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No. There is no set of three consecutive even integers with a sum of 40.
4, 6, 8, 10
No.Let the four numbers be (n-1), n, (n+1), (n+2).Their sum is 4n + 2 = 2(2n + 1)If 2000 is the sum of four consecutive integers, then:2(2n+1) = 2000⇒ 2n + 1 = 1000⇒ 2n = 999but 999 is odd, not even and so n cannot be an integer; therefore 2000 is not the sum of four consecutive integers.
There are two consecutive even integers. The numbers are 118 and 120.
The sum is four.