BdMO 2017 National Round Secondary 10

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Kazi_Zareer
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BdMO 2017 National Round Secondary 10

Unread post by Kazi_Zareer » Fri Feb 10, 2017 9:27 pm

$p$ is an odd prime. The integer $k$ is in the range $1 \leq k \leq p-1.$ 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 $a_1 + a_2 + \dots + a_{p-1}.$
We cannot solve our problems with the same thinking we used when we create them.

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Thanic Nur Samin
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Re: BdMO 2017 National Round Secondary 10

Unread post by Thanic Nur Samin » Tue Feb 14, 2017 6:34 pm

This was also the problem 10 for higher secondary.

Let $1\le u\le p-1$ be an integer. We claim that there is exactly one number $k$ so that $u|pk+1$ and $1\le k\le u$.

Now, consider the solutions of the linear diophantine equation $ux-py=1$. 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 $um=pk+1\Rightarrow u|pk+1$, and the $k$ is unique.

So, for each integer $u$ so that $1\le u\le p-1$, there is exactly one integer $k$ so that $u|pk+1$. But the sum actually counts the number of such pairs $(u,k)$. So, the sum is $\boxed{p-1}$.
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samiul_samin
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Re: BdMO 2017 National Round Secondary 10

Unread post by samiul_samin » Mon Feb 11, 2019 1:11 pm

Kazi_Zareer wrote:
Fri Feb 10, 2017 9:27 pm
$p$ is an odd prime. The integer $k$ is in the range $1 \leq k \leq p-1.$ 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 $a_1 + a_2 + \dots + a_{p-1}.$
This is Japan Mathematical Olympiad finals 2016 question no $1$ .

prottoy das
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Re: BdMO 2017 National Round Secondary 10

Unread post by prottoy das » Thu Feb 14, 2019 11:32 am

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!!!!

prottoy das
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Re: BdMO 2017 National Round Secondary 10

Unread post by prottoy das » Fri Feb 15, 2019 11:52 pm

now the question is what's the exact answer? Some say it is $p-1$ & some say it is $p-2$!!!!!!! I think the answer is $p-1$

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samiul_samin
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Re: BdMO 2017 National Round Secondary 10

Unread post by samiul_samin » Sat Feb 16, 2019 4:19 pm

prottoy das wrote:
Fri Feb 15, 2019 11:52 pm
now the question is what's the exact answer? Some say it is $p-1$ & some say it is $p-2$!!!!!!! I think the answer is $p-1$
$P-1$
You are right. :D

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