all even integers can be represented by 2n where n is an integer
since odd integers are always 1 more or 1 less than even numbers, an odd integer can be represented as 2n+1 or 2n-1
we want to find 3 consecutive odd integers that have a sum of more than 35 and no more than 55
so solve 35<sum≤55
consecutive odd numbers are 2 apart (3,5,7, etc)
so the 3 odd numbers are 2n-1, 2n+1, 2n+3
their sum is 2n-1+2n+1+2n+3=6n+3
so solve 35<6n+3≤55 for n
minus 3 from everybody
32<6n≤52
divide everybody by 6
5 and 1/3<n≤8 and 1/3
n has to be an integer so the smallest value of n is 6 and the largest allowed is 8
if n=6, then the integers are 2(6)-1, 2(6)+1, 2(6)+3 or 11, 13, 15
if n=7, then the integers are 2(7)-1, 2(7)+1, 2(7)+3 or 13, 15, 17
if n=8, then the integers are 2(8)-1, 2(8)+1, 2(8)+3 or 15, 17, 19
so the 3 odd integers can be any 3 consecutive integers in the set {11,13,15,17,19}
the 3 odd integers can be
a. 11,13,15
b. 13,15,17
c. 15,17,19
(all are sets of consecutive odd integers that satisfy the condition)
