33, 34, 35, 36
Here are two ways to solve this: Intuitively: three consecutive even integers with a sum of 42 must be centered around 42/3=14. The answer is 12, 14, and 16, which can be verified by adding them: 12+14+16=42 Or algebraically: With N the lowest even integer: N+N+2+N+4=42 3N=36 N=12 So the answer is 12, 14, and 16.
The negative part is just a trick because any number squared, whether positive or negative, will turn out to be a positive number. But, to satisfy the question... -6*-6=36 -8*-8=64 36+64=100 tada. -6 and -8 would be the answer.
integers are either 1 and 8, 2 and 7, 3 and 6 or 4 and 5 Squares are 1 + 63, 4 + 49, 9 + 36 and 16 + 25 Only 4 + 49 sums to 53 so your answer is 2 and 7
Let the first lowest number be x, then x + x + 1 + x + 2 + x + 3 = 138 4x + 6 = 138 4x = 138 - 6 4x = 132 x = 33 and the number are 33, 34, 35, and 36, add them and you will get 138
11,12,13
The integers are -14, -12 and -10.
10 12 14 are the consecutive even integers which add up to 36
The integers are -37, -36 and -35. Also, using consecutive even integers: -38, -36 and -34.
-13, -12, & -11.
10 12 14
The two consecutive integers whose sum is 73 are 36 and 37.
36
The answer is 12 because I took 36 divided by 3 and got 12
35 and 36
36 and 38
The consecutive odd integers for 114 are 37, 38 and 39.