Question

Find the values of the letters in the following and give reasons for the steps involved.
$$\begin{array}{r}\underset{―}{\begin{array}{ll}A& B\\ \times & 3\end{array}}\\ \underset{―}{CAB}\end{array}$$

Answer 1

Hint: Here, we have to use the concept of multiplication to find the values. We will first find the value of B by using the multiplication table of 3 and choosing the value based on the given condition. Then, we will find the value of A using the same multiplication table and applying the condition. We will then substitute the value of A and B in the given multiplication to find the value of C.

Complete step-by-step answer:
We are given that when $$B$$ is multiplied by 3, we have to get the unit digit again as $$B$$.
Now, we have to check the multiplication table of 3 by assigning the digits for $$B$$ which satisfies the given criteria.
 $$\begin{array}{l}0\times 3=0\\ 1\times 3=3\\ 2\times 3=6\\ 3\times 3=9\\ 4\times 3=12\\ 5\times 3=15\\ 6\times 3=18\\ 7\times 3=21\\ 8\times 3=24\\ 9\times 3=27\end{array}$$
By the multiplication table, we get that
When 0 and 5 multiplied by 3, we will again get the unit digit as 0 and 5.
So, the value of $$B$$ should be 0 or 5.
If the value of $$B$$ is 0, then we will not be having any carry-over which has to be added to the next digit.
 If the value of $$B$$ is 5, then we will have carry-over which has to be added to the next digit.
Since, we have a condition that when $$A$$ is multiplied with 3, we have to get the unit digit again as $$A$$.
So, the value of $$B$$ is 5 should be neglected.
Thus the value of $$B$$ is 0.
$$\begin{array}{r}\underset{―}{\begin{array}{ll}A& 0\\ \times & 3\end{array}}\\ \underset{―}{CA0}\end{array}$$
Next, we are given that when $$A$$ is multiplied by 3, we have to get the unit digit again as$$A$$.
Now, we have to check the multiplication table of 3 by assigning values of $$A$$ which satisfies the given criteria.
 $$\begin{array}{l}0\times 3=0\\ 1\times 3=3\\ 2\times 3=6\\ 3\times 3=9\\ 4\times 3=12\\ 5\times 3=15\\ 6\times 3=18\\ 7\times 3=21\\ 8\times 3=24\\ 9\times 3=27\end{array}$$
By the multiplication table, we get that
When 0 and 5 multiplied by 3, we will again get the unit digit as 0 and 5.
So, the value of $$A$$ should be 0 or 5.
If the value of $$A$$ is 0, then we will not be having any carry-over which has to be added to the next digit.
If the value of $$A$$ is 5, then we will be having any carry-over which has to be added to the next digit.
Since, we have a condition that when $$A$$ is multiplied with 3, we have to get $$C$$ and $$A$$.
When 0 is multiplied with 3, we will get 00, but we don’t have both $$C$$ and $$A$$ equal.
Thus the value of $$A$$ is 5
$$\begin{array}{r}\underset{―}{\begin{array}{ll}5& 0\\ \times & 3\end{array}}\\ \underset{―}{150}\end{array}$$
Thus, the value of $$C$$ is 1.
Therefore, the value of the letters are $$A=5,B=0$$ and $$C=1$$.

Note: We know that when multiplication is done, multiplication has to be started with the unit’s digit, then multiplication has to be followed to ten’s digit. Here, we might make a mistake by taking the value of $$B$$ as 5 because if we multiply 5 by 3 then we will get a product as 15. So, 1 will be the carry over to $$A$$, then multiplying 3 to $$A$$ and adding 1 to it will not give us the desired result. So, we need to choose the digit carefully and according to the given condition.