## Secondary and Higher Secondary Marathon

For students of class 11-12 (age 16+)
Fahim Shahriar
Posts: 138
Joined: Sun Dec 18, 2011 12:53 pm

### Re: Secondary and Higher Secondary Marathon

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].
$24*2=48$

$6+36+48=90$.
Name: Fahim Shahriar Shakkhor
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 $50$.
Name: Fahim Shahriar Shakkhor
Notre Dame College

sm.joty
Posts: 327
Joined: Thu Aug 18, 2011 12:42 am
Location: Dhaka

### Re: Secondary and Higher Secondary Marathon

Solution - $13$
(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.........

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Secondary and Higher Secondary Marathon

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 $N$.

এই সমস্যাটা সম্ভবত এইবারের 'উদ্ভাস' আয়োজিত 'জুনিয়র বাংলাদেশ' এ আসছিল(যদিও নিশ্চিত নই)।
বড় ভালবাসি তোমায়,মা

sm.joty
Posts: 327
Joined: Thu Aug 18, 2011 12:42 am
Location: Dhaka

### 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 $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
হার জিত চিরদিন থাকবেই
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 $n^2$.
Name: Fahim Shahriar Shakkhor
Notre Dame College

sm.joty
Posts: 327
Joined: Thu Aug 18, 2011 12:42 am
Location: Dhaka

### Re: Secondary and Higher Secondary Marathon

হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Secondary and Higher Secondary Marathon

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 $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
$\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$
$\Rightarrow (x-n)(y-n)=n^2$, 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 $\angle ASB=90^{\circ}$.
বড় ভালবাসি তোমায়,মা

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
$\frac{1}{0}$

SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm

### Re: Secondary and Higher Secondary Marathon

@Nadim vai: You haven't mentioned the source...
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$