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put this solution on YOUR website! the sum of four consecutive even integer is 2 more than five times the first integer. Find the smallest integer.
First (smallest) consecutive even integer = n
Second consecutive even integer = n+2
Third consecutive even integer = n+4
Fourth consecutive even integer = n+6
n+(n+2)+(n+4)+(n+6) = 5n+2
n+n+2+n+4+n+6 = 5n+2
4n+12 = 5n+2
10 = n
Checking: The 4 consecutive even integers are 10, 12, 14, and 16.
Their sum is 10+12+14+16 = 52.
Five times the first is 5×10 = 50
2 more than 50 is 52.
So the answer is correct.
Edwim
