Question 168829
find 4 consecutive intergers such that the sum of the first three exceeds the 4th by 18. 
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Let x = first of 4 consecutive integers
then 
x+1 = second of 4 consecutive integers
x+2 = third of 4 consecutive integers
x+3 = fourth of 4 consecutive integers
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Then from "sum of the first three exceeds the 4th by 18" we get:
x+(x+1)+(x+2) = (x+3)+18
3x+3 = x+21
2x+3 = 21
2x = 18
x = 9
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Answer:
9, 10, 11, and 12