There are three consecutive integers. Let's call the middle one x. The one before it is x - 1 and the one after it is x + 1. Adding them together equals 3x. That means that whatever they equal has to be a multiple of 3. Since 176 is not divisible by 3, you will not be able to find three consecutive integers with that total.
im not 100 percent sure but i thing its: (x+1)+(x+2)+(x+3)=96
Let x be the first integer. Then the sum of the three consecutive integers is x + (x+1) + (x+2), which equals 3x + 3. We are given that this sum is 43, so we can write the equation 3x + 3 = 43. Solving this equation, we find that x = 13. Therefore, the three consecutive integers are 13, 14, and 15.
There is no set of three consecutive integers for 106.
The sum of the squares of the three consecutive integers 11, 13, 15 = 515
Since they are consecutive and three, they must be in their 20s (81/3 = 20 with a rest 0f 21) So its just finding out which 3 integers sum up to 21 ... and thats ..... 6,7,8
There is no set of three consecutive integers for 187.
im not 100 percent sure but i thing its: (x+1)+(x+2)+(x+3)=96
Three consecutive integers have a sum of 12. What is the greatest of these integers?
Let x be the first integer. Then the sum of the three consecutive integers is x + (x+1) + (x+2), which equals 3x + 3. We are given that this sum is 43, so we can write the equation 3x + 3 = 43. Solving this equation, we find that x = 13. Therefore, the three consecutive integers are 13, 14, and 15.
There is no set of three consecutive integers for 106.
There is no set of three consecutive integers whose sum is 71.
The sum of three consecutive integers is -72
9240 is the product of the three consecutive integers 20, 21, and 22.
If three consecutive integers have the sum of 96, then the problem can be expressed with the equation... N + (N+1) + (N+2) = 96 ...Simplify that and solve and you get... 3N + 3 = 96 3N = 93 N = 31 ... so the three integers are 31, 32, and 33.
The sum of the squares of the three consecutive integers 11, 13, 15 = 515
Since they are consecutive and three, they must be in their 20s (81/3 = 20 with a rest 0f 21) So its just finding out which 3 integers sum up to 21 ... and thats ..... 6,7,8
There must be three consecutive integers to guarantee that the product will be divisible by 6. For the "Product of three consecutive integers..." see the Related Question below.