**SQUARE A 3 DIGIT NUMBER WHICH ARE NEAR THE NO WHOSE LAST TWO DIGITS ARE 00 (2)

if the given no is more 
RULE:-
we have to find (ABC)^2
first, subtract ABC - A*100= BC
the first part of the result will be (ABC+BC)*A
and the last two digits will be square of the subtraction result
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:-
508^2=?
508-500=8
the first part of the result will be (508+8)*5=516*5=2580
and the last two digits will be 8^2=64
so the result is 258064