Find the number of triangles with Exactly one side odd and perimeter = 203

• Ans: 101
  
  We can presume that, one side has odd length and sum of the lengths of other two sides is even formed by sum of two even lengths.
  Combinations of the sides A & (B+C) can be (1, 202), (3,200), (5,198), ..... (201,2)
  Now to calculate number of triangles, we can consider the series of increasing length 1, 3, 5, ...., 201 which has 'n' terms, where first term a=1, last term=201 & difference d=2
  Therefore, Last term=a + (n-1)d
  => 201 = 1+(n-1)2
  201 = 2n - 1
  n = 101
  Hence, required number of triangles=101
• 5 years agoHelpfull: Yes(0) No(1)
• total soln.
  a + b + c = 203
  a + b > c
  b + c > a
  c + a > b
  
  
  203/2 = 101.5
  
  so we can have the max value of the sides = 101
  
  (101 - a) + (101 - b) + (101 - c) = 203
  a + b + c = 100
  102c2
  we want distinct soln.
  
  a b c
  0 0 100
  1 1 98
  ....
  50 50 0
  
  51 solutions will be of type aab
  and repeated 3 times.
  (102c2 - 51*3)/3! + 51 = 884
  
  all odd we need to exclude that...
  2a + 1 + 2b + 1 + 2c + 1 = 203
  a + b + c = 100
  100/2 = 50
  (50 - a) + (50 - b) + (50 - c) = 100
  a + b + c = 50
  52c2
  
  0 0 47
  ...
  25 25 0
  26 sol
  26 * 3 = 78
  
  (52c2 - 26 * 3)/3! + 26 = 234
  
  884 - 234 = 650
• 5 years agoHelpfull: Yes(0) No(0)