Beginner's Marathon
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: Beginner's Marathon
Problem $11$
For a positive integer $n$, let $f(n)$ denote the greatest odd divisor of $n$. Prove that for any integer $x$, $f(x+1)+f(x+2)+........f(2x) = x^2$.
For a positive integer $n$, let $f(n)$ denote the greatest odd divisor of $n$. Prove that for any integer $x$, $f(x+1)+f(x+2)+........f(2x) = x^2$.
Frankly, my dear, I don't give a damn.
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: Beginner's Marathon
$\text{Solution to Problem 11}$
Lemma 1: $f(k)=f(2k)$
Proof: Trivial
Lemma 2: $f(2k+1)=2k+1$
Proof: Trivial
We will use induction. For $n=1$, it is true.
Now for $g(m)=f(m+1)+f(m+2)+\cdots+f(2m)=m^2$
We will be done if we can proof
$g(m+1)=f(m+2)+f(m+3)+\cdots+f(2m)+f(2m+1)+f(2m+2)=(m+1)^2$
Now,$f(m+1)+f(m+2)+\cdots +f(2m)=m^2$
$\Rightarrow f(m+2)+\cdots +f(2m)+f(2m+2)=m^2$ [Lemma 1 on $f(m+1)$]
$\Rightarrow f(m+2)+\cdots +f(2m)+f(2m+1)+f(2m+2)=m^2+2m+1$ [Lemma 2]
That means $g(m+1)=m^2+2m+1=(m+1)^2$
$[DONE]$
Lemma 1: $f(k)=f(2k)$
Proof: Trivial
Lemma 2: $f(2k+1)=2k+1$
Proof: Trivial
We will use induction. For $n=1$, it is true.
Now for $g(m)=f(m+1)+f(m+2)+\cdots+f(2m)=m^2$
We will be done if we can proof
$g(m+1)=f(m+2)+f(m+3)+\cdots+f(2m)+f(2m+1)+f(2m+2)=(m+1)^2$
Now,$f(m+1)+f(m+2)+\cdots +f(2m)=m^2$
$\Rightarrow f(m+2)+\cdots +f(2m)+f(2m+2)=m^2$ [Lemma 1 on $f(m+1)$]
$\Rightarrow f(m+2)+\cdots +f(2m)+f(2m+1)+f(2m+2)=m^2+2m+1$ [Lemma 2]
That means $g(m+1)=m^2+2m+1=(m+1)^2$
$[DONE]$
Last edited by Thamim Zahin on Mon Apr 10, 2017 7:15 pm, edited 1 time in total.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: Beginner's Marathon
$\text{Problem 12}$
For every positive integer $k$ let $f(k)$ be the largest integer such that $2^{f(k)} \mid k$. For every positive integer $n$ determine : $f(1) + f(2) +f(3)+f(4)+f(5)+\cdots+ f(2^n)$.
Or, $\displaystyle\sum_{i=1}^{2^n} f(i)=?$
For every positive integer $k$ let $f(k)$ be the largest integer such that $2^{f(k)} \mid k$. For every positive integer $n$ determine : $f(1) + f(2) +f(3)+f(4)+f(5)+\cdots+ f(2^n)$.
Or, $\displaystyle\sum_{i=1}^{2^n} f(i)=?$
Last edited by Thamim Zahin on Tue Apr 11, 2017 10:49 pm, edited 3 times in total.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
Re: Beginner's Marathon
Mod note: This solution was for a wrong version of problem 12, asking for sum of $f(n)$ where $n=2^k$ only. The correct solution is below.
Last edited by Zawadx on Tue Apr 18, 2017 7:11 pm, edited 2 times in total.
Reason: Fixed LaTeX issues
Reason: Fixed LaTeX issues
Re: Beginner's Marathon
$\text{SOLUTION TO PROBLEM 12}$
$f(1) + f(2) +..... f(2^n) = 2^0 + 2^1+...... 2^{n1}$
PROOF: lemma 1:$ f(2k +1) = 0$
proof: trivial
Lemma : $f(2^m + 1) + f(2^m + 2) +........f(2^{m+1}) = 2^m$
Proof : We use induction. Base case m = 1 is true.
Now, let $g(m) = f(2^m + 1) + f(2^m + 2) +........f(2^{m+1}) = 2^m$
For $f(2\times 2^m + 1) + f(2\times2^m + 2) +........f(2\times2^m + 2^m) = f(2(2^m + 1)) + f(2(2^m + 2)) +........f(2(2^{m+1})) = 2^m + 2^m$
Here, we multiply 2 with every even number from $2^{m+1}$ to $2^{m+2}$, As there as $2^m$ even numbers between them, we get $f(2^{m + 1} + 1) + f(2^{m + 1} + 2) +........f(2^{m+2}) = 2^{m+1}$
Therefore, our result follows.
$f(1) + f(2) +..... f(2^n) = 2^0 + 2^1+...... 2^{n1}$
PROOF: lemma 1:$ f(2k +1) = 0$
proof: trivial
Lemma : $f(2^m + 1) + f(2^m + 2) +........f(2^{m+1}) = 2^m$
Proof : We use induction. Base case m = 1 is true.
Now, let $g(m) = f(2^m + 1) + f(2^m + 2) +........f(2^{m+1}) = 2^m$
For $f(2\times 2^m + 1) + f(2\times2^m + 2) +........f(2\times2^m + 2^m) = f(2(2^m + 1)) + f(2(2^m + 2)) +........f(2(2^{m+1})) = 2^m + 2^m$
Here, we multiply 2 with every even number from $2^{m+1}$ to $2^{m+2}$, As there as $2^m$ even numbers between them, we get $f(2^{m + 1} + 1) + f(2^{m + 1} + 2) +........f(2^{m+2}) = 2^{m+1}$
Therefore, our result follows.
Last edited by dshasan on Sat Apr 15, 2017 12:51 am, edited 1 time in total.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
 Charles Caleb Colton
 Charles Caleb Colton
Re: Beginner's Marathon
$\text{PROBLEM 13 :}$
Let $AD, BE, CF$ be concurrent cevians in a triangle, meeting at $P$. Prove that $\dfrac{PD}{AD} + \dfrac{PE}{BE} + \dfrac{PF}{CF} =1$.
Let $AD, BE, CF$ be concurrent cevians in a triangle, meeting at $P$. Prove that $\dfrac{PD}{AD} + \dfrac{PE}{BE} + \dfrac{PF}{CF} =1$.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
 Charles Caleb Colton
 Charles Caleb Colton

 Posts: 57
 Joined: Sun Dec 11, 2016 2:01 pm
Re: Beginner's Marathon
Let us denote by $ABC$,the area of $\bigtriangleup ABC$. From geometric proportion,$\frac{PD}{AD} = \frac{PBC}{ABC}$ as the bases of $\bigtriangleup ABC$ and $\bigtriangleup PBC$ are the same.This case works for $\bigtriangleup PAC$ and $\bigtriangleup PAB$ as well.So,we concurr that $\frac{PD}{AS} + \frac{PE}{BE} + \frac{PF}{CF} = \frac{PBC}{ABC} + \frac{PAC}{ABC} + \frac{PAB}{ABC} = 1$.
Last edited by Zawadx on Tue Apr 18, 2017 7:06 pm, edited 2 times in total.
Reason: Fixed LaTex
Reason: Fixed LaTex

 Posts: 57
 Joined: Sun Dec 11, 2016 2:01 pm
Re: Beginner's Marathon
Problem 14:Let the incircle of triangle ABC touch BC at D. Let,DT be a diameter of the circle.If AT meets BC at M,prove that BD=CM.
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: Beginner's Marathon
$\text{Solution to 14}$
Consider the dilation with center $A$ that carries the incircle to an excircle. The diameter $DT$ of the incircle must be mapped to the diameter of the excircle that is perpendicular to $BC$. It follows that $T$ must get mapped to the point of tangency between the excircle and $BC$. Since the image of $T$ must lie on the line $AT$, it must be $M$. That is, the excircle is tangent to $BC$ at $M$. Then, it follows easily that $BD = CM$.
Consider the dilation with center $A$ that carries the incircle to an excircle. The diameter $DT$ of the incircle must be mapped to the diameter of the excircle that is perpendicular to $BC$. It follows that $T$ must get mapped to the point of tangency between the excircle and $BC$. Since the image of $T$ must lie on the line $AT$, it must be $M$. That is, the excircle is tangent to $BC$ at $M$. Then, it follows easily that $BD = CM$.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: Beginner's Marathon
$\text{Problem 15}$
Let n be an integer greater than $2$. Prove that among the fractions
$$\frac{1}{n},\frac{2}{n}\cdots ,\frac{n1}{n}$$
an even number are irreducible.
Let n be an integer greater than $2$. Prove that among the fractions
$$\frac{1}{n},\frac{2}{n}\cdots ,\frac{n1}{n}$$
an even number are irreducible.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.