Solve three consecutive odd integers have a sum more than 35 and no more than 55

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
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 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)

Integers 13, 15, and 17 equal 45, if that's what you're asking ...