35 and 36
35 and 36
The integers are 67, 69 and 71.
Let x represent the first integer. The second consecutive integer is then x + 1. The equation can be written as x + (x + 1) = 71.
* 71 * 73 * 75
219/3 = 73 so the three are 73-2 = 71, 73 and 73+2 = 75.
71+72=143
There is no set of three consecutive integers whose sum is 71.
35 and 36
The integers are 67, 69 and 71.
Let x represent the first integer. The second consecutive integer is then x + 1. The equation can be written as x + (x + 1) = 71.
* 71 * 73 * 75
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.
40+31 41+30 42+29 43+28 44+27 45+26 these are some pairs of numbers that equal 71, although there 35 ways.
They are 8 and 9
219/3 = 73 so the three are 73-2 = 71, 73 and 73+2 = 75.
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.
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.