solve this mathematically without assumption or by hit and trial method ?

let B = 10x + y

then, A = (x+y)^2

now According to question,

x^2 + y^2 + 2xy - 10x - y = 27

rearranging the eqn we get:

x^2 + y^2 + 25 + 2xy - 10x - 10y = 52 - 9y

or

(x+y-5)^2 = 52 - 9y

or

x+y-5 = (52-9y)^0.5

now x,y lies in the set {0,1,2,3,4,5,6,7,8,9}

so x+y-5 must be an integer..

and hence (52 - 9y) must be a perfect square..

hence we get y=3,4(corresponding values of (52 - 9y are 25 and 16)...

if y=3

x=7

if y=4

x=5

Hence there are two such values of B

they are 73 and 54.

Hope you got this man.

plz vote as best answer and complete my double century. :)

If we denote B with xy, the value of B is 10*x + y (living in the decimal system). x and y are digits between 0, 1, 2...9, x not being zero (for B to be two digit). Knowing

A = (x+y)^2, A = 27+B = 27 + 10*x + y we have

(x+y)^2 = 27 + 10*x + y = 27 + 9 *x + (x+y),

or (x+y)^2 - (x+y) = 27 + 9*x

if introduce x+y=z that is

z*(z-1) = 9*(3+x)

As 1=<x=<9 and bearing in mind that z is a positive, this yields

6.52 =< z =< 10.9. More, z should be such an integer, that z*(z-1) is divided by 9. That gives us it might be 9 or 10.

From the above if z = 9 x = 5 y = 4

' if z = 10 x = 7 y = 3

Thus there are two solutions B = 54 or B = 73.

Let the tenth digit of the number B = x

and the unit's digit of the number B = y

Hence number B= 10x+y

Number A =square of the sum of two digits of B =(x+y)²

A−B=(x+y)²−(10x+y)=27

and (x+y)²−10x−y=27

x can be ≤ 9 and y ≤9

x+y ≤ 18

(x+y)² ≥ 27 hence x+y ≥ 6

B=A−27 since B is a two digit number A≤126

and since A is a perfect square hence A= 121 , 100, 81, 64, 49, 36,25,16

Since A > by 27

Hence A = 121 , 100, 81, 64, 49, 36

and B = 94, 73, 54 ,22,

Since (x+y)² =

(x+y)²=(9+4)² =169

(x+y)²=(7+3)² =100

(x+y)²=(5+4)² =49

(x+y)²=(2+2)² =16Because B < A

Hence B can Be 94 or 73

Since 169 −27 > 100

Hence B can be 73 or 54

Heat And Trial Method