Question

Find the values of the letters in the following and give reasons for the steps involved.
 \[\begin{array}{r}\underline {\begin{array}{*{20}{l}}A&B\\ \times &6\end{array}} \\\underline {BBB} \end{array}\]

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 6 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.

Complete step-by-step answer:
We are given that when \[B\] is multiplied by 6, we have to get the unit digit again as\[B\].
Now, we have to check the multiplication table of 6 by assigning the digits for\[B\] which satisfies the given criteria.
\[\begin{array}{l}0 \times 6 = 0\\1 \times 6 = 6\\2 \times 6 = 12\\3 \times 6 = 18\\4 \times 6 = 24\\5 \times 6 = 30\\6 \times 6 = 36\\7 \times 6 = 42\\8 \times 6 = 48\\9 \times 6 = 54\end{array}\]
By the multiplication table, we can conclude that when 0, 2, 4, 6, 8 multiplied by 6, we will again get the unit digit as 0, 2, 4, 6, 8
So, the value of \[B\] should be 0 or 2 or 4 or 6 or 8.
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 2, 4, 6, 8, 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 6, we have to get the unit digit as \[B\].
We are given that \[6A = BB\] .
If \[B = 0\], then the product also will be 0. Therefore this value is not possible.
If \[B = 2\], then we have a carry-over of 1 on the next step.
\[\begin{array}{l}6A + 1 = BB\\ \Rightarrow 6A + 1 = 22\end{array}\]
Subtracting 1 from both sides, we get
\[\begin{array}{l} \Rightarrow 6A = 22 - 1\\ \Rightarrow 6A = 21\end{array}\]
The integer value of \[A\] is not possible, so this value of \[B\] is not possible.
If \[B = 4\], then we have a carry-over of 2 on the next step.
\[\begin{array}{l}6A + 2 = BB\\ \Rightarrow 6A + 2 = 44\end{array}\]
Simplifying the equation, we get
\[\begin{array}{l} \Rightarrow 6A = 44 - 2\\ \Rightarrow 6A = 42\end{array}\]
Dividing both side by 6, we get
\[ \Rightarrow A = \dfrac{{42}}{6} = 7\]
This value of \[A\] and \[B\] is possible.
If \[B = 6\], then we have an carry-over of 3 on the next step
\[\begin{array}{l}6A + 3 = BB\\ \Rightarrow 6A + 3 = 66\end{array}\]
Subtracting 3 from both sides, we get
\[\begin{array}{l} \Rightarrow 6A = 66 - 3\\ \Rightarrow 6A = 63\end{array}\]
The integer value of \[A\] is not possible, so this value of \[B\] is not possible.
If \[B = 8\], then we have a carry-over of 4 on the next step.
\[\begin{array}{l}6A + 4 = BB\\ \Rightarrow 6A + 4 = 88\end{array}\]
Simplifying the equation, we get
\[\begin{array}{l} \Rightarrow 6A = 88 - 4\\ \Rightarrow 6A = 84\end{array}\]
Dividing both sides by 6, we get
\[ \Rightarrow A = \dfrac{{84}}{6} = 14\]
This value of \[A\] and \[B\] is possible.
So, the value of \[B\] will be 4 or 8.
But the value of \[A = 14\] is a two digit number which is not possible, so the corresponding value of \[B = 8\] is not possible.
Thus, the value of \[B\] is 4 and the corresponding value of \[A\] is 7.
\[\begin{array}{r}\underline {\begin{array}{*{20}{l}}7&4\\ \times &6\end{array}} \\\underline {444} \end{array}\]
Therefore, the values of the letters are \[A = 7\] and \[B = 4\].

Note: We know that when multiplication is done, multiplication has to start with the unit’s digit, then multiplication has to be followed to ten’s digit. Here, the value of A can be a whole number only and not a fraction. Also, its value can’t be a 2 digit number or 3 digit number, it can be only one-digit number. So, we need to take only one digit whole number for both letters.