Dhaka Secondary 2009/10

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Secondary 2009/10

Unread post by BdMO » Tue Feb 01, 2011 11:50 pm

Find three consecutive odd whole numbers such that the sum of their squares is a four digit number whose digits are all the same.

photon
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Re: Dhaka Secondary 2009/10

Unread post by photon » Sun Mar 27, 2011 3:29 pm

let the odds are $2n-1,2n+1,2n-3$
now,
$(2n-1)^2+(2n+1)^2+(2n-3)^2=1111x$ [x is odd natural digit as sum of 3 odd squares is odd]
or,$12n(n-1)+11=11.101x$
or,$12n(n-1)=11(101x-1)$...............(1)
so,24 divides 12n(n-1)
so 24 divides 101x-1
let,$24k=101x-1$[k belongs to N]
or,$24k+1=101x$
of course,x is not divisible by 3.x can be 1,5,7
now either x is in $3m+1$ or in $3m-1$ form.
as $101$ is in $3m-1$ form, and dividing $101x$ by 3 remainder is 1,$x$ is in $3m-1$ form.
because $(3m-1)(3m-1)=3m(m-1)+1$
so,x is 5=3.2-1
from 1,we can get,n=22,-21
$41^2+43^2+45^2=5555$
also approve for $-41,-43,-45$
:)
Try not to become a man of success but rather to become a man of value.-Albert Einstein

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