Question

 The smallest no. that must be subtracted from 2800 so that the obtained no.

is a perfect square

Answer 1

In order to solve this question, we'll follow a strategy to minimise time requirement.

First of all we have to find the closest perfect square on both the limits of 2800.

i.e. we all $50^{2}$ = 2500;

$51^{2}$ = 2601;

$52^{2}$ = 2704;

$53^{2}$ = 2809;

We can clearly see that $52^{2}$ < 2800 < $53^{2}$

but, we have to subtract something so we'll choose 2704 as closest.

Therefore, the smallest no. that must be subtracted from 2800 so that the obtained no.  is a perfect square = 96. [Ans]

Answer 2

First of all find the closest perfect square on both the limits of 2800 that we have = 2500; 2601; 2704; 2809;

We can clearly see that 2809 is greater than 2800  and thus we have to subtract something so we'll choose 2704 as closest.

Therefore, the smallest no. that must be subtracted from 2800 so that, obtained number is a perfect square = 96