Dhaka Secondary 2009/2

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Secondary 2009/2

Unread post by BdMO » Fri Jan 21, 2011 6:12 pm

If $(DBC)^2=BCABC$ find the value of $D$?

Sudip Deb
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Re: Dhaka Secondary 2009/2

Unread post by Sudip Deb » Tue Jan 25, 2011 7:15 pm

D = 2 . Because DBC = 276

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leonardo shawon
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Re: Dhaka Secondary 2009/2

Unread post by leonardo shawon » Tue Jan 25, 2011 9:10 pm

what is DBC? Where did it come?
Ibtehaz Shawon
BRAC University.

long way to go .....

Sudip Deb new
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Re: Dhaka Secondary 2009/2

Unread post by Sudip Deb new » Wed Jan 26, 2011 10:00 am

FroM (DBC)^2

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Moon
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Re: Dhaka Secondary 2009/2

Unread post by Moon » Tue Feb 01, 2011 11:48 pm

Sudip: Please post complete solution (or at least an outline on how to solve it). Posting just the answer is strongly discouraged as we can not learn anything from a number, can we?
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

Shifat
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Re: Dhaka Secondary 2009/2

Unread post by Shifat » Thu Aug 25, 2011 2:03 am

Moon vai I am posting a soultion,theoritical though,ar pura puri observation er upor prove kora,(observation gular nirghat prove ase but ami janina)so viewers please look for a better solution -

ami amar koyekta observation mention kori( shobari egula jana so skip korleo chole):-
1.last digit 5 hole shetar square(je kono power) e last digit alltime 5 hoy and second last digit hoy 2
2. last digit 6 hole shetar khetreo last digit (1) follow kore but second last digit e ei sequence ta paoa jay
a) second last digit jodi 0 hoy tahole square er por 2nd last digit hobe 3(ofcourese)
b)last two digit 16 hole square er por 56 hoy
c)last two digit 26 hole 76 paoa jai
d)last two digit 36 hole 96 paoa jai ... blah blah blah
point hoilo last two digit jodi (0,1,2,3,4)6 type hoi tahole 2nd last digit hoy 3,5,7,9,1.....(2)
same happenns when(5,6,7,8,9)type hoy tahole loop akare again 3,5,7,9,1 ashe(3)...(program e deksi,really ashe, but kan ashe thikmoton akhono prove kore dekhate pari nai) eita apnara try koren pls....(ei kothagular aktao ami exma hall e liktam na)
$COME$ $TO$ $THE$ $SOLUTION$-
DBC three digit but BCABC two digit , and $(300)^2=90000, 32\times300=9600 + 256= 9856$, kajei number ta 316 xceed korte parbe na.so D(max)=3
metioned above, $(DBC)^2=BCABC$ er mane hoilo c should be either 5 or 6...
case 1
if $C=5$ then
$B=2$ hobe so the equation will be like $(D25)^2= 25A25, 25^2=625$, so $A=6$
Akhon D= 1 hoile B er value 1,2,3 er moddhe, B=2 kajei D er value 1 hobe. Akhon D= 1
hole $(125)^2$ er small part - $10000+2.100.25= 10000+5000=15000+$ ja ashe tai.....
So D= 1 hole B= 2 hobe na, so c=5 hobe na
Case 2-
if c=6 then
b= 1,3,5,7,9 er kisu akta... (2) and (3) explore kore pai
b= 7 hole $BC^2$ er second last digit 7 orfe last 2 digit $76$, kajei $BCABC= 76A76$
akhon D=1 hole B= 1,2,3 , but b=7, D er value 2 hole B er value $[2^2,3^2)=4,5,6,7,8$
So $D= 2$ hole or $DBC =276$ holei all condition fulfil hoy....
SO the result is $D=2$..........

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nafistiham
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Re: Dhaka Secondary 2009/2

Unread post by nafistiham » Fri Jan 06, 2012 7:12 pm

A little information.
if $N=.........25$ then, $N^n=.........25$
and, if $N=..........76$ then $N^n=.........76$
these are the only two numbers of two digits like this.

from 'আমি তপু' by Muhammad Zafar Iqbal
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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