Since the sum is -22, the 4 integers are all negative. I first began with smaller integers. If you choose -4, -5, -6 , -7. The sum is -22. To check, add -4 + -5 + -6 + -7. I am unsure of your level, but there is a way to solve it algebraically. I choose n to be the first number. The next number is also negative, and that will be n - 1. The next number will be n - 2 and then the fourth consecutive integer is n - 3. The sum will be n + (n - 1) + (n - 2) + (n - 3) = -22 4n - 6 = -22 4n = -16 n = -4 Therefore the first number is -4, then -5, -6 and finally -7.
The numbers are 51, 53, 55 and 57.
-2, -1, 0, 1
No two consecutive integers have a sum of 2012.
The numbers are 93 and 95.
The numbers are -187 and -186.
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.
The numbers are 51, 53, 55 and 57.
-2, -1, 0, 1
The numbers are 82, 84, 86 and 88.
This is impossible with positive integers. However, four numbers separated by a difference of 1 with a sum of -92 are: -21.5 -22.5 -23.5 -24.5
The numbers are 32, 33, 34 and 35.
The numbers are -7, -6, -5 and -4.
The numbers are -11, -10, -9 and -8.
The numbers are -12, -11, -10, and -9.
The numbers are 22, 23, 24 and 25.
No two consecutive integers have a sum of 2012.
The numbers are 55, and 57. Two consecutive integers have an odd sum.