There is no set of three consecutive integers whose sum is 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.
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).
71 + 72 + 73 = 216
They are 8 and 9
* 71 * 73 * 75
35 and 36
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.
Let x represent the first integer. The second consecutive integer is then x + 1. The equation can be written as x + (x + 1) = 71.
The numbers are 69, 70, and 71.
The integers are 67, 69 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 two consecutive odd integers be ( x ) and ( x + 2 ). The product of the integers is ( x(x + 2) ) and their sum is ( x + (x + 2) = 2x + 2 ). The equation based on the problem statement is ( x(x + 2) = 7(2x + 2) + 71 ). Solving this gives ( x^2 + 2x = 14x + 14 + 71 ), leading to ( x^2 - 12x - 85 = 0 ), which factors to ( (x - 17)(x + 5) = 0 ). Thus, the integers are 17 and 19.