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Three consecutive odd integers have a sum of 33

User Haehn
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2 Answers

2 votes
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.
User Parth Verma
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4 votes
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.
User Claviska
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