Find the least number which must be added to the following number so as to get a perfect square:
$6412$

52

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149

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63

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112

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Solution

The correct option is B $149$
$6412$ lies between two perfect square nos. i.e $80^{2} = 6400$ and $81^{2} = 6561$
Here we need $161 \times 1 = 161$
Required difference $= 161 - 12 = 149$
So, $6412 + 149 = 6561$ is a perfect square.

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Similar questions

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 the following number so as to get a perfect square:6412

The correct option is B 1496412 lies between two perfect square nos. i.e 802=6400 and 812=6561Here we need 161×1=161Required difference =161−12=149So, 6412+149=6561 is a perfect square.

Find the least number which must be added to the following number so as to get a perfect square:
$6412$

The correct option is B $149$
$6412$ lies between two perfect square nos. i.e $80^{2} = 6400$ and $81^{2} = 6561$
Here we need $161 \times 1 = 161$
Required difference $= 161 - 12 = 149$
So, $6412 + 149 = 6561$ is a perfect square.