Question

Show that the square of any number in the form $xxx..376$ will also be in the form $yyyy..376$.
${{\left( xxx..376 \right)}^{2}}=yyy....376$

Answer 1

Hint: To solve this question, we should use the multiplication of numbers. We are given a number $xxx..376$ and we are asked about its square. The multiplication of the numbers $abcd$ and $pqrs$is done using the long multiplication process. It is
$\begin{align}
  & \underline{\begin{align}
  & abcd\text{ }\times \\
 & pqrs \\
\end{align}} \\
 & \ \ \ s\left( abcd \right)\ + \\
 & \ r\left( abcd \right)\otimes + \\
 & q\left( abcd \right)\otimes \otimes + \\
 & . \\
 & . \\
\end{align}$

Complete step-by-step solution:
For example, the multiplication of 345 and 11 can be done as
$\begin{align}
  & \underline{\begin{align}
  & 345\ \times \\
 & \ \ 11 \\
\end{align}} \\
 & \ \underline{\begin{align}
  & \ \ 1\left( 345 \right) \\
 & 1\left( 345 \right)\otimes \\
\end{align}} \\
 & \ \underline{\begin{align}
  & \ \ 345+ \\
 & 345\otimes \\
\end{align}} \\
 & \ 3795 \\
\end{align}$
We need to get the last three digits in the square of $xxx..376$. By using the above method, we can get the required digits.
We are given a number whose last three digits are 376. The number is $xxx..376$. We are asked about the last three digits of the given number. To get this, we use the method of long multiplication. The multiplication of the numbers $abcd$ and $pqrs$is done using the long multiplication process. It is
$\begin{align}
  & \underline{\begin{align}
  & abcd\text{ }\times \\
 & pqrs \\
\end{align}} \\
 & \ \ \ s\left( abcd \right)\ + \\
 & \ r\left( abcd \right)\otimes + \\
 & q\left( abcd \right)\otimes \otimes + \\
 & . \\
 & . \\
\end{align}$
For example, the multiplication of 345 and 11 can be done as
$\begin{align}
  & \underline{\begin{align}
  & 345\ \times \\
 & \ \ 11 \\
\end{align}} \\
 & \ \underline{\begin{align}
  & \ \ 1\left( 345 \right) \\
 & 1\left( 345 \right)\otimes \\
\end{align}} \\
 & \ \underline{\begin{align}
  & \ \ 345+ \\
 & 345\otimes \\
\end{align}} \\
 & \ 3795 \\
\end{align}$
The gist of the above process is that we add an empty space after every single multiplication of the number $abcd$ and the digits in $pqrs$. This process can be applied to any number of digits. Let us consider the given number $xxx..376$. Squaring it, we get
\[\begin{align}
  & \\
 & \underline{\begin{align}
  & xxx..376\ \ \times \\
 & xxx..376 \\
\end{align}} \\
 & \ \underline{\begin{align}
  & \ \ \ \ \ \ 6\left( xxx..376 \right) \\
 & \ \ \ 7\left( xxx..376 \right)\otimes \\
 & 3\left( xxx..376 \right)\otimes \otimes \\
\end{align}} \\
 & \ \underline{\begin{align}
  & \ \ \ \ \ \ yyy...2256+ \\
 & \ \ \ yyy....2632\otimes + \\
 & \ yyy....1128\otimes \otimes \ \ \ \ \ \ \ \\
\end{align}} \\
 & \ yyy.....141376 \\
\end{align}\]
We can observe that the last three digits are 376.
$\therefore $Hence, we proved that ${{\left( xxx..376 \right)}^{2}}=yyy....376$.

Note: Another way to prove the given statement is by writing the given number as
$xxx....376=xxx...000+376$
Squaring on both sides, we get
${{\left( xxx....376 \right)}^{2}}={{\left( xxx...000+376 \right)}^{2}}$
We can use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and calculate the last three digits in the R.H.S to prove that the last three digits turn out to be 376.