Find three consecutive odd integers such that twice the sum of the three numbers is 33 or more than the third number

Celeste Barajas

Celeste Barajas

Answered question

2022-11-03

Find three consecutive odd integers such that twice the sum of the three numbers is 33 or more than the third number

Answer & Explanation

barene55d

barene55d

Beginner2022-11-04Added 23 answers

Let the three consecutive odd numbers be: ( 2 n - 1 ) , ( 2 n + 1 ) , ( 2 n + 3 ) for some n
We are told that twice the sum of these numbers is at least equal to 33 + ( 2 n + 3 )
Thus, considering the limiting case, 2 ( 2 n - 1 + 2 n + 1 + 2 n + 3 ) = ( 2 n + 3 ) + 33
2 ( 6 n + 3 ) = ( 2 n + 3 ) + 33
12 n + 6 = 2 n + 36
10 n = 30
n = 3
Replacing n=3 into our sequence of odd numbers { 5 , 7 , 9 }
To check: 5+7+9=21
2 × 21 = 42
42 - 9 = 33

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