In a hotel there are 5 rooms which are square in shape. on their floor square tiles of size 1X1 are fitted. on the diagonals of the room black tiles are fitted and remaining are white. 101,111,121,131,141 these are the no. of black tiles of all five rooms. now find which No's of black tiles are not possible out of the given no?

• for n*n sq. room there will be (2n-1) black tiles if n is odd no.
  & if n is even, 2n black tiles will be there of size 1*1
  
  101 = 51*2-1
  111 = 56*2 -1(not possible as n is even), 112 is possible for 56*56
  121 = 61*2-1
  131 = 66*2-1 (not possible)
  141 = 71*2-1
• 9 years agoHelpfull: Yes(39) No(4)
• @Soumen Maity, it is ques of elitmus held on 5th july. 101,111,121,131,141 are the no. of black tiles of each of the five rooms. we have to find which no's from this five no. can not be possible for black tiles as all rooms are square and black tiles are on the diagonals of the floor of the rooms.
• 9 years agoHelpfull: Yes(1) No(1)
• 101,121,131
• 9 years agoHelpfull: Yes(0) No(4)
• Sry, I didn't understand the question..... how you calculating the no. of tiles(like 101,111,121...)... is it any kind of dimension???...... someone pls explain
• 9 years agoHelpfull: Yes(0) No(0)
• 111 and 131
• 9 years agoHelpfull: Yes(0) No(0)
• Rooms are in square shape. Black tiles make the diagonal.
  According to euclid...
  a^2 + b^2 = c^2 (ex. 3^2 +4^2 = 5^2 pytho...theorem) only exists if it follow 1(modula 4). Fundamental of number theory.
  all a,b,c should be Positive Integer.
  Options that does not follow rule not a part of the answer.
• 9 years agoHelpfull: Yes(0) No(1)
• ans is=4*R root(1+pi)
  
  using the pathagorus theoram.
  
  we will find the radius of the circle.
  R sq.=L/2 sq.+B/2 sq.
  using (a+b) sq.= a sq.+ b sq. + 2 a.b
  
  get ans.
• 7 years agoHelpfull: Yes(0) No(0)