Secondary and Higher Secondary Marathon

For students of class 11-12 (age 16+)
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Fahim Shahriar
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Re: Secondary and Higher Secondary Marathon

Unread post by Fahim Shahriar » Tue Dec 04, 2012 12:27 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].
$24*2=48$

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

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Fahim Shahriar
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Re: Secondary and Higher Secondary Marathon

Unread post by Fahim Shahriar » Tue Dec 04, 2012 12:52 pm

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 $50$. :D :) :D
Name: Fahim Shahriar Shakkhor
Notre Dame College

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sm.joty
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Re: Secondary and Higher Secondary Marathon

Unread post by sm.joty » Tue Dec 04, 2012 1:05 pm

Solution - $13$
(may be I'm correct :? )
At first proof a nice lemma,
the product of all divisors of a number $N$ is
"$N^{k/2}$"
where $k$ is the number of divisors.

if we consider only the proper divisors, then
according to the condition we get,
$\frac{N^{k/2}}{N}=N$
so
$k=4$
Now
$k=4=(3+1)=(1+1)(1+1)$

if we consider $k=(3+1)$ we will have
N=2,3
if we consider $k=(1+1)(1+1)$
there are only $13$ numbers who have only two prime with power $1$ and less than $50$
$(2.3),(2.5),(2.7),(2.11),(2.13),(2.17),(2.19),(2.23),$
$(3.5),(3.7),(3.11),(3.13),$
$(5.7)$

SO, ans $ = 13+2=15$
nice solution by Nadim
Fahim's solve is like the official one. :)
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Tahmid Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Tahmid Hasan » Tue Dec 04, 2012 1:43 pm

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 $N$.

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sm.joty
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Re: Secondary and Higher Secondary Marathon

Unread post by sm.joty » Wed Dec 05, 2012 6:31 pm

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 $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
Source : Putnam 1960
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Fahim Shahriar
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Re: Secondary and Higher Secondary Marathon

Unread post by Fahim Shahriar » Thu Dec 06, 2012 11:04 am

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 $n^2$. :D
Name: Fahim Shahriar Shakkhor
Notre Dame College

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sm.joty
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Re: Secondary and Higher Secondary Marathon

Unread post by sm.joty » Thu Dec 06, 2012 6:57 pm

New problem please . :D
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Tahmid Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Tahmid Hasan » Thu Dec 06, 2012 7:07 pm

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 $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
Source : Putnam 1960
$\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$
$\Rightarrow (x-n)(y-n)=n^2$, 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 $\angle ASB=90^{\circ}$.
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Nadim Ul Abrar
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Re: Secondary and Higher Secondary Marathon

Unread post by Nadim Ul Abrar » Mon Dec 17, 2012 5:51 pm

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
$\frac{1}{0}$

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SANZEED
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Re: Secondary and Higher Secondary Marathon

Unread post by SANZEED » Mon Dec 17, 2012 6:52 pm

@Nadim vai: You haven't mentioned the source... :|
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