Find the value of A , B and C in the following multiplication:
A B*5= C A B
A=
B=
C=
Dear student,
The multiplication of B and 5 is giving a number whose ones digit is B again.
This is possible when B=5 or B=0
In case of B=5 , the product , B×5=5×5=25
2 will be carry for next step
We have 5×A+2=[CA] which is possible for A=2 or 7
Then Multiplications will be
→25×5=125 (for A=2)→75×5=375 (for A=7)
If B=0
B×5=0×5=0 There will be no carry in this step
In the next step , 5×A=[CA]
It is possible only when A=5 or A=0
But A cannot be 0 as AB is a two digit number
Hence A can be 5 only
Then multiplication will be: 50×5=250
Hence the possible value for A,B and C are:
A=5,B=0 and C=2
Regards