এই খানে একদিক থেকে কাউন্ট করসস। উল্টা দিক থেকেও ত্রিভুজ গুনতে হবে।Tahmid wrote:$$\sum_{k=1}^{k=n}\frac{k(k+1)}{2}$$
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- Tue Feb 11, 2014 11:23 am
- Forum: Secondary Level
- Topic: Large = Large+big+small+little
- Replies: 5
- Views: 4534
Re: Large = Large+big+small+little
- Mon Feb 10, 2014 11:16 pm
- Forum: Secondary Level
- Topic: Large = Large+big+small+little
- Replies: 5
- Views: 4534
Large = Large+big+small+little
একটি সমবাহু ত্রিভুজের প্রতিটি বাহুর উপর $$n$$ টি সমদূরবর্তী বিন্দু আছে। এর ফলে প্রতিটি বাহু $$n+1$$ ভাগে বিভক্ত। এখন এই সমবাহু ত্রিভুজে সর্বমোট কতটি সমবাহু ত্রিভুজ বিদ্যমান?
- Mon Feb 10, 2014 10:16 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies: 25
- Views: 16075
Re: warm-up problems for national BdMO'14
Solution to 9.
$$p^2+2p-8=(p+1)^2-9$$
$$=(p+1-3)(p+1+3)=(p+4)(p-2)$$
but it is a prime.
So,
$$p-2=1$$
$$\boxed{p=3}$$.
Now if $$p=3$$ both $$p+2,p^2+2p-8$$ are prime.
$$p^2+2p-8=(p+1)^2-9$$
$$=(p+1-3)(p+1+3)=(p+4)(p-2)$$
but it is a prime.
So,
$$p-2=1$$
$$\boxed{p=3}$$.
Now if $$p=3$$ both $$p+2,p^2+2p-8$$ are prime.
- Mon Feb 10, 2014 5:48 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies: 25
- Views: 16075
Re: warm-up problems for national BdMO'14
Let $$x=\sqrt{1+8N}$$ and $$y=\sqrt{9+8N}$$ $$y^2-x^2=8$$ So, $$(y+x)=4,8$$ and $$(y-x)=2,1$$ Solving these we get $$(x,y)=(1,3)$$ Now, $$ \dfrac{1+\sqrt{1+8N}}{2}=\dfrac{1+x}{2}=\dfrac{2}{2}=1$$. $$ \dfrac{1+\sqrt{9+8N}}{2}=\dfrac{1+y}{2}=\dfrac{4}{2}=2$$. And there exists no positive integer $$a$$...
- Mon Feb 10, 2014 11:07 am
- Forum: Junior Level
- Topic: Last goes first
- Replies: 3
- Views: 3579
Re: Last goes first
$$10^5a+10^4b+10^3c+10^2d+10e+2=$$ $$10^5*6+10^4*3a+10^3*3b+10^2*3c+10*3d+3e$$
Solving this you will get the previous number $$857142$$ and the new number $$\boxed{285714}$$.
Solving this you will get the previous number $$857142$$ and the new number $$\boxed{285714}$$.
- Sun Feb 09, 2014 9:24 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies: 25
- Views: 16075
Re: warm-up problems for national BdMO'14
8.
$$\binom{n}{n-1}n=3n$$
So, $$n=3$$
now,
$$(3+1)^3=2*3^k+3*3+1$$
$$=> 64-10=2*3^k$$
$$=>2*3^k=54$$
$$k=3$$
$$(n,k)=(3,3)$$
$$\binom{n}{n-1}n=3n$$
So, $$n=3$$
now,
$$(3+1)^3=2*3^k+3*3+1$$
$$=> 64-10=2*3^k$$
$$=>2*3^k=54$$
$$k=3$$
$$(n,k)=(3,3)$$
- Sun Feb 09, 2014 8:33 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies: 25
- Views: 16075
Re: warm-up problems for national BdMO'14
$$log_2(log_{2^a}(log_{2^b}(2^{1000})))=0$$
$$=> log_{2^a}(log_{2^b}(2^{1000}))=1$$
$$=> log_{2^b}(2^{1000})=2^a$$
$$=> 2^{1000}=(2^b)^{2^a}$$.
$$a=[1,3]$$. So,
$$(a,b)= (1,500),(2,250), (3,125)$$.
So value of all $$a+b=881$$
$$=> log_{2^a}(log_{2^b}(2^{1000}))=1$$
$$=> log_{2^b}(2^{1000})=2^a$$
$$=> 2^{1000}=(2^b)^{2^a}$$.
$$a=[1,3]$$. So,
$$(a,b)= (1,500),(2,250), (3,125)$$.
So value of all $$a+b=881$$
- Sun Feb 09, 2014 3:48 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National Secondary 2010/10
- Replies: 1
- Views: 1953
National Secondary 2010/10
In a set of $$131$$ natural numbers, no number has a prime factor greater than $$42$$. Prove that it
is possible to choose four numbers from this set such that their product is a perfect square.
is possible to choose four numbers from this set such that their product is a perfect square.
- Sun Feb 09, 2014 3:31 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies: 25
- Views: 16075
Re: warm-up problems for national BdMO'14
$$\lambda _n = \sqrt {3n^2+2n+2}$$
$$=>(\lambda _n)^2 = 3n^2+2n+2$$
$$=>3(\lambda _n)^2 = 9n^2+6n+6$$
$$=>3(\lambda _n)^2 = (3n+1)^2+5$$
$$L.H.S \equiv 0,3,4\pmod{8}$$
$$R.H.S\equiv 5,6,9\pmod{8}$$.
L.H.S and R.H.S cannot be same.
$$\therefore \lambda _n \neq \sqrt {3n^2+2n+2} $$
$$=>(\lambda _n)^2 = 3n^2+2n+2$$
$$=>3(\lambda _n)^2 = 9n^2+6n+6$$
$$=>3(\lambda _n)^2 = (3n+1)^2+5$$
$$L.H.S \equiv 0,3,4\pmod{8}$$
$$R.H.S\equiv 5,6,9\pmod{8}$$.
L.H.S and R.H.S cannot be same.
$$\therefore \lambda _n \neq \sqrt {3n^2+2n+2} $$
- Sun Feb 09, 2014 3:14 pm
- Forum: Junior Level
- Topic: Acute angled circular
- Replies: 13
- Views: 10081
Re: Acute angled circular
Raiyan speaking , I don't understand , if the points are A , B , C , D , E , F and G . Everyone says 14 or 28 are possible . But taking all types of triangle , we get 343 types . So , there might be much more types . Please try to think the ans carefully . If you kindly explain what you are saying....