SOLUTION: find four consecutive integers such that the sum of the first three is 54 more than the fourth.

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Question 563489: find four consecutive integers such that the sum of the first three is 54 more than the fourth.
Found 3 solutions by mananth, josmiceli, Maths68:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Let the consequtive numbers be x,x+1,x+2,x+3
x+x+1+x+2=x+3+54
3x+3=x+57
3x-x=57-3
2x=54
/2
x=27
27,28,29,30

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the integers +n+, +n+%2B+1+, +n+%2B+2+, and +n+%2B+3+
given:
+n+%2B+n+%2B+1+%2B+n+%2B+2++=+n+%2B+3+%2B+54+
+3n+%2B+3+=+n+%2B+57+
+2n+=+54+
+n+=+27+
+n+%2B+1+=+28+
+n+%2B+2+=+29+
+n+%2B+3+=+30+
The integers are 27,28,29, and 30
check:
+3n+%2B+3+=+n+%2B+57+
+3%2A27+%2B+3+=+27+%2B+57+
+81+%2B+3+=+84+
+84+=+84+
OK

Answer by Maths68(1474) About Me  (Show Source):
You can put this solution on YOUR website!
Let
First integer = x%2B1
Second integer=x%2B2
Third integer = x%2B3
Fourth integer = x%2B4

Given
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The sum of the first three is 54 more than the fourth.
%28x%2B1%29%2B%28x%2B2%29%2B%28x%2B3%29=%28x%2B4%29%2B54
x%2B1%2Bx%2B2%2Bx%2B3=x%2B4%2B54
3x%2B6=x%2B58
3x-x=58-6
2x=52
Divide by 2 both sides of the above equation
2x%2F2=52%2F2
cross%282%29x%2Fcross%282%29=cross%2852%29%2Fcross%282%29
x=26


First integer = x%2B1=26%2B1=27
Second integer=x%2B2=26%2B2=28
Third integer = x%2B3=26%2B3=29
Fourth integer = x%2B4=26%2B4=30



Check
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The sum of the first three is 54 more than the fourth.
27%2B28%2B29=30%2B54
84=84