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

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

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)

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step-by-step explanation:

two? xd

step-by-step explanation: