BdMO 2017 National Round Secondary 10
 Kazi_Zareer
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BdMO 2017 National Round Secondary 10
$p$ is an odd prime. The integer $k$ is in the range $1 \leq k \leq p1.$ Let $a_k$ be the number of divisors of $kp + 1 $ that are greater than or equal to $k$ and less than $p.$ Find the value of $a_1 + a_2 + \dots + a_{p1}.$
We cannot solve our problems with the same thinking we used when we create them.
 Thanic Nur Samin
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 Joined: Sun Dec 01, 2013 11:02 am
Re: BdMO 2017 National Round Secondary 10
This was also the problem 10 for higher secondary.
Let $1\le u\le p1$ be an integer. We claim that there is exactly one number $k$ so that $upk+1$ and $1\le k\le u$.
Now, consider the solutions of the linear diophantine equation $uxpy=1$. Since $(u,p)=1$, there exists infinitely many solutions. Let $x_0,y_0$ be one of them. So, the infinite family of solutions is given by $(x_t,y_t)=(x_0+pt,y_0+ut)$. Note that the terms of $y_t$ forms an AP with difference $u$, so it can hold exactly one value in the interval $[1,u]$. Let that be $k$, and the correspoinding $x$ be $m$.
So, we get $um=pk+1\Rightarrow upk+1$, and the $k$ is unique.
So, for each integer $u$ so that $1\le u\le p1$, there is exactly one integer $k$ so that $upk+1$. But the sum actually counts the number of such pairs $(u,k)$. So, the sum is $\boxed{p1}$.
Let $1\le u\le p1$ be an integer. We claim that there is exactly one number $k$ so that $upk+1$ and $1\le k\le u$.
Now, consider the solutions of the linear diophantine equation $uxpy=1$. Since $(u,p)=1$, there exists infinitely many solutions. Let $x_0,y_0$ be one of them. So, the infinite family of solutions is given by $(x_t,y_t)=(x_0+pt,y_0+ut)$. Note that the terms of $y_t$ forms an AP with difference $u$, so it can hold exactly one value in the interval $[1,u]$. Let that be $k$, and the correspoinding $x$ be $m$.
So, we get $um=pk+1\Rightarrow upk+1$, and the $k$ is unique.
So, for each integer $u$ so that $1\le u\le p1$, there is exactly one integer $k$ so that $upk+1$. But the sum actually counts the number of such pairs $(u,k)$. So, the sum is $\boxed{p1}$.
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Because destroying everything mindlessly isn't cool enough.
 samiul_samin
 Posts: 1004
 Joined: Sat Dec 09, 2017 1:32 pm
Re: BdMO 2017 National Round Secondary 10
This is Japan Mathematical Olympiad finals 2016 question no $1$ .Kazi_Zareer wrote: ↑Fri Feb 10, 2017 9:27 pm$p$ is an odd prime. The integer $k$ is in the range $1 \leq k \leq p1.$ Let $a_k$ be the number of divisors of $kp + 1 $ that are greater than or equal to $k$ and less than $p.$ Find the value of $a_1 + a_2 + \dots + a_{p1}.$

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Re: BdMO 2017 National Round Secondary 10
So sorry that it was an stolen problem. The 9th problem of National BDMO 2017, secondary, was also an stolen problem. This means at least two problems of the national olympiad was stolen!!!!

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 Location: Sylhet
Re: BdMO 2017 National Round Secondary 10
now the question is what's the exact answer? Some say it is $p1$ & some say it is $p2$!!!!!!! I think the answer is $p1$
 samiul_samin
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 Joined: Sat Dec 09, 2017 1:32 pm
Re: BdMO 2017 National Round Secondary 10
$P1$prottoy das wrote: ↑Fri Feb 15, 2019 11:52 pmnow the question is what's the exact answer? Some say it is $p1$ & some say it is $p2$!!!!!!! I think the answer is $p1$
You are right.