Question

please solve as soon as possible ☺️​

Answer 1

Answer:

Let the number be II, III, IIII, ....

To disprove that last two digits cannot be odd for any square

As the last digit is are the last digit of number must be either 1(0)9

Let number be xy. then y = 1(0)9

when y = 1 then tens digit is (x+x) (i.e even)

when y = 9 then Here digits is 9x + 8 + 9x

=18x+8 (i.e even)

So we can say that a square cannot end

with tow odd digits.

∴II,III,IIII,... cannot be a square