# BA*B3=57A

Please find below the solution to the asked query:

Given : BA $\times $ 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 $\times $ 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 $\times $ B3 = ( 10 B + 5 ) ( 10 B + 3) = 575 , So

100 B

^{2}+ 30 B + 50 B + 15 = 575

100 B

^{2}+ 80 B + 15 = 575

100 B

^{2}+ 80 B = 560

10 ( 10 B

^{2}+ 8 B ) = 560

10 B

^{2}+ 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 $\times $ 23 = 575 , A = 5 and B = 2 ( Ans )**

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