Question

The value of A is: -
   A A A
+ B B B
+ C C C
B A A C
(a) 9
(b) 7
(c) 6
(d) 1

Answer 1

Hint: Use the properties of place values of a digit in a number to add the unit digits, tens, and hundred digits separately. Consider the case of the carry process in the given sum and form three linear equations in A, B and C. Solve these equations for the value of A to get the answer.

Complete step by step answer:
Here, we have been provided with the sum: -
   A A A
+ B B B
+ C C C
B A A C
We have to find the value of A.
Now, we can see that here A, B, and C are actually single-digit numbers, i.e. face value, and their place values are different. In the number AAA, three A’s are present at different places namely – ones, tens, and hundreds of places. Similar is the case with the numbers BBB and CCC. In the resultant sum, we have the number BAAC in which the digit B is at thousand places, two A’s are at a hundred and tens place and C is present at one’s place. So, let us consider the sum of these numbers.
We can see that at the units place we have a sum of digits A, B and C whose resultant is C. Since, none of the digits can be 0 so this sum is only possible if we have a carry.
Therefore, we have,
\[\Rightarrow \] A + B + C = C + 10
This is because the unit digits add up to form a ten in the case of carry. So, we have,
\[\Rightarrow \] A + B = 10 – (1)
Now, this ten will add up with tens of place digits and form a hundred in the case of carry. So, considering tens place digits, we have,
\[\Rightarrow \] 10A + 10B + 10C + 10 = 10A + 100
\[\Rightarrow \] 10B + 10C + 10 = 100
\[\Rightarrow \] B + C + 1 = 10
\[\Rightarrow \] B + C = 9 – (2)
Finally, this hundred will add up with hundred place digits and form a thousand in the case of carry. So, considering hundred place digits, we have,
\[\Rightarrow \] 100A + 100B + 100C + 100 = 100A + 1000B
\[\Rightarrow \] 100B + 100C + 100 = 1000B
\[\Rightarrow \] B + C + 1 = 10B
\[\Rightarrow \] 9B = C + 1 – (3)
Now, we have three variables A, B, and C, and three equations, so solving these equations for the value of A, we get,
\[\Rightarrow \] A = 9
Hence, option (a) is the correct answer.

Note:
One may note that we have solved the question using properties of the number system and face value, place value of the digits. There can be an algebraic method also to solve the question. What we will do is we will write AAA as 100A + 10A + A = 111A and similarly, BBB as 111A and CCC as 111C. We will take their sum to equate with 1000B + 100A + 10A + C = 1000B + 110A + C. Now, we will find the value of A in terms of B and C and form such a combination that it will give the value of A as a single integer from it will give the value of A as a single integer from 1 to 9. This will be done by selecting the values of B and C by hit and trial method.