Prothom Alo 4/1/2012 page 6 column 5

Latest News, Announcements, and Forum Rules
User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:
Prothom Alo 4/1/2012 page 6 column 5

Unread post by Phlembac Adib Hasan » Wed Jan 11, 2012 9:34 pm

In an advertisement titled "A difficult sum" Mr. Alamgeer Hossain gave a math problem and told that it may be included in Guiness Book of World. He also claimed nobody was able to solve that till then and if anyone can solve it ,then he or she must announce it before 31/1/2012.The problem and my solution are given below. I am asking everyone to solve it and to post solution here or contact with Mr. Alamgeer Hossain:01920779889

Problem: তিন অঙ্ক বিশিষ্ট একটি সংখ্যার একক ও শতক স্থানীয় অঙ্কদ্বয়ের পরস্পর স্থান বিনিময়ে প্রাপ্ত সংখ্যাটি মূল সংখ্যা থেকে বিয়োগ করলে যে ধনাত্মক বিয়োগফল পাওয়া যায় তার অঙ্ক ত্রয়ের সমষ্টি অপেক্ষা মূল সংখ্যার অঙ্ক ত্রয়ের সমষ্টি 1 বেশি। বিয়োগফলটির একক স্থানীয় অঙ্কটি ইহার শতক স্থানীয় অঙ্ক অপেক্ষা 1 বেশি এবং বিয়োগফলের শতক স্থানীয় অঙ্কটি মূল সংখ্যার একক স্থানীয় অঙ্ক থেকে এক বেশি হলে সংখ্যাটি নির্ণয় করুন।
(বি.দ্র. x,y,z ধরিয়া সহসমীকরণের মাধ্যমে সমাধান করিতে হইবে এবং সমাধানের কোথাও x,y,z-এর মান 1,2,3,4,5,6 ইত্যাদি ধরা যাইবে না)

My solution:
Let the number is $100x+10y+z$.So after interchanging first and last digit the number will become $100z+10y+x$.
After subtraction the answer will be ,
$100x+10y+z-(100z+10y+x)$
$=100x-100z+10y-10y-x+z$
$=100(x-z)-100+90+10-x+z$
$=100(x-z-1)+10*9+(10-x+z)$
It means from the right, the digits are $(10-x+z),9$ and $(x-z-1)$
From the first condition we get,
$(x-z-1)+9+(10-x+z)+1=x+y+z$
$\Rightarrow x-x-z+z-1+1+9+10=x+y+z$
$\Rightarrow x+y+z=19$........................................(i)
From second condition,
$10-x+z=(x-z-1)+1$
$\Rightarrow 10-x+z=x-z$
$\Rightarrow 2(x-z)=10$
$\Rightarrow x=z+5$............................................(ii)
From third condition,
$x-z-1=z+1$
$\Rightarrow x=2z+2$...........................................(iii)
From (ii) and (iii),
$z+5=2z+2$
$\Rightarrow 2z-z=5-2$
$\Rightarrow z=3$
After putting value of $z$ into (ii) we find,
$x=z+5$
$\Rightarrow x=3+5$
$\Rightarrow x=8$
After putting value of $x,z$ into (i) we get,
$x+y+z=19$
$\Rightarrow 8+y+3=19$
$\Rightarrow y+11=19$
$\Rightarrow y=8$
So the number is,
$100*8+10*8+3$
$=800+80+3$
$=883$
Ans:$883$.
Last edited by Phlembac Adib Hasan on Thu Jan 12, 2012 10:41 am, edited 1 time in total.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
Masum
Posts:592
Joined:Tue Dec 07, 2010 1:12 pm
Location:Dhaka,Bangladesh

Re: Prothom Alo 4/1/2012 page 6 column 5

Unread post by Masum » Wed Jan 11, 2012 11:02 pm

What did he say? I am so curious. Was the title "A difficult sum?" In english?
One one thing is neutral in the universe, that is $0$.

sourav das
Posts:461
Joined:Wed Dec 15, 2010 10:05 am
Location:Dhaka
Contact:

Re: Prothom Alo 4/1/2012 page 6 column 5

Unread post by sourav das » Wed Jan 11, 2012 11:57 pm

Solution:
Let $n=(abc)_{10}$
$p=(abc)_{10}-(cba)_{10}=99(a-c)$
If $a-c=1$then $p=99$ contradiction;
So let $p=100q+10m+j$
According to problem:
$j=q+1..(i)$
$c=q-1..(ii)$
$a+b+c=q+m+j+1\Rightarrow b=q+m-a+3$
Now,
$99(a-q+1)=99q+10m+2q+1$ So, $99|10m+2q+1$
Only possible case $10m+2q+1=99$ So, $5m+q=49$ As must be $m\geq 8$ ;$q\in\{ 9,4 \}$.
If $q=9$ then $j=10$ contradiction. Setting values, $q=4,m=9$: ans is 883
Adib, are you sure about the problem and article. Can anyone give any link of that article? Adib, do you call him?You should be the first one.....CONGRATZZZZ
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

User avatar
nafistiham
Posts:829
Joined:Mon Oct 17, 2011 3:56 pm
Location:24.758613,90.400161
Contact:

Re: Prothom Alo 4/1/2012 page 6 column 5

Unread post by nafistiham » Thu Jan 12, 2012 12:45 am

actually, it was not an article, just an advertisement in bangla named একটি কঠিন অংক "A difficult sum" .it also included Mr. alamgir hosen's stamp size photo.the advertisement him as the inventor of $3$ triple s problem like this one.as it was an advertisement you can not see in the website.i would post the picture of the column if i had not cut and given it to Adib.

I don't think something like mathematics should be used for business like this.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

sourav das
Posts:461
Joined:Wed Dec 15, 2010 10:05 am
Location:Dhaka
Contact:

Re: Prothom Alo 4/1/2012 page 6 column 5

Unread post by sourav das » Thu Jan 12, 2012 6:32 am

I think all kind of business with education should be banned.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:

Re: Prothom Alo 4/1/2012 page 6 column 5

Unread post by Phlembac Adib Hasan » Thu Jan 12, 2012 10:26 am

sourav das wrote:Solution:
Can anyone give any link of that article? Adib, do you call him?You should be the first one.....CONGRATZZZZ
Thanks.Sorry,I can't give you its link because the online copy of Prothom Alo doesn't publish all the advertisements.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
FahimFerdous
Posts:176
Joined:Thu Dec 09, 2010 12:50 am
Location:Mymensingh, Bangladesh

Re: Prothom Alo 4/1/2012 page 6 column 5

Unread post by FahimFerdous » Thu Jan 12, 2012 11:27 am

The problem is actually an exercise and it's super easy. I did it just when I saw it. I don't know what Mr. Alamgir thinks of himself.
Your hot head might dominate your good heart!

sakibtanvir
Posts:188
Joined:Mon Jan 09, 2012 6:52 pm
Location:24.4333°N 90.7833°E

Re: Prothom Alo 4/1/2012 page 6 column 5

Unread post by sakibtanvir » Thu Jan 12, 2012 8:29 pm

:lol: The problem and the article is a funny joke...... :lol: :lol: :lol: :lol:

User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:

Re: Prothom Alo 4/1/2012 page 6 column 5

Unread post by Phlembac Adib Hasan » Thu Jan 12, 2012 8:35 pm

FahimFerdous wrote:The problem is actually an exercise and it's super easy. I did it just when I saw it. I don't know what Mr. Alamgir thinks of himself.
I also agree with you.Actually I needed only 10 minutes.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: Prothom Alo 4/1/2012 page 6 column 5

Unread post by *Mahi* » Thu Jan 12, 2012 11:42 pm

sakibtanvir wrote::lol: The problem and the article is a funny joke...... :lol: :lol: :lol: :lol:
I agree :)
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

Post Reply