Latest self Aptitude Question SOLUTION: Find a four digit number which is a perfect sqaure and its first two digit are same and last two digit are also same.

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(#M40020109) SELF question Number system Keep an EYE

Find a four digit number which is a perfect sqaure and its first two digit are same and last two digit are also same.

sanjay kumar saha (10 years ago)

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ABHISHEK KUMAR MISHRA  (10 years ago)

7744
i.e 88*88=7744

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FAISAL RIZWAN  (10 years ago)

7744 which i s a perfect square of 88

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sanjay kumar saha  (10 years ago)

if square of number 1 to 25 is known than it is very easy to calculate square of number greater than 25.

example find square of number 39?
39-25=14; 25-14=11 square of 39 = 11^2+14*100=1521
above rule is applicable of number greater than 25 and upto 50
example find square of number 78?
78-50=28; 50-28=22 square of 78 = 22^2+28*200=6084
above rule is applicable for number greater than 50 and upto 100
now see carefully last two digit of square of any number greater than 25 is remain same and it is equal to last two digit of number less than 25.
see square of number 1 to 25 only 12^2=144 whose last two digit is same i.e 44
now apply above rule in reverse:- 50-12=38; 50+38=88 and 25-12=13;25+13=38
that means only square of 38 and 88 is the number whose last two digit is same.
find square of 88:- 88-50=38; 50-38=12
square of 88 = 12^2+38*200=7744
find square of 38:- 38-25=13; 25-13=12
square of 38 = 12^2+13*100= 1444
hence answer is 7744.

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ashiwani kumar singh  (10 years ago)

(7744)^1/2=88

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Chandni Maurya  (10 years ago)

ans=7744
88^2=7744

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