AB*BA=(B+1)CBFind the value of A,B,C.

We have  :  AB $×$ BA  = ( B  +  1 )CB

Here we have two digit number multiplied by same two digit with different place value , that gives three digit number .

So, we check

20 $×$02  =  40   ( That only gives us 2 digit number )

19 $×$ 91  = 1729 ( That gives four digit number )

18 $×$ 81  = 1458 ( That gives four digit number )

17 $×$ 71  = 1207 ( That gives four digit number )

16 $×$ 61  = 976 ( That gives three digit number ) , But we can't write 9 = B + 1  where B  = 6

15 $×$ 51  = 765 ( That gives three digit number ) , But we can't write 7 = B + 1  where B  = 5

14 $×$ 41  = 574 ( That gives three digit number ) , Here we can  write 5 = B + 1  where B  =4

SO, we can say

A  =  1  ,  B  = 4  and C  = 7                               (  Ans )

And
13 $×$ 31  = 403 ( That gives three digit number ) , Here we can  write 4 = B + 1  where B  = 3

SO, we can say

A  =  1  ,  B  = 3  and C  = 0                               (  Ans )

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