SOLUTION: find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18.

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Question 544329: find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18.
Found 2 solutions by stallingsc, MathTherapy:
Answer by stallingsc(1) About Me  (Show Source):
You can put this solution on YOUR website!
3,21, 19, 61
3+21+19= 43
43+18=61

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18.

Let the first of the four consecutive odd integers = F

Then others are: F + 2, F + 4, and F + 6

We therefore have: F + F + 2 + F + 4 = F + 6 + 18

3F + 6 = F + 24

3F - F = 24 - 6

2F = 18

F, or first of the three odd integers = 18%2F2, or highlight_green%289%29

Other three are: highlight_green%2811_13_and_15%29

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