Question

Find the least number which must be added to 6203 to obtain a perfect square also find the square root of a number so obtained

Answer 1

$6203 + x = y {}^{2} \\ let \: x \: be \: 38 \\ 6203 + 38 = 6241 \\ \sqrt{6241 } = {79 } \\ hence \: the \: number \: to \: be \: added \: is \: 38 \\ and \: th e \: perfect \: square \: is \: 6241 \: nd \: the \: perfect \: square \: nuber \: is \: 79$

Answer 2

            78
          ---------
7 |       62 03 
           49
          ----------
148 |   13 03
           11 84
          -----------
             19
          -----------

78^2 = 6084.

We observe that 78^2 < 6203.

79^2 = 6241.

We observe that 79^2 > 6203.

Hence the number to be added to 6203 is 6241 - 6203 = 38.

6203 + 38 = 3241

                   = 79 * 79

                   = 79.


Therefore 38 should be added to 6203 to obtain a perfect square.


Hope this helps!