Question

Find the values of the letters in the following and give reasons for the steps involved.
\[\begin{array}{r}\underline {\begin{array}{*{20}{l}}A&B\\ \times &3\end{array}} \\\underline {CAB} \end{array}\]

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.
\[\begin{array}{r}\underline {\begin{array}{*{20}{l}}A&0\\ \times &3\end{array}} \\\underline {CA0} \end{array}\]
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
\[\begin{array}{r}\underline {\begin{array}{*{20}{l}}5&0\\ \times &3\end{array}} \\\underline {150} \end{array}\]
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.