Simply consider what ARE the sums of four consecutive non-negative integers: 0+1+2+3 = 6
1+2+3+4 = 10
2+3+4+5 = 14
3+4+5+6 = 18
.... You can see that it's any number of the form: 6 plus a (non-negative integer) multiple of 4.
A more mathematical proof is to consider an arbitrary starting number n, then: n+(n+1)+(n+2)+(n+3) = 4*n + 6 (which is the same as concluded above).
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.
33, 34, 35, 36
The sum of the first four non-negative, consecutive, even integers is 20.
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.
6,7,8,9
The sum is four.
There are four consecutive odd integers: 81, 83, 85 and 87.
-4, -2, 0, and 2 are the four consecutive even integers. When you add them up they equal -4.