Math help!!!
I heard that tutorvista math offers online help for mathematics. Does anyone know the exact website? Thanx in advance.
I am really struggling with it.
I am really struggling with it.