Search found 75 matches
- Sun Mar 02, 2014 12:08 pm
- Forum: Junior Level
- Topic: RMO-2010/3
- Replies: 9
- Views: 10711
RMO-2010/3
Find the number of $$4 \text{ digit numbers (in base 10)}$$ having non-zero digits and which are divisible by $$4$$ but not by $$8$$.
- Sun Mar 02, 2014 11:57 am
- Forum: Secondary Level
- Topic: Congruent Rectangles
- Replies: 1
- Views: 2575
Congruent Rectangles
A $$6 \times 6 \text{ square}$$ is dissected into $$9 \text{ rectangles}$$ by lines parallel to its sides
such that all these rectangles have only integer sides. Prove that $$ \text{ there
are always two congruent rectangles}$$.
such that all these rectangles have only integer sides. Prove that $$ \text{ there
are always two congruent rectangles}$$.
- Tue Feb 25, 2014 5:06 pm
- Forum: Secondary Level
- Topic: BDIOI -2013/7
- Replies: 1
- Views: 2820
BDIOI -2013/7
Find of the value of $$F(n,k)$$
$$F(n,k) = 2F(n-1,k) + F(n-1,k-1)$$ where, $$F(0,k) = 1$$, $$F(n,0) = 1$$
$$F(n,k) = 2F(n-1,k) + F(n-1,k-1)$$ where, $$F(0,k) = 1$$, $$F(n,0) = 1$$
- Wed Feb 19, 2014 11:50 am
- Forum: Secondary Level
- Topic: Even + Odd = Odd
- Replies: 3
- Views: 3712
Even + Odd = Odd
Let n be an odd integer greater than $$1$$. Prove that the sequence
$$ \binom{n}{1}, \binom{n}{2},...., \binom{n}{\frac{n-1}{2}}$$
contains an odd number of odd numbers.
$$ \binom{n}{1}, \binom{n}{2},...., \binom{n}{\frac{n-1}{2}}$$
contains an odd number of odd numbers.
- Tue Feb 18, 2014 10:16 pm
- Forum: Combinatorics
- Topic: IMO - 1964 - 4
- Replies: 2
- Views: 3062
Re: IMO - 1964 - 4
First mark 3 languages as p,q,r and a person as A. এখন A কমপক্ষে ৬ জনকে একই বিষয়ে চিঠি লেখে ,ধরি এটি p । এই ৬ জনের মধ্যে যদি ২ জন p বিষয়ে চিঠি লেখে তাহলে হয়েই গেল। যদি তা না হয়, তাহলে ৬ জনের মধ্যে আরেকজন B নিই। বাকী ৫ জনের মধ্যে B কমপক্ষে ৩ জনকে একই বিষয়ে চিঠি লেখে, ধরি এটি q. এই ৩ জনের মধ্যে ২...
- Tue Feb 18, 2014 6:02 pm
- Forum: Secondary Level
- Topic: Altitudes
- Replies: 2
- Views: 2948
Re: Altitudes
My solution was a bit small
$$\angle APQ=\angle C=\angle BFD=180^{\circ}-\angle AFQ$$.So $$APQF$$ cyclic.
$$\angle AQP=\angle AFP= \angle C$$
$$\angle APQ= \angle AQP$$ $$=>AQ=AP$$
$$\angle APQ=\angle C=\angle BFD=180^{\circ}-\angle AFQ$$.So $$APQF$$ cyclic.
$$\angle AQP=\angle AFP= \angle C$$
$$\angle APQ= \angle AQP$$ $$=>AQ=AP$$
- Tue Feb 18, 2014 5:55 pm
- Forum: Secondary Level
- Topic: An easy problem !!
- Replies: 9
- Views: 6938
Re: An easy problem !!
Edited
- Tue Feb 18, 2014 5:39 pm
- Forum: Secondary Level
- Topic: An easy problem !!
- Replies: 9
- Views: 6938
Re: An easy problem !!
$$N=2m$$. $$N(N^2+20)=8m(m^2+5)$$. (1.1)If $$m \equiv 0 \pmod2$$ and $$m \equiv 0 \pmod3 $$ then $$8m(m^2+5)$$ is divisible by 48. (1.2)If $$m \equiv 0 \pmod2$$ and $$m \equiv 1,2 \pmod3 $$ then $$m^2+5 \equiv 0 \pmod 3$$. So, $$8m(m^2+5)$$ is divisible by 48. (2.1) If $$m \equiv1 \pmod2$$ and $$m \...
- Tue Feb 18, 2014 5:14 pm
- Forum: Secondary Level
- Topic: An easy problem !!
- Replies: 9
- Views: 6938
Re: An easy problem !!
I think the question is wrong. It would be find all $$N$$ such that: $$N$$ একটি জোড় সংখ্যা ,$$68 | N(N^2 + 20)$$.
Now,
$$N=2m$$.
$$N(N^2+20)=4m(2m^2+10)$$.
So,$$ 17|m$$.
$$m=17k$$
$$N=34k$$
Now,
$$N=2m$$.
$$N(N^2+20)=4m(2m^2+10)$$.
So,$$ 17|m$$.
$$m=17k$$
$$N=34k$$
- Tue Feb 18, 2014 10:27 am
- Forum: Secondary Level
- Topic: Altitudes
- Replies: 2
- Views: 2948
Altitudes
Let $$ABC$$ be an acute angle triangle with $$D,E,F$$ the feet of altitude line of $$BC,CA,AB$$ respectively. One of the intersection points of the EF and the circumcircle is P. BP and DF meet at point Q. Show that, $$AP=AQ$$