COUNT THE NUMBER OF STUDENTS

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MATHPRITOM
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COUNT THE NUMBER OF STUDENTS

Unread post by MATHPRITOM » Mon Jan 02, 2012 10:22 pm

In a Olympiad 13 boys & d girls have participated.Each student has got same marks & their total number is $d^2+10d+17$. Find out the total number of students.

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Phlembac Adib Hasan
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Re: COUNT THE NUMBER OF STUDENTS

Unread post by Phlembac Adib Hasan » Tue Jan 03, 2012 10:43 am

First thanks to Pritom Vaia for giving such a lovely problem. :D
Answer: $14,28,56$

Our problem says $ m(d+13)=d^2+10d+17$ where $m$ is the mark that each student got.

$\Rightarrow m(d+13)=(d+13)^2-16d-152 $

$ \Rightarrow 16d+152 \equiv 0 (mod (d+13))$

$ \Rightarrow 16d+152 \equiv 16d+208 (mod (d+13))$

$\Rightarrow 152 \equiv 208 (mod (d+13)) $

$\Rightarrow 56 \equiv 0 (mod (d+13)) $

$\Rightarrow d+13=1,2,4,7,8,14,28,56$

As $d+13 \ge 13$, $d+13=14,28,56$
Is it self-made?
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Masum
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Re: COUNT THE NUMBER OF STUDENTS

Unread post by Masum » Fri Jan 13, 2012 2:38 pm

Phlembac Adib Hasan wrote:First thanks to Pritom Vaia for giving such a lovely problem. :D
Answer: $14,28,56$

Our problem says $ m(d+13)=d^2+10d+17$ where $m$ is the mark that each student got.

$\Rightarrow m(d+13)=(d+13)^2-16d-152 $

$ \Rightarrow 16d+152 \equiv 0 (mod (d+13))$

$ \Rightarrow 16d+152 \equiv 16d+208 (mod (d+13))$

$\Rightarrow 152 \equiv 208 (mod (d+13)) $

$\Rightarrow 56 \equiv 0 (mod (d+13)) $

$\Rightarrow d+13=1,2,4,7,8,14,28,56$

As $d+13 \ge 13$, $d+13=14,28,56$
Is it self-made?
In such problems try to use divisibility rather than congruence, though both are same. Because using the sign $|$ is more convenient and helpful, also makes a better sense I think. After all, the solution remains much clearer.
Then you would have to write only: \[d+13|d^2+10d+17,d^2+13d\Rightarrow d+13|3d-17,3d+39\Rightarrow d+13|56\]
Doesn't it look better?
One one thing is neutral in the universe, that is $0$.

MATHPRITOM
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Re: COUNT THE NUMBER OF STUDENTS

Unread post by MATHPRITOM » Fri Jan 13, 2012 3:13 pm

No....... it,s a problem form Kaikobad sir's book ....& I also solved the problem using function..... I think by this problem,one could be able to understand the real beauty of function ...

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*Mahi*
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Re: COUNT THE NUMBER OF STUDENTS

Unread post by *Mahi* » Fri Jan 13, 2012 8:44 pm

MATHPRITOM wrote:No....... it,s a problem form Kaikobad sir's book ....& I also solved the problem using function..... I think by this problem,one could be able to understand the real beauty of function ...
$f(-13) \equiv 0$?
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Re: COUNT THE NUMBER OF STUDENTS

Unread post by MATHPRITOM » Mon Jan 30, 2012 1:26 pm

*Mahi* wrote:
MATHPRITOM wrote:No....... it,s a problem form Kaikobad sir's book ....& I also solved the problem using function..... I think by this problem,one could be able to understand the real beauty of function ...
$f(-13)=0$?
No, since ,here d is the number of girls .. we can't think d= -13. here,each student got same point.so,the total score is a function of total students. It was my thought & then, I approached with my thought.

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*Mahi*
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Re: COUNT THE NUMBER OF STUDENTS

Unread post by *Mahi* » Mon Jan 30, 2012 6:00 pm

MATHPRITOM wrote:
*Mahi* wrote: $f(-13) \equiv 0$?
No, since ,here d is the number of girls .. we can't think d= -13. here,each student got same point.so,the total score is a function of total students. It was my thought & then, I approached with my thought.
Residue of $f(x)$ while divided by $x-a$ is $f(a)$, again $P(x)=x^2+10x+17$ is divisible by $x+13$ so $x+13|f(x) \Rightarrow x+13|f(-13) =54$.
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