Call the lowest of the even integers n. Then from the problem statement, n + n + 2 + n + 4 = n + 6 =244. Collecting like terms results in 4n + 12 = 244, or 4n = 244 - 12 = 232, or n = 232/4 or 58. The consecutive even integers are therefore, 58, 60, 62, and 64.
21, 22, 23
The numbers are 51, 53, 55 and 57.
-2, -1, 0, 1
The sum of two consecutive integers will always be an odd number.
The sum is four.
58, 60, 62, 64
The smallest is 121.
The sum of the first four non-negative, consecutive, even integers is 20.
Call the lowest of the even integers n. Then from the problem statement, n + n + 2 + n + 4 = n + 6 =244. Collecting like terms results in 4n + 12 = 244, or 4n = 244 - 12 = 232, or n = 232/4 or 58. The consecutive even integers are therefore, 58, 60, 62, and 64.
The smallest is -1
If n is the smallest of the four integers, their sum is 4n+6.
No. The sum of four consecutive integers is two odd numbers plus two even numbers which is an even number. 2001 is an odd number, therefore it cannot be the sum of four consecutive numbers.
There are none.
There is a set of four consecutive even integers whose sum is four. The set is: -2, 0, 2 and 4.
14, 15, 16 and 17.
They are odd consecutive integers: 21, 23, 25 and 27.