# BA*B3=57A

Dear Student,

Given :  BA $×$ B3  =  57A

First we see that unit's place of the number we get when 3 is multiplied by A, must be A. This can happen in one case only and that is When A =  5 As :

5 $×$ 3  =  15 , Here we see the unit's digit is same as 5. So A = 5.

Now ,  BA = ( 10B + A ) = ( 10 B + 5 )  and B 3 = ( 10 B + 3)

And we get

B5 $×$ B3 = ( 10 B + 5 ) ( 10 B + 3) =  575 , So

100 B2 + 30 B + 50 B + 15 =  575

100 B2 + 80 B + 15 = 575

100 B2  + 80 B = 560

10 ( 10 B2 + 8 B ) =  560

10 B2 + 8 B = 56

Here we used hit and trial method and substitute B  =  2 and get

10 ( 2 )2   + 8 ( 2 ) = 56

10 ( 4 ) + 8 ( 2 ) = 56

40 + 16 =  56

56 = 56

So,

B  =  2

Therefore,

25 $×$ 23 =  575 , A  = 5  and B  =  2                                             ( Ans )