Secondary and Higher Secondary Marathon
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
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 .
$\displaystyle \frac {n^5}{5}+\frac {n^4}{2}+\frac {n^3}{3}−\frac {n}{30}$
is always an integer.
Source : An NT textbook .
$\frac{1}{0}$
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: Secondary and Higher Secondary Marathon
$\frac {n^5} {5} + \frac {n^4} {2} + \frac {n^3} {3} - \frac {n} {30}$
$= \frac {6n^5+15n^4+10n^3-n} {30}$
$= \frac {n*(n+1)*(2n+1)*(3n^2+3n-1)} {30}$ ; 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. ^_^
$= \frac {6n^5+15n^4+10n^3-n} {30}$
$= \frac {n*(n+1)*(2n+1)*(3n^2+3n-1)} {30}$ ; 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
Notre Dame College
Re: Secondary and Higher Secondary Marathon
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.
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.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
Re: Secondary and Higher Secondary Marathon
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.
$1+2+3+...+n = \frac{n(n+1)}{2}$
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.
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.........
Re: Secondary and Higher Secondary Marathon
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
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
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.........
- 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
If the Ans is $90$ . Then I'm ready to post my solution ..
joty ভাই , confirmation দেন ।
joty ভাই , confirmation দেন ।
$\frac{1}{0}$
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: Secondary and Higher Secondary Marathon
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
Notre Dame College
Re: Secondary and Higher Secondary Marathon
Good job, Nadim.Nadim Ul Abrar wrote:If the Ans is $90$ . Then I'm ready to post my solution ..
joty ভাই , confirmation দেন ।
Ans is $90$
Now post your solution and a new problem
And Shahrier, you also can post your solution
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.........
- 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
Just find the possible patterns for this six patterns of central $2 \times 2$ array .
এডিট : আমার ২০০ তম পোস্ট ।
Then consider their possible rotations and Count .এডিট : আমার ২০০ তম পোস্ট ।
$\frac{1}{0}$
- 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
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 .
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 .
$\frac{1}{0}$