Find the values of the letters in each of the following and give reasons for the steps involved


The multiplication of 3 and B gives a number whose ones digit is B again

Therefore, B must be 0 or 5


Let B is 5


Multiplication of first step = 3 × 5 = 15


Now,


1 will be a carry for the next step


3 × A + 1 = CA


This is not valid for any value of A


Hence, B must be 0 only.


If B = 0, then there won’t be any carry for the next step


We will get, 3 × A = CA


That is, the one's digit of 3 × A should be A


This is possible when A = 5 or 0


But, A cannot be 0 as AB is a two-digit number


Therefore, A must be 5 only.


The multiplication is as follows:



Hence, the values of A, B, and C are 5, 0, and 1 respectively


5
1