Question 56851:  Find three consecutive odd integers such that the sum of the smallest and 4 times the largest is 61. 
 Answer by Scriptor(36)      (Show Source): 
You can  put this solution on YOUR website! Call your first odd number 2n+1 , the second one 2n+3 and the last one 2n+5
 
Since the sum of the smallest and 4 times the largest is 61, we get that:
 
(2n+1) + 4(2n+5) = 61
 
2n + 1 + 8n + 20 = 61
 
10n = 40
 
n =4  => 2n+1 = 9
 
Answer: 9,11,13
 
Note: we could have solved it with x instead of 2n+1, but I wanted to show clear that it is an odd number. 
  
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