Let x represent the first number x+1 represent the second number x+2 represent the third number x+3 represent the fourth number x + (x+1) + (x+2) + (x+3) = 2 4x + 6 =2 << group the like terms 4x + 6 - 6 = 2-6 <<subtract 6 on both sides 4x = -4 4x/4 = -4/4 << divide by 4 x=-1 first number : -1 second number: - 1 + 1 = 0
third number : -1 + 2= 1 fourth : -1 + 3= 2
The numbers are 32, 33, 34 and 35.
The numbers are -7, -6, -5 and -4.
The numbers are 55, and 57. Two consecutive integers have an odd sum.
There are no such numbers. If S is the sum of any 4 consecutive integers then S = 2 (mod 4) In other words, any four consecutive integers add up to an even number that is NOT dvisible by 4.
The numbers are 27, 28, 29 and 30.
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.