Problems

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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always 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.
tushar7
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Problems

Unread post by tushar7 » Sat Dec 11, 2010 12:57 am

Since no has not posted any problems ...............i am giving some

$1.$ what is the unit digit of $13^{2010}$ ?
$2.$ $m$ points are taken in each sides of a ragular $n$ gon .what is the total number of straight lines that can be drawn using all those lines ?

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Zzzz
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Re: Problems

Unread post by Zzzz » Sat Dec 11, 2010 5:25 am

What means unit digit?
Every logical solution to a problem has its own beauty.
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HandaramTheGreat
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Re: Problems

Unread post by HandaramTheGreat » Sat Dec 11, 2010 8:36 am

Zzzz wrote:What means unit digit?
একক স্থানীয় অংক

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rakeen
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Re: Problems

Unread post by rakeen » Sat Dec 11, 2010 4:07 pm

1. It should be 9!
2. I know thatthis prob has to be solved by combinatorix. But ami ekhono combinatorix sikhi nai, so....
r@k€€/|/

tushar7
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Re: Problems

Unread post by tushar7 » Sat Dec 11, 2010 5:33 pm

rakeen wrote:1. It should be 9!
2. I know thatthis prob has to be solved by combinatorix. But ami ekhono combinatorix sikhi nai, so....
its better to post the logic behind your solution

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Zzzz
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Re: Problems

Unread post by Zzzz » Sat Dec 11, 2010 7:22 pm

Probably Rakeen will give solution of problem 1.

Here is solution to problem 2:
If we choose any two points that are not on same side of the $n\ gon$ then we will find a straight line.
Lets number the sides of $n\ gon$ with $1,2,3,...,n$.
Start with side $1$. For each of the $m$ points of this side, we have $(n-1)m$ points. So in total $(n-1)m^2$ lines.
For each point of side #$2$, we have $(n-2)m$ points. So $(n-2)m^2$ lines.
Thus, total number of line is\[(n-1)m^2+(n-2)m^2+(n-3)m^2+...+1\cdot m^2\] \[=\frac{n(n-1)}{2}\cdot m^2\]
Every logical solution to a problem has its own beauty.
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AntiviruShahriar
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Re: Problems

Unread post by AntiviruShahriar » Mon Dec 13, 2010 12:25 pm

1.ans is 9
logic 13^{fi(10)}_=1(mod10)
13^4 _=1(mod10)
13^2008 _=1(mod10)
13^2008+2_=13^2_=9(mod10)
ঠিক আছে?
2.solution টাও বুঝি নাই।কারণ আমি combinatorics অতো ভাল বুঝি না। :) :) :) :) :) :) :) :) :)
এখানে _= হচ্ছে is congruent to

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Moon
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Re: Problems

Unread post by Moon » Mon Dec 13, 2010 12:58 pm

শাহরিয়ার, ফোরামে স্বাগতম। :)
আশা করি লক্ষ্য করেছ যে এই ফোরামে সমীকরণ লেখার চমৎকার সুবিধ আছে। তুমিও ইচ্ছা করলে কিভাবে LaTeX দিয়ে লিখতে হয় এই টপিকটা পড়ে সমীকরণ লিখতে পার। :)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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rakeen
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Re: Problems

Unread post by rakeen » Mon Dec 13, 2010 1:15 pm

Ami congurance ekhono sikhi nai. Ei jonno ami ebhabe ber korsi:

$3^1$ er ekok sthanio onko 3
$3^2$ er ekok sthanio onko 9
$3^3$ er ekok sthanio onko 7
$3^4$ er ekok sthanio onko 1

and 2010 is divided by 2, not by 4. So the answer is 9!
r@k€€/|/

AntiviruShahriar
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Re: Problems

Unread post by AntiviruShahriar » Mon Dec 13, 2010 1:38 pm

ভাই exclamatory sign use কইরো না প্লিজ।এইটায় ৩৬২৮৮০ বুঝায়।@রাকীন।

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