Question

Three consecutive odd integers have a sum of 33

Answer 1

Let the odd integers be:

x,  x + 2,  x + 4          Since consecutive odd integers differ by 2.

Sum = 33

x + (x + 2) + (x + 4) = 33

x + x + 2 + x + 4 = 33

x + x + x + 2 + 4 = 33

3x + 6 = 33

3x = 33 - 6

3x = 27          Divide by 3

x = 27/3

x = 9

Recall the odd integers were, x, x + 2, x + 4

= 9,   9 + 2,  9 + 4

= 9,  11,  13

The three consecutive integers are 9, 11, & 13.

I hope this helps.

Answer 2

Three consecutive integers that add up to 33 are 2 above and lower than a middle number.  This means that the average of these 3 numbers will be the same.  Therefore, 33/3=11 as the average.  Therefore 11 is the middle number, 9 is 2 below (a consecutive odd integer) and 13 is 2 above (a third consecutive odd integer).  Therefore the 3 numbers are 9, 11, and 13.