BDMO Secondary National 2021 #3

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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Mehrab4226
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BDMO Secondary National 2021 #3

Unread post by Mehrab4226 » Sat Apr 10, 2021 12:42 pm

মনে করো, \(r\) একটা ধনাত্মক বাস্তব সংখ্যা। \([r]\) দিয়ে আমরা \(r\)-এর পূর্ণসাংখ্যিক অংশ বোঝাই আর \(\{r\}\) দিয়ে আমরা \(r\)-এর ভগ্নাংশিক অংশটা বোঝাই। যেমন যদি \(r=32.86\) হয়, তাহলে \(\{r\}=0.86\) এবং \([r]=32\)। এমন সব ধনাত্মক সংখ্যা \(r\)-এর যোগফল কত যদি \(25\{r\}+[r]=125\) হয়?

Let $r$ be a positive real number. Denote $[r]$ by the integer part of $r$ and by $\{r\}$ the fractional part of $r$. For example, if $r=32.86$, then $\{r\}=0.86$ and $[r]=32$ . What is the sum of all positive numbers $r$ satisfying
\[ 25\{r\} + [r]=125\]
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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Pro_GRMR
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Re: BDMO Secondary National 2021 #3

Unread post by Pro_GRMR » Sat Apr 10, 2021 5:58 pm

Mehrab4226 wrote:
Sat Apr 10, 2021 12:42 pm
Let $r$ be a positive real number. Denote $[r]$ by the integer part of $r$ and by $\{r\}$ the fractional part of $r$. For example, if $r=32.86$, then $\{r\}=0.86$ and $[r]=32$ . What is the sum of all positive numbers $r$ satisfying
\[ 25\{r\} + [r]=125\]
As $\{r\}$ is a fraction,
$0 \leq \{r\} < 1$
$\Longrightarrow 0 \leq 25\{r\} < 25$

Also Notice as $25\{r\} = 125 - [r]$,
$25\{r\}$ is an integer.

By taking different integer values of $25\{r\}$ from 0 to 24 we see that $r = 101.04, 102.08, \dots, 124.96, 125$
Summing the partially arithmetic series we get the answer of $\frac{101.04+124.96}{2} \times 24 + 125 = \boxed{2837}$
"When you change the way you look at things, the things you look at change." - Max Planck

Pritom12345
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Re: BDMO Secondary National 2021 #3

Unread post by Pritom12345 » Sun Apr 11, 2021 9:21 pm

Isn't 125 a real number?

I mean it fulfills the condition:: [125] + {0} = 125.0 and there no condition like {r} != 0

so, why won't we add that number?

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gwimmy(abid)
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Re: BDMO Secondary National 2021 #3

Unread post by gwimmy(abid) » Mon Apr 12, 2021 11:17 am

Pritom12345 wrote:
Sun Apr 11, 2021 9:21 pm
Isn't 125 a real number?

I mean it fulfills the condition:: [125] + {0} = 125.0 and there no condition like {r} != 0

so, why won't we add that number?
It is considered in the solution
Umm....the healer needs healing...

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Pro_GRMR
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Re: BDMO Secondary National 2021 #3

Unread post by Pro_GRMR » Fri Apr 16, 2021 11:01 pm

There were some mistakes. The forum doesn't give me edit access anymore.
This is the correct version
As $\{r\}$ is a fraction,
$0 \leq \{r\} < 1$
$\Longrightarrow 0 \leq 25\{r\} < 25$

Also Notice as $25\{r\} = 125 - [r]$,
$25\{r\}$ is an integer.

By taking different integer values of $25\{r\}$ from 0 to 24 we see that $r = 101.96, 102.92, \dots, 124.04, 125$
Summing the partially arithmetic series we get the answer of $\frac{101.96+124.04}{2} \times 24 + 125 = \boxed{2837}$
The answer is the same but Notice the decimal value of $r$ was wrong.
"When you change the way you look at things, the things you look at change." - Max Planck

Lelu
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Re: BDMO Secondary National 2021 #3

Unread post by Lelu » Mon Apr 19, 2021 7:56 pm

\begin{align*}
&25\left\{r\right\}+\left[r\right]=125\\
\Rightarrow&25\left\{r\right\}=125-\left[r\right]
\end{align*}
As $125-\left[r\right]$ is an integer, $25\left\{r\right\}$ must be an integer too.
If $\left\{r\right\}=0$, then $r=125$.
Otherwise, suppose $\left\{r\right\}=\dfrac{p}{q}$ where $p$ and $q$ are natural numbers, $p<q$ and $gcd\left(p, q\right)=1$. Since $\dfrac{25p}q$ is an integer, $q$ is a divisor of $25$. But $q$ cannot be equal to $1$ because then there will be no value for $p$. So, $q=5, 25$. Let us find the sum of the possible values of $r$ in two cases, each dealing with one value of $q$.
  • If $q=5$ then,
    \begin{align*}
    &\left[r\right]=125-\frac{25p}5 \\
    \Rightarrow&\left[r\right]=125-5p
    \end{align*}
    Here $p=1, 2, 3, 4$. Now the sum of the possible values of $\left[r\right]$ in this case is:
    \begin{align*}
    &125\cdot4-5\left(1+2+3+4\right)\\
    =&500-50\\
    =&450
    \end{align*}
    And the sum of the possible values of $\left\{r\right\}$ in this case is:
    \begin{align*}
    &\frac{1+2+3+4}5\\
    =&2
    \end{align*}
    So the sum of the possible values of $r$ in this case is:
    \begin{align*}
    &450+2\\
    =&452
    \end{align*}
  • If $q=25$ then,
    \begin{align*}
    &\left[r\right]=125-\frac{25p}{25} \\
    \Rightarrow&\left[r\right]=125-p
    \end{align*}
    Here $1\leq p\leq24$. But $p$ is not equal to $5, 10, 15$ and $20$ because $p$ and $q$ are co-prime. So the sum of the possible values of $\left[r\right]$ in this case is:
    \begin{align*}
    &125\cdot20-\left[\left(1+2+\dots+24\right)-\left(5+10+15+20\right)\right]\\
    =&2500-\left(300-50\right)\\
    =&2250
    \end{align*}
    The sum of the possible values of $\left\{r\right\}$ in this case is:
    \begin{align*}
    &\frac{\left(1+2+\dots+24\right)-\left(5+10+15+20\right)}{25}\\
    =&\frac{300-50}{25}\\
    =&10
    \end{align*}
    So the sum of the possible values of $r$ in this case is:
    \begin{align*}
    &2250+10\\
    =&2260
    \end{align*}
And as stated before, $r=125$ when $\left\{r\right\}=0$
So the sum of all possible values of $r$ is:
\begin{align*}
&125+2260+452\\
=&\fbox{2837}
\end{align*}

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