Secondary and Higher Secondary Marathon

For students of class 11-12 (age 16+)
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Nadim Ul Abrar
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Re: Secondary and Higher Secondary Marathon

Unread post by Nadim Ul Abrar » Sun Nov 18, 2012 5:27 pm

Prove that if $n$ is a natural number,
$\displaystyle \frac {n^5}{5}+\frac {n^4}{2}+\frac {n^3}{3}−\frac {n}{30}$
is always an integer.

Source : An NT textbook .
$\frac{1}{0}$

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Fahim Shahriar
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Re: Secondary and Higher Secondary Marathon

Unread post by Fahim Shahriar » Mon Nov 19, 2012 12:52 am

$\frac {n^5} {5} + \frac {n^4} {2} + \frac {n^3} {3} - \frac {n} {30}$

$= \frac {6n^5+15n^4+10n^3-n} {30}$

$= \frac {n*(n+1)*(2n+1)*(3n^2+3n-1)} {30}$ ; it will be a integer if the numerator is divisible by $30$.

Surely one of $n$ and $(n+1)$ is divisible by $2$.
One of $n$, $(n+1)$ and $(2n+1)$ is divisible by $3$.

When $n≡1,3 (mod 5)$; $(3n^2+3n-1)$ is divisible by $5$.
When $n≡2 (mod 5)$; $(2n+1)$ is divisible by $5$.
When $n≡4 (mod 5)$; $(n+1)$ is divisible by $5$.

So it is always an integer. ^_^
Name: Fahim Shahriar Shakkhor
Notre Dame College

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SANZEED
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Re: Secondary and Higher Secondary Marathon

Unread post by SANZEED » Mon Nov 19, 2012 11:49 am

Problem $\boxed {11}$:
Prove that the sum $1^k+2^k+3^k+...+n^k$ where $n$ is an arbitrary integer and $k$ is odd, is divisible by $1+2+3+...+n$.

Source: The USSR Olympiad Problem Book-The Divisibility of Integers.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

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sm.joty
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Re: Secondary and Higher Secondary Marathon

Unread post by sm.joty » Wed Nov 28, 2012 8:00 pm

we know
$1+2+3+...+n = \frac{n(n+1)}{2}$
Let,
$S=1^k+2^k+3^k+...+n^k$

We know that for any odd $n$
$a+b|a^n+b^n$
so $1+(n-1)|1^k+(n-1)^{k}$
$2+(n-2)|2^k+(n-2)^{k}$
.
.
.
so all terms of this sequence is divisible by $n$
Now,
$1+(n+1)-1|1^k+n^k$
$2+(n+1)-2|2^k+(n-1)^k$
.
.
.
so $S$ is also divisible by $n+1$
we conclude that
$n(n+1)|S$
so,
$\frac{2S}{n(n+1)}=c$
where $c$ is a natural number.
so
$\frac{S}{n(n+1)/2}=c$
so
S is divisible by $\frac{n(n+1)}{2}$
so are done.
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sm.joty
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Re: Secondary and Higher Secondary Marathon

Unread post by sm.joty » Wed Nov 28, 2012 8:09 pm

Problem $\boxed {12}$
How many different $4 \times 4$ arrays whose entries are all $1$'s and $-1$'s have the property that the sum of entries in each row is $0$ and sum of entries in each column is $0$

Source: AIME 1997
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Nadim Ul Abrar
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Re: Secondary and Higher Secondary Marathon

Unread post by Nadim Ul Abrar » Sun Dec 02, 2012 11:04 pm

If the Ans is $90$ . Then I'm ready to post my solution .. :?
joty ভাই , confirmation দেন ।
$\frac{1}{0}$

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Fahim Shahriar
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Re: Secondary and Higher Secondary Marathon

Unread post by Fahim Shahriar » Mon Dec 03, 2012 10:48 am

Nadim Ul Abrar wrote:If the Ans is $90$ . Then I'm ready to post my solution .. :?
joty ভাই , confirmation দেন ।


Yes. It's $90$.
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 » Mon Dec 03, 2012 1:55 pm

Nadim Ul Abrar wrote:If the Ans is $90$ . Then I'm ready to post my solution .. :?
joty ভাই , confirmation দেন ।
Good job, Nadim. :D
Ans is $90$
Now post your solution and a new problem :)
And Shahrier, you also can post your solution
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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 03, 2012 5:28 pm

Just find the possible patterns for this six patterns of central $2 \times 2$ array .
চরকি.PNG
চরকি.PNG (7.19KiB)Viewed 19100 times
Then consider their possible rotations and Count .





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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 03, 2012 5:35 pm

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 .
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