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Three consecutive integers will always total a multiple of three. Specifically, they will always add up to three times the middle number.

Divide 48 by 3. Choose the the integers on either side of that.

15 + 16 + 17 = 48

The three numbers could be described as x, x + 1 and x - 1. That adds up to 3x.

3x = 48

x = 16

x + 1 = 17

x - 1 = 15

Q: How do you solve the equation for three consecutive integers whose sum is 48?

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Three consecutive integers are [X], [X+1], and [X+2].x + (x+1) + (x+2) = 963x + 3 = 963x = 93x = 31The numbers are 31, 32, and 33.

You can find those by trial and error. You can also write an equation for the three consecutive integers, and solve it. If the first number is "n", the others are "n + 1" and "n + 2". By solving the equation for "n", you get the first of the three numbers.

Your question is not well formed, but i assume you mean 3 consecutive integers that sum to -363. If that is the case solve the following equation: (n-1) + (n) + (n+1) = -363 to give you the middle integer.

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.

Divide by 3. The quotient is the middle of the three numbers. A harder way of doing it, but probably the expected way, is n + (n + 1) + (n + 2) = 33 Solve for n (n is the first of your three numbers).

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That isn't possible; three consecutive integers, or three consecutive positive integers, always have a sum that is a multiple of 3. In general, you can solve this quickly by trial and error. In this case, you will quickly find that a certain set of three consecutive integers will give you a sum that is TOO LOW, while the next-higher even integers will give you a sum that is TOO HIGH. You can also write an equation and solve it: n + (n + 2) + (n + 4) = 32. If you solve it, you will find that the solution is fractional, not integral.

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.

Three consecutive integers are [X], [X+1], and [X+2].x + (x+1) + (x+2) = 963x + 3 = 963x = 93x = 31The numbers are 31, 32, and 33.

You can find those by trial and error. You can also write an equation for the three consecutive integers, and solve it. If the first number is "n", the others are "n + 1" and "n + 2". By solving the equation for "n", you get the first of the three numbers.

Your question is not well formed, but i assume you mean 3 consecutive integers that sum to -363. If that is the case solve the following equation: (n-1) + (n) + (n+1) = -363 to give you the middle integer.

It isn't possible to solve that. The sum of three consecutive integers is always a multiple of 3 (try it out with a few small numbers); 125 is not a multiple of 3, ergo, the problem has no solution.

There is no set of three consecutive integers for 187.

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.

You can do this by trial-and-error. Or, give the lowest of the four consecutive integers a name, like "x". The three other integers will then be "x+2", "x+4", and "x+6". So, you have to solve the equation:x + (x + 2) + (x + 4) + (x + 6) = 4The answer is the lowest of the four consecutive even integers.You can do this by trial-and-error. Or, give the lowest of the four consecutive integers a name, like "x". The three other integers will then be "x+2", "x+4", and "x+6". So, you have to solve the equation:x + (x + 2) + (x + 4) + (x + 6) = 4The answer is the lowest of the four consecutive even integers.You can do this by trial-and-error. Or, give the lowest of the four consecutive integers a name, like "x". The three other integers will then be "x+2", "x+4", and "x+6". So, you have to solve the equation:x + (x + 2) + (x + 4) + (x + 6) = 4The answer is the lowest of the four consecutive even integers.You can do this by trial-and-error. Or, give the lowest of the four consecutive integers a name, like "x". The three other integers will then be "x+2", "x+4", and "x+6". So, you have to solve the equation:x + (x + 2) + (x + 4) + (x + 6) = 4The answer is the lowest of the four consecutive even integers.

Divide by 3. The quotient is the middle of the three numbers. A harder way of doing it, but probably the expected way, is n + (n + 1) + (n + 2) = 33 Solve for n (n is the first of your three numbers).

There is no set of three consecutive integers for 106.