Question

Find the digit A in
\[\begin{array}{*{20}{c}}
  {}&2&2&2&2 \\
  {}&{}&{}&9&9 \\
   + &{}&{}&{}&9 \\
   + &{}&A&A&A \\
\hline
  {}&2&9&9&A
\end{array}\]

Answer 1

Hint:The given sum can be written as $2222+99+9+AAA=299A$. We replace the numbers which has the unknown digit A with their decimal expansion using the fact that the decimal expansion of an $n$ digits number with digits ${{d}_{n}},{{d}_{n-1}},....{{d}_{1}},{{d}_{0}}$ say $N={{d}_{n}}{{d}_{n-1}}....{{d}_{1}}{{d}_{0}}$ is ${{d}_{n}}\times {{10}^{n}}+{{d}_{n-1}}\times {{10}^{n-1}}+....{{d}_{1}}\times {{10}^{1}}+{{d}_{0}}\times {{10}^{0}}$.

Complete step by step answer:
The given puzzle is

\[\begin{array}{*{20}{c}}
  {}&2&2&2&2 \\
  {}&{}&{}&9&9 \\
   + &{}&{}&{}&9 \\
   + &{}&A&A&A \\
\hline
  {}&2&9&9&A
\end{array}\]

We observe that there are four numbers that are being added. The first number is 2222, the second number is 99, the third number is 9, the fourth number is AAA. The sum of the four numbers is written at the bottom as 299A. Here we are asked to find the unknown digit A.
We know that form the expansion in number system of a number with $n$ digits is denoted as $N={{d}_{n}}{{d}_{n-1}}....{{d}_{1}}{{d}_{0}}$ where ${{d}_{n}},{{d}_{n-1}},....{{d}_{1}},{{d}_{0}}$ are the digits with base 10 can be expanded as
\[\begin{align}
  & N={{d}_{n}}{{d}_{n-1}}....{{d}_{1}}{{d}_{0}} \\
 & ={{d}_{n}}\times {{10}^{n}}+{{d}_{n-1}}\times {{10}^{n-1}}+....{{d}_{1}}\times {{10}^{1}}+{{d}_{0}}\times {{10}^{0}} \\
\end{align}\]
We call the digit ${{d}_{0}}$ the digit at the unit place, ${{d}_{1}}$ the digit at the hundredth place and the digit ${{d}_{2}}$ as the digit at the thousandth place and so on. We can expand the fourth number from the top AAA with unknown digit A with decimal expansion. We see that the number cosmists of three digits. So We expand for $n=3$ and have ,
\[AAA=A\times {{10}^{3-1}}+A\times {{10}^{2-1}}+A\times {{10}^{0}}=111A\]
Similarly we can expand the sum ate bottom 299A with decimal expansion as it also includes the unknown digit A . So We expand for $n=3$ and get ,
\[299A=2\times {{10}^{4-1}}+9\times {{10}^{3-1}}+9\times {{10}^{2-1}}+A\times {{10}^{0}}=2990+A\]
We can write the given addition in a line as
\[2222+99+9+AAA=299A\]
We replace the numbers which has the unknown digit A with their decimal expansion and get ,
\[\begin{align}
  & 2222+99+9+111A=2990+A \\
 & \Rightarrow 2330+111A=2990+A \\
\end{align}\]
We subtract A both side of the equation to get ,
\[\begin{align}
  & \Rightarrow 2330+110A=2990 \\
 & \Rightarrow 110A=2990-2330=660 \\
\end{align}\]
We divide both side of the equation to get by 110 and get $A=6.$

Note:
We note that the question is silent about the number system where the given numbers belong to. It cannot be a binary or octal number system as 9,A are involved. It can be Hexadecimal but in hexadecimal $A=10$ not a digit . So we reject that. The binary expansion is $N={{d}_{n}}{{d}_{n-1}}....{{d}_{1}}{{d}_{0}}={{d}_{n}}\times {{2}^{n}}+{{d}_{n-1}}\times {{2}^{n-1}}+....{{d}_{1}}\times {{2}^{1}}+{{d}_{0}}\times {{2}^{0}}$.