Search found 110 matches
- Sun Feb 23, 2014 4:09 am
- Forum: National Math Olympiad (BdMO)
- Topic: Bdmo National 2014: Junior 6
- Replies: 6
- Views: 5826
Re: Bdmo National: Junior 6
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- Sat Feb 22, 2014 8:54 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Bdmo National 2014: Junior 6
- Replies: 6
- Views: 5826
Re: Bdmo National: Junior 6
I have solved the problem in this way , If we colour $14*14$ grid as a cheese board then there will be same number of black and white part. there are $14*14=196$ parts in the grid and a "T" shaped block has 4 parts. so, if we want to fill up the $14*14$ grid with "T" shaped block we need $\frac{196}...
- Sat Feb 22, 2014 5:36 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Bdmo National 2014: Junior 6
- Replies: 6
- Views: 5826
Bdmo National 2014: Junior 6
Is it possible to completely cover a $14*14$ grid by "T" shaped blocks from the diagram such that no block overlaps any other bolcks? Explain your answer with logic.
- Sat Feb 22, 2014 12:46 am
- Forum: Test Forum
- Topic: Bdmo 2014 national: junior 04
- Replies: 4
- Views: 9996
Re: Bdmo 2014 national: junior 04
let, $x$ is a 5 digit number. If we place 1 at the right most of $x$ we get the original 6 digit number. so the original 6 digit number is $10x+1$ again if we place 1 at the left most of $x$ we get the new 6 digit number. so the new 6 digit number is ${10}^{5}+x$ given the new number is $\frac{1}{3}...
- Thu Feb 20, 2014 8:37 pm
- Forum: Algebra
- Topic: IMO - 1987 - 4
- Replies: 9
- Views: 6033
Re: IMO - 1987 - 4
sadman sakib.......... the IMO compendium নামে একটা বই আছে । সেখানে 1959-2004 পর্যন্ত IMO এর প্রশ্ন পাবে । IMO shortlist এর problem গুলাও আছে।
- Thu Feb 20, 2014 1:24 pm
- Forum: Geometry
- Topic: Jumping from cosine to sine
- Replies: 3
- Views: 2616
Re: Jumping from cosine to sine
solved by cosine rule : $3cosA+cosB+cosC=3$ $\Leftrightarrow cosB+cosC=3-3cosA$ $\Leftrightarrow \frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}+\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=3-3(\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc})$ $\Leftrightarrow \frac{b{a}^{2}+b{c}^{2}-{b}^{3}+c{a}^{2}+c{b}^{2}-{c}^{3}}{2abc}=3(\frac...
- Wed Feb 19, 2014 3:51 pm
- Forum: Secondary Level
- Topic: An easy problem !!
- Replies: 9
- Views: 6928
Re: An easy problem !!
$N=2n$ $N(N^{2}+20)=2n(4n^{2}+20)=8n(n^{2}+5)$ now $48\mid N(N^{2}+20)=48\mid 8n(n^{2}+5)=6\mid n(n^{2}+5)$ so,we need to prove that $6$ divides $n(n^{2}+5)$ among two integers $n$ and $n^{2}+5$ one must be even . that means $2$ divides $n(n^{2}+5)$ $n(n^{2}+5)=n^{3}+5n$ now,$n\equiv 0/1/2 (mod3)$ $...
- Tue Feb 11, 2014 1:26 pm
- Forum: Secondary Level
- Topic: Large = Large+big+small+little
- Replies: 5
- Views: 4537
Re: Large = Large+big+small+little
সেইটা আজকে সকালে বুঝতে পারলাম। এখন জোড় এবং বিজোড় এর জন্য আলাদা আলাদা উত্তর আসছে। n=জোড় হলে, মোট সমবাহু ত্রিভুজ : $\sum_{k=1}^{k=n+1}\frac{k(k+1)}{2}+\sum_{k=1}^{k=\frac{n}{2}}\frac{2k(2k+1)}{2}$ n=বিজ়োড় হলে, মোট সমবাহু ত্রিভুজ : $\sum_{k=1}^{k=n+1}\frac{k(k+1)}{2}+\sum_{k=1}^{k=\frac{n+1}{2}}\f...
- Tue Feb 11, 2014 11:52 am
- Forum: Higher Secondary Level
- Topic: find the prime
- Replies: 3
- Views: 3752
Re: find the prime
@jamil
....i'm sorry....it was just printing mistake now it's ok.
....i'm sorry....it was just printing mistake now it's ok.
- Tue Feb 11, 2014 4:05 am
- Forum: Secondary Level
- Topic: giant prime
- Replies: 3
- Views: 3373
Re: giant prime
$111....1111$ [119, 1's] $=10^{118}+10^{117}+10^{116}+......+10^{2}+10^{1}+10^{0}$ $=\frac{10^{119}-10^{0}}{10-1}$ $=\frac{(10^{7})^{17}-1^{17}}{9}$ $=\frac{(10^{7}-1)*(something)}{9}$ $=\frac{9999999*(something)}{9}$ $=1111111*(something)$ so, 111...111 [119, 1's] is a product of two integers. so i...