If (AB)^2=CAD, where A,B,C & D are distinctive integers what is the value of X
I was wondering how to find x! And I think you mean "distinct" integers, not "distinctive" integers. There is no single solution to this.
You are given (AB)^2= CAD. Of course, (AB)^2= A^2B^2 so A= (CD)/B^2. Since CD/B^2 is an integer, CD must be divisible by B^2.
To get as small as possible take B= 3, C= 2(9)= 18 and D= 1. Then A= 18(1)/9= 2. AB= 6 so (AB)^2= 36 and CAD= (18)(2)(1)= 36.
Yet another possibility is to take B= 2, C= 2(2)= 4 and D= 2(3)= 6. Then A= 24/4= 12. AB= 2(12)= 24 and (AB)^2= 25^2= 576. CAD= 4(12)6= 48(6)= 576. Ycan take B to be any integer, take C and D to be that integer multiplied by different integers, (or either C or D to be B^2 multiplied by another number, while the other is 1) and then solve A= CD/B^2.
For example, if I choose C to be 2(3)= 6 rather than 4 and D= 2(5)= 10, A= (6)(10)/2^2= 60/5= 15. Now AB= 300 and (AB)^2= 900. Of course, CAD= (6)(15)(10)= (90)(10)= 900 also.