How many two-digit numbers are there which, when subtracted from the number formed by reversing its digits as well as when added to the number formed by reversing its digits, result in a perfect square?
option
a) 1
b) 2
c) 3
d) 4

• Only one 56,65
  56+65=121
  65-56=9
• 10 years agoHelpfull: Yes(15) No(11)
• 10x+y be the 2 digit number then
  10x+y+10Y+x(reverse) == 11(x+y)
  and 10x+y - (10y+x)== 9(x-y)
  solving x and y trail error method.. only one satisfies both the equations 6 and 5
  ans: 1
  
• 10 years agoHelpfull: Yes(15) No(3)
• @DANISH But how can u say only one. Please explain.
• 10 years agoHelpfull: Yes(9) No(1)
• Let the number xy = 10x + y
  Given that, 10x+y - (10y - x) = 9(x-y) is a perfect square
  So x-y can be 1, 4, 9.  -------- (1)
  So given that 10x+y +(10y +x) = 11(x+y) is a perfect square.
  So x+y be 11. options are (9,2), (8,3),(7,4),(6,5) hence only ( 6,5 ) satisfy the condition. .therefore ans. Is 1
• 8 years agoHelpfull: Yes(3) No(0)
• there is only one such number 65. which gives 65+56=121=11^2 & 65-56=3^2.
• 8 years agoHelpfull: Yes(0) No(0)