Question

ABA*A=CBD5; find A+B+C+D

Answer 1

Answer: here all A are same and the product of 2 A must have digit 5 at last. So, 5*5 produce 25

now remainder 2 is remaining.. so if we use B as 9 then 5*9+2 will be 47

again 4 is remainder now multiply last digit 5*5 and add remainder 5*5+4=29

Step-by-step explanation:

595

*5

--------

2975

So A=5

B=9

C=2

D=7

total=23

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