Secondary and Higher Secondary Marathon
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Using mobile.
$(12)$
1st row can be arranged in 6 ways. So is the 2nd. The first two rows can be arranged in $6*6=36$ ways.
1)When 1st & 2nd row have same numbers in each column[6 ways], 3rd & 4th can be solved in only [1 way].
$6*1=6$
2)When 1st & 2nd row have different numbers in each column[6 ways], 3rd & 4th have [6 ways].
$6*6=36$ ways
3)In other 24 arranging ways of the first two rows, the 3rd & 4th have [2 ways].
$24*2=48$
$6+36+48=90$.
$(12)$
1st row can be arranged in 6 ways. So is the 2nd. The first two rows can be arranged in $6*6=36$ ways.
1)When 1st & 2nd row have same numbers in each column[6 ways], 3rd & 4th can be solved in only [1 way].
$6*1=6$
2)When 1st & 2nd row have different numbers in each column[6 ways], 3rd & 4th have [6 ways].
$6*6=36$ ways
3)In other 24 arranging ways of the first two rows, the 3rd & 4th have [2 ways].
$24*2=48$
$6+36+48=90$.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: Secondary and Higher Secondary Marathon
Solution: 13 [Product-Perfect]
A number becomes product-perfect if
1) It is the product of two different primes -» $6,10,14,15,21,22,26,33,34,35,38,39,46$
OR
2)It is the cube of any prime number -»$8,27$
So there is $15$ Product-Perfect Numbers below $50$.
A number becomes product-perfect if
1) It is the product of two different primes -» $6,10,14,15,21,22,26,33,34,35,38,39,46$
OR
2)It is the cube of any prime number -»$8,27$
So there is $15$ Product-Perfect Numbers below $50$.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
Re: Secondary and Higher Secondary Marathon
Solution - $13$
(may be I'm correct )
nice solution by Nadim
Fahim's solve is like the official one.
(may be I'm correct )
Fahim's solve is like the official one.
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
- Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: Secondary and Higher Secondary Marathon
এই সমস্যাটা সম্ভবত এইবারের 'উদ্ভাস' আয়োজিত 'জুনিয়র বাংলাদেশ' এ আসছিল(যদিও নিশ্চিত নই)।Nadim Ul Abrar wrote:P $13$
A number ($ \geq2$), is called product-perfect if it is equal to the product of all of its proper divisors. For example, $6=1×2×3$, hence $6$ is product-perfect. How many product-perfect numbers are there below $50$?
Note: A proper divisor of a number $N$ is a positive integer less than $N$ that divides $N$.
Source : Facebook .
বড় ভালবাসি তোমায়,মা
Re: Secondary and Higher Secondary Marathon
No One post any problem !!!!!!
Ok ,so I think, I can post one.
Problem - $\boxed {14}$
Let $n$ is a natural number.How many solution are there in ordered positive integers pairs $(x,y)$ to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
Ok ,so I think, I can post one.
Problem - $\boxed {14}$
Let $n$ is a natural number.How many solution are there in ordered positive integers pairs $(x,y)$ to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: Secondary and Higher Secondary Marathon
Solution- \boxed {14}
As $x,y,n$ are positive integers, we can observe that n is less than x,y. Let $x = n+k$.
$\frac {1}{x} + \frac {1}{y} = \frac {1}{n}$
$\frac {x+y}{xy} = \frac {1}{n}$
$nx + ny = xy$
$nx = y(x-n)$
$n(n+k) = y(n+k-n)$
$n^{2}+nk = yk$
$\frac {n^{2}} {k} + n = y$
$y$ will be a positive integer if $k$ divides $n^2$. That means $k$ has to be a divisor of $n^2$.
So the number of solutions of $(x,y)$ are same as the number of divisors of $n^2$.
As $x,y,n$ are positive integers, we can observe that n is less than x,y. Let $x = n+k$.
$\frac {1}{x} + \frac {1}{y} = \frac {1}{n}$
$\frac {x+y}{xy} = \frac {1}{n}$
$nx + ny = xy$
$nx = y(x-n)$
$n(n+k) = y(n+k-n)$
$n^{2}+nk = yk$
$\frac {n^{2}} {k} + n = y$
$y$ will be a positive integer if $k$ divides $n^2$. That means $k$ has to be a divisor of $n^2$.
So the number of solutions of $(x,y)$ are same as the number of divisors of $n^2$.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
Re: Secondary and Higher Secondary Marathon
New problem please .
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
- Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: Secondary and Higher Secondary Marathon
$\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$sm.joty wrote:No One post any problem !!!!!!
Ok ,so I think, I can post one.
Problem - $\boxed {14}$
Let $n$ is a natural number.How many solution are there in ordered positive integers pairs $(x,y)$ to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
$\Rightarrow (x-n)(y-n)=n^2$, the rest is easy.
Problem-$15$:In a square $ABCD$, let $P$ be a point in the side $CD$, different from $C$ and $D$. In the triangle $ABP$, the altitudes $AQ$ and $BR$ are drawn, and let $S$ be the intersection point of lines $CQ$ and $DR$. Show that $\angle ASB=90^{\circ}$.
Source:Cono Sur Olympiad 2012.
বড় ভালবাসি তোমায়,মা
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
Re: Secondary and Higher Secondary Marathon
problem 16 :
In triangle $ABC$, the centroid is $G$ and $N$ is the midpoint of $CA$. The line through $G$ parallel to $BC$ meets $AB$ at $X$. Prove that $\angle AXC = \angle NGC$ if and only if angle $ACB$ is a right angle
In triangle $ABC$, the centroid is $G$ and $N$ is the midpoint of $CA$. The line through $G$ parallel to $BC$ meets $AB$ at $X$. Prove that $\angle AXC = \angle NGC$ if and only if angle $ACB$ is a right angle
$\frac{1}{0}$
Re: Secondary and Higher Secondary Marathon
@Nadim vai: You haven't mentioned the source...
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$