Question

Find the least number which must be added to $6412$ so as to get a perfect square. Also find the square root of the perfect square so obtained.

• A. $147;81$
• B. $147;83$
• C. $149;81$
• D. $149;83$

Answer 1

The correct option is C $149;81$

$6412$
$\mid 80\overline{8\mid \overline{64}\overline{12}}\mid -64\overline{160\mid 012}\mid -00\overline{\mid 12}$
Since remainder $=12$
$\Rightarrow {80}^2<6412<{81}^2$
$\therefore$ Required number to be added to make it a perfect square is
${81}^2-6412=6561-6412=149$
$\therefore 6412+149=6561$
and $\sqrt{6561}=81$.