Question

Find the least number, that must be subtracted from 16142 to obtain a perfect square. Also, find the
square root of the perfect square.​

Answer 1

To solve this problem, we need the closest least perfect square.

The square root is more than 400 and less than 401. So, the desired number is 400.

16142 must be subtracted by 142 to get 400². Also, its square root is 400.

Answer 2

Answer:

We know that, if we subtract the remainder from the number, we get a perfect square.

$\Large\begin{array}{r |l| l}</p><p></p><p>\sf1& \sf\bar1\bar{61}\bar{42} &\sf127\\</p><p></p><p> & \sf1 \\</p><p></p><p>\cline{1-2} \sf22 &\:\:\sf61 \\</p><p></p><p>& \:\:\sf44 \\</p><p></p><p>\cline{1-2} \sf247 &\:\:\sf1742\\</p><p></p><p>&\:\:\sf1729 \\\cline{2-2} &\:\:\:\:\sf13\end{array}$

Here, we get the remainder 13. Therefore 13 must be subtracted from 16142 to get a perfect square.

⇒ 16142 - 13

⇒ 16129

⠀

$\therefore\:\underline{\textsf{Least number that must be subtracted is \textbf{13}}}.$

$\rule{200}{1}$

• Perfect Square :

$\Large\begin{array}{r |l| l}</p><p></p><p>\sf1& \sf\bar1\bar{61}\bar{29} &\sf127\\</p><p></p><p> & \sf1 \\</p><p></p><p>\cline{1-2} \sf22 &\:\:\sf61 \\</p><p></p><p>& \:\:\sf44 \\</p><p></p><p>\cline{1-2} \sf247 &\:\:\sf1729\\</p><p></p><p>&\:\:\sf1729 \\\cline{2-2} &\:\:\:\:\sf{xx}\end{array}$

⠀

$\therefore\:\underline{\textsf{It is perfect square of \textbf{127}}}.$