The first degit of the square of the two digit number is 8. How many number is satisfied the above condition and how?

• ans=29,90,91,92,93,94 .
  because 29^2=841
  90^2=8100
  91^2=8281
  92^2=8464
  93^2=8649
  94^2=8836
  
• 12 years agoHelpfull: Yes(9) No(6)
• No. Number is satisfied:
  Any two digit No. if squared, the unit digit cannot become 8.
  For example,
  UNIT DIGIT OF
  THE TWO DIGIT
  NO.(IF) 1 2 3 4 5 6 7 8 9 0
  UNIT DIGITS
  OF THE SQUARE:1 4 9 6 5 6 9 4 1 0
• 12 years agoHelpfull: Yes(7) No(5)
• 6 two digits numbers are possible.
  
  They are,
  29,90,91,92,93,94
  
  29^2=841
  90^2=8100
  91^2=8281
  92^2=8464
  93^2=8649
  94^2=8836
• 12 years agoHelpfull: Yes(5) No(4)
• there are only 5 two digit number who satisfy the given condition. They are 90,91,92,93,94. As it is given that first digit should be 8 means MSB is 8. when we multiply 9 with 9 we get MSB as 8. For two digit is second digit should be less than 5 oterwise MSB increases by 1 becomes 9.
• 12 years agoHelpfull: Yes(4) No(1)
• 6.
  explaination-
  29,90,91,92,93,94
• 9 years agoHelpfull: Yes(3) No(2)