7.
\(\;\; \begin{array} \hline \;\;\;A & B\\ \times& 6\\ \hline B\;\;B & B\\ \hline \end{array} \)


Answer :

We can observe that: Possible values of BBB are 111, 222, 333, etc.
Let us divide these numbers by 6 and check the nummber which is completely divisible.

111 ÷ 6 =18, remainder 3 so, 111 is rejected.

222 ÷ 6 = 37, remainder 0 but the quotient 37 is not of the form A 2. So,222 is rejected.

333 ÷ 6 = 55, remainder 3 so, 333 is rejected. 444 ÷ 6 = 74, remainder 0 and also the quotient 74 is of the form A 4.

So, we can say the number B=4 is satisfying.Let's check and find value of A

\( \begin{array} \hline \;\;\;A & 4\\ \times& 6\\ \hline 4\;\;4 & 4\\ \hline \end{array} \)

We know this is only possible if A=7

Hence, we have got A=7 and B=4

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