Top Answer

There is no set of three consecutive integers whose sum is 71.

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0* 71 * 73 * 75

35 and 36

The equation would be 35+36=71.

Divide the sum of the three consecutive odd integers by 3: 183/3 = 61. The smallest of these integers will be two less than 61 and the largest will be two more than 61, so the three consecutive odd integers will be 59, 61, 63.

The numbers are 71, 72 and 73.

The integers are 67, 69 and 71.

The numbers are 69, 70, and 71.

35 and 36

219/3 = 73 so the three are 73-2 = 71, 73 and 73+2 = 75.

71+72=143

In order to do any consecutive integer problem, you should think of the numbers as x's. Here's what I did to solve this problem (and what you can do to solve future problems): (x) + (x +1) + (x +2) + (x + 3) = 290 4x + 6 = 290 4x = 284 x = 71, therefore the four numbers are 71, 72 (71 + 1), 73 (71 + 2), and 74 (71 + 3).

Let the first integer be x. then x + (x+1) + (x+2) + (x+3) = 290 4x + 6 = 290 4x = 284 x = 71 so the nanswer is 71+72+73+74 = 290

They are are 70, 71 and 72.

They are 70, 71 and 72.

The numbers are 69, 70 and 71.

They are: 71+(-24) = 47

The sum of the digits of 171 is 9.

Let x = first integer2nd integer = x + 13rd integer = x + 2x + (x +1) + (x + 2) = x - 713x + 3 = x - 713x = x - 742x = -74x = -371st Integer = -372nd Integer = -363rd Integer = -35(-37) + (-36) + (-35) = -108-108 + 71 = -37

There are no two consecutive integers that add up to 72. You can prove it this way: Let our numbers be represented by the variables "a" and "b". We are told: a = b + 1 a + b = 72 So we can use substitution to solve for either variable: (b + 1) + b = 72 2b + 1 = 72 2b = 71 b = 35.5 But 35.5 is not an integer, so this condition can not be met.

71 + 72 + 73 = 216

69, 70, 71

The numbers are 69, 71 and 73.

The sum of 7 plus 71 is 77.

The complex conjugate pair 2.5 - i*0.5*sqrt(71) and 2.5 + i*0.5*sqrt(71) where i is the imaginary number representing the square root of -1.

71 is a single number and so there is no "between" possible!

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