Search found 65 matches
- Tue Feb 07, 2017 5:04 pm
- Forum: Divisional Math Olympiad
- Topic: BDMO REGIONAL 2015
- Replies: 4
- Views: 16815
Re: BDMO REGIONAL 2015
We do know either $p$ or $q$ is $5$.So if we take the square of both,we can ensure.Thus,The answer is $4$.
- Tue Feb 07, 2017 5:00 pm
- Forum: Divisional Math Olympiad
- Topic: BDMO REGIONAL 2015
- Replies: 5
- Views: 5232
Re: BDMO REGIONAL 2015
I think the statement should be "There exists a pair such
that the sum is $126$"
that the sum is $126$"
- Tue Feb 07, 2017 3:52 pm
- Forum: Divisional Math Olympiad
- Topic: BDMO REGIONAL 2015
- Replies: 3
- Views: 3975
Re: BDMO REGIONAL 2015
Let's draw a line such that it goes through $E$ and perpendicular to $AB$ and $CD$.It intersects $AB$ at $F$ and $CD$ at $G$.We also can say that $BF = CG , AF = DG$. We can form four equations: $DE^2 = EG^2 + DG^2...(i)$ $CE^2 = EG^2 + CG^2...(ii)$ $AE^2 = EF^2 + AF^2...(iii)$ $BE^2 = EF^2 + BF^2.....
- Mon Feb 06, 2017 11:01 pm
- Forum: Divisional Math Olympiad
- Topic: BDMO REGIONAL 2015
- Replies: 5
- Views: 5232
Re: BDMO REGIONAL 2015
Solution:
Note that there exists a number for each number such that the sum is $126$,except $2$.Thus if we take
$14$ number,we can ensure that the sum of any two number is $126$.
Note that there exists a number for each number such that the sum is $126$,except $2$.Thus if we take
$14$ number,we can ensure that the sum of any two number is $126$.
- Mon Feb 06, 2017 10:54 pm
- Forum: Geometry
- Topic: BDMO regional 2015
- Replies: 4
- Views: 4535
Re: BDMO regional 2015
Problem 10:
viewtopic.php?f=19&t=3844
viewtopic.php?f=19&t=3844
- Sun Feb 05, 2017 10:12 pm
- Forum: Junior Level
- Topic: Find $x$
- Replies: 4
- Views: 3828
Re: Find $x$
I missed the point.Thanks to dhasan.
Correct solution:
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
If we check the factors of $6$,we will find
that the solution is $x \equiv 7$.
Correct solution:
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
If we check the factors of $6$,we will find
that the solution is $x \equiv 7$.
- Sat Feb 04, 2017 10:47 pm
- Forum: Junior Level
- Topic: Find $x$
- Replies: 4
- Views: 3828
Re: Find $x$
Check my solution:
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
But if we check the factors of $6$,we will find
that there is no solution of $x$.
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
But if we check the factors of $6$,we will find
that there is no solution of $x$.
- Sat Feb 04, 2017 10:39 pm
- Forum: Junior Level
- Topic: Find $x$
- Replies: 4
- Views: 3828
Find $x$
Find $x$ such that $x^2 - 3x + 7 \equiv 2 (mod 11)$
- Sat Feb 04, 2017 6:57 pm
- Forum: Geometry
- Topic: Circle is tangent to circumcircle and incircle
- Replies: 3
- Views: 3872
Re: Circle is tangent to circumcircle and incircle
Will bisector $\angle BAC$ intersect both $DE$ and $DF$?
- Tue Jan 31, 2017 5:56 pm
- Forum: Secondary Level
- Topic: Find (m , n)
- Replies: 1
- Views: 2294
Re: Find (m , n)
My solution : The equation can be re-written lilke this : $(m+1)(m-1) = 5.2^n$ Notice that either $(m+1)$ or $(m-1)$ can not be $5$.Because if $(m+1)$ or $(m-1)$ is $5$ then the other will also be an odd number.Thus,the L.H.S is odd.But the R.H.S. is even and so it's impossible. So, we can say that ...