**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

**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

** SOME DIVISION RULES FOR REPEATION OF THE NUMBERS

if a number is repeated 2 or its multiplied no of times then it will be divided by 11
if a number is repeated 6 or its multiplied no of times then it will be divided by 3/7/11/13/37
if two number is repeated 3 or its multiplied no of times then it will be divided by 3/7/13/37
if 0 is repeated at-least 2 times at the last then it will be divided by 4
if 0 is repeated at-least 3 times at the last then it will be divided by 8

(we know the every rules but when the combination is used then we forgot about these formulas)

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EX:-

#) 44444....4 here 4 is repeated 429 times.if it is divided by 1144.then what will be the remainder?
Ans:- here 1144=8*11*13
now 4444....4 if it is repeated 426 times and the last three digits are 000
then this will be divided by 1144 or (11*13*8) as 426 divided by 6 and as last three digits are 0 so it will be divided by 8
so remainder is 444

#) 1212...12 here 12 is repeated 283 times.if it is divided by 1036. then what will be the remainder?
Ans:- here 1036= 4*7*37
now 1212...12 if it is repeated 282 times and the last two digits are 00
then this will be divided by 1036 or (4*7*37) as 282 is divided by 3 and as last two digits are 0 so it will be divided by 4
so remainder is 12

**SQUARE A NUMBER WHICH ARE NEAR 100(2)

if the given no is more than 100 
first take that no
now first subtract 100 from that no
the first part of the result will be (the given no+the subtraction result )
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


.......................................................................................
EX:-
108^2=?
108-100=8

here first part is 108+8=116
and last two digit will be 8^2=64
so the ans will be=11664

116^2=?
116-100=16
here first part is 116+16=132
and last two digit will be 16^2=256
132+2=134
so the ans will be=13456

**SQUARE A NUMBER WHICH ARE NEAR 100(1)

if the given no is less than 100 
first take that no
now first subtract that number from 100
the first part of the result will be (the given no-the subtraction result )
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

.......................................................................................
EX:-
97^2=?
100-97=3
so first part will be 97-3= 94
and last two digit will be 3^2=09
so the ans will be 9409

84^2=?
100-84=16
so the first part will be 84-16=68
and the last two digit will be 16^2= 256
68+2=70
so the ans will be 7056