Find the least number which must be added to each of the following numbers so as to get a perfect square.
$6412$

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Solution

The square root of $6412$ gives a remainder of divisor $80$ and remainder as $12$
The next number is $81$ ; $81^{2}$ = $6561$
The number to be added to $6412 = 6561 - 6412 = 149$
$\sqrt{6561}$ $= 81$

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

(i) 525 (ii) 1750

(iii) 252 (iv) 1825

(v) 6412

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Find the least number which must be added to each of the following numbers so as to get a perfect square. 6412

The square root of 6412 gives a remainder of divisor 80 and remainder as 12The next number is 81; 812 = 6561The number to be added to 6412=6561−6412=149√6561 =81

Find the least number which must be added to each of the following numbers so as to get a perfect square.
$6412$

The square root of $6412$ gives a remainder of divisor $80$ and remainder as $12$
The next number is $81$ ; $81^{2}$ = $6561$
The number to be added to $6412 = 6561 - 6412 = 149$
$\sqrt{6561}$ $= 81$