Dhaka Secondary 2009/10
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Find three consecutive odd whole numbers such that the sum of their squares is a four digit number whose digits are all the same.
Re: Dhaka Secondary 2009/10
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 $-41,-43,-45$
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 $-41,-43,-45$
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