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Answer :
We can observe that A×A gives A as the unit's digit to the multiplication, that means it's only possible if A=1, 6 and 5.
Checking for all the case:
For A=1: \( \;\;\;\begin{array}
\hline
\;\;\;1 & 1\\
\;\;×& 1\\
\hline
\;\;1 & 1\\
\hline
\end{array}
\)
We can see clearly that it is not similar with the given multiplication so A\(\neq\)1.
For A=5: \( \;\;\;\begin{array}
\hline
\;\;\;1 & 5\\
\;\;×& 5\\
\hline
\;\;7 & 5\\
\hline
\end{array}
\)
We can see clearly that it is not similar with the given multiplication so A\(\neq\)5.
For A=6: \( \;\;\;\begin{array}
\hline
\;\;\;1 & 6\\
\;\;×& 6\\
\hline
\;\;9 & 6\\
\hline
\end{array}
\)
We can see clearly that it is similar with the given multiplication so A=6.