BA*B3=57A
Dear Student,
Please find below the solution to the asked query:
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 )
Hope this information will clear your doubts about Playing with numbers.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Please find below the solution to the asked query:
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 )
Hope this information will clear your doubts about Playing with numbers.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards