such that the s f the first and second is 31 less than 3 times the third. find the integers

Find  three consecutive odd integers such that the sum of the first and second is 31 less than 3 times the third.

Let teh 3 numbers be n1, n2, n3.

And n3 = n2 + 2, n2 = n1 + 2  (since they are consecutive (odd) integers)

We are told that,

n1 + n2 = n3 - 31

n1 + (n1 + 2) = (n1 + 4) + 31

n1 = 33

n2 = 35

n3 = 37

by Level 11 User (81.5k points)
n1 + n2 = 33 + 35 = 68
n3 - 31 = 37 - 31 = 6   ????
Even n3 + 31 is wrong.
The problem says (3 * n3) - 31
(3 * 37) - 31 = 80
three consecutive odd integers such that the sum of the first and second is 31 less than 3 times the third. find the integers

The numbers are n, n+2 and n+4

(3 * (n+4)) - 31 = n + (n+2)

3n + 12 - 31 = 2n + 2

3n - 19 = 2n + 2

3n - 2n = 2 + 19

n = 21

n+2 = 23

n+4 = 25

n + (n+2) = 21 + 23 = 44

(3 * (n+4)) - 31 = (3 * 25) - 31 = 75 - 31 = 44

The three numbers are 21, 23 and 25  (sorry about that, Fermat)
by Level 11 User (78.4k points)
Well spotted CWA. Thanks for giving the correct solution.