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.
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......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
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.
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.