Number system: Product of a $3$ digit and a $2$ digit number

Question

There is a three digit number $abc$ and a two digit number $xy$ such that the product of $abc$ with $xy$ is same as the product of $cba$ and $yx$,if $x$,$y$ can be similar digits, then find number of such pairs of $P$ and $Q$?

a) 99

b) 900

c) 810

d) cannot be determined

Answer 1

Suggeston: If you write out abc as 100a+10b+c and similarly with the others, you can write an explicit equation that your numbers need to satisfy. Then if you impose the conditions that all variables are naturals, $0\lt a\lt 10$ and so on, you can look for a solution. A computer search would be easy. Without a computer, looking for divisibility requirements is your friend on problems like this.

Added: None of the proposed answers is correct. I get 843 possibilities with $x\le y$. 810 of them are of the form xx * aba. The others are ( 12 , 231 ) ( 12 , 462 ) ( 12 , 693 ) ( 13 , 341 ) ( 13 , 682 ) ( 14 , 451 ) ( 15 , 561 ) ( 16 , 671 ) ( 17 , 781 ) ( 18 , 891 ) ( 23 , 352 ) ( 24 , 231 ) ( 24 , 462 ) ( 24 , 693 ) ( 25 , 572 ) ( 26 , 341 ) ( 26 , 682 ) ( 27 , 792 ) ( 28 , 451 ) ( 34 , 473 ) ( 35 , 583 ) ( 36 , 231 ) ( 36 , 462 ) ( 36 , 693 ) ( 39 , 341 ) ( 39 , 682 ) ( 45 , 594 ) ( 46 , 352 ) ( 48 , 231 ) ( 48 , 462 ) ( 48 , 693 ) ( 68 , 473 ) ( 69 , 352 )

So if we count these both ways the total would be 810+2*33=876