Bangladesh IMO TST 1: 2011/ NT-2 (P 3)

[Moon]

Problem 3:
$N$ is a $5$ digit number, of which the first and the last digits are nonzero. $N$ is a palindromic product if

• $N$ is a palindrome (it reads the same way from the left to right or right to left such as $12321$).
• $N$ is a product of two positive integers, of which the first, when read from left to right, is equal to the second, when read from right to left. For example, $20502$ is a palindromic product, since $102\cdot 201=20502$ and $20502$ itself is a palindrome.

Determine all palindromic products with $5$ digits.

[Masum]

Let the $5$-digit number be $10^4a+10^3b+10^2c+10b+a=(100d+10e+f)(100f+10e+d)$
$\Longrightarrow 1010b+100c=10001(df-a)+1010e(d+f)+100(f^2+d^2+e^2)$
Note that the equation is symmetric over $d,f$.So it is enough to take only the pairs $(d,f)$ or $(f,d)$.
Now $a\ge df$ or $a<df$ whatever,the side in which $10001.|df-a|$ exists,will be larger.So $a=df$,and the equation is reduced to $101(b-ed-ef)=10(f^2+e^2+d^2-c)$ which implies that $10| |b-e(d+f)|\le 9<10$.Then $b=e(d+f),c=d^2+f^2+e^2\le 9$.Because none of $d,f$ is $0$,so $e,f,d< 3$.Let's deal with the cases.
$e=0$.In this case we have from the last equation,$10|b$ or $b=0$.Also $f^2+d^2=c\le 9\Longrightarrow f=d=2,f=1,d=2,f=d=1.$.We get solutions $102,101,202$
Now see when $e=1$.We shall avoid the solutions we have already found.Then,$f^2+d^2\le 8$ and we get $f=d=1,f=1,d=2,f=d=2$.Therefore the solutions we get are $111,112,212$
See what if $e=2.$Then $f^2+d^2\le 5$,we conclude that $f=1,d=2$ and the solution found is $122$

So all such $Palindromic\ Products$ are $20502,10201,40804,12321,23632,44944,26962$.