**SQUARE A 3 DIGIT NUMBER WHICH ARE NEAR THE NO WHOSE LAST TWO DIGITS ARE 00 (1)

if the given no is less 
RULE:-
we have to find (ABC)^2
first, subtract that no from it's next nearest no whose last two digits are 0
suppose it becomes X
(  that means X=[(A+1)*100-ABC]  )
then, first part of the result will be (ABC - X)*(A+1)
and last two digits of the result will be X^2
if the square is more than two digit then put last two digits at last and add the extra with first part

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EX:-
(298)^2=?
now it's next nearest no is 300
so 300-298=2
so first part will be =(298-2)*(2+1)=296*3=888
and last two digits will be 2^2= 04
so ans is 88804

482^2=?
it's next nearest no is 500
so 500-482=18
so first part will be= (482-18)*(4+1)=464*5=2320
and the last two digits will be 18^2=3 24
so the ans is 232 (0+3) 24=232324