Sum of squares of digit is 50 and left digit is greater than right digit. Find the highest product of digit.

• ans- 7
  let digits of no. be x nd y i.e xy
  as x^2+y^2=50 nd x>y
  for x=7 nd y=1 both condition satisfied
  so 7*1=7
• 10 years agoHelpfull: Yes(7) No(1)
• mine ans will be 7
  
• 10 years agoHelpfull: Yes(4) No(0)
• ansr 7...and the numbr is 71 which satisfy both th gvn condtn
• 10 years agoHelpfull: Yes(2) No(1)
• ans-12
  since 6^2+2^2=50
  6>2
  6*2=12
• 10 years agoHelpfull: Yes(1) No(5)
• Only 7 n 1 satisfy the condition.
  hence number will be 71.
• 10 years agoHelpfull: Yes(0) No(2)
• (7*7)+(1*1)=50
  7*1=7
  my ans is 7.
• 10 years agoHelpfull: Yes(0) No(1)
• 71 bcoz thesqaure is maximum 50..
  it means maximum of one digit can be 7 (square=49) and at that time other digit is 1.
  so it may maximum be
  71 since leftdigit greater than right
• 10 years agoHelpfull: Yes(0) No(0)
• 1^2+2^2#50
  1+9#50
  .........
  1+49=50
  =>no. is 71
  =>highest product=7
• 10 years agoHelpfull: Yes(0) No(0)