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

• 7744
  i.e 88*88=7744
• 11 years agoHelpfull: Yes(1) No(0)
• 7744 which i s a perfect square of 88
• 11 years agoHelpfull: Yes(1) No(0)
• 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.
• 11 years agoHelpfull: Yes(1) No(0)
• (7744)^1/2=88
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• ans=7744
  88^2=7744
• 11 years agoHelpfull: Yes(0) No(0)