Find three consecutive even integers such that the sum of the smallest and largest is 36
16,18,20
2n ,2n+2 , 2n+4
2n + 2n+4 = 36
4n+4=36
4n=36-4
4n=32
n=8
2n= 16
2n+2= 18
2n+4=20
Why did the even integers go to therapy? They were feeling odd! Anyway, let's solve this problem. Let's call the smallest even integer x. The next even integer will be x + 2, and the largest even integer will be x + 4. According to the problem, we know that the sum of the smallest and largest is 36. So, we can set up the equation x + (x + 4) = 36. Simplifying this equation, we get 2x + 4 = 36. Subtracting 4 from both sides of the equation, we have 2x = 32. Finally, dividing both sides by 2, we find that x = 16. Therefore, the three consecutive even integers are 16, 18, and 20.
To find three consecutive even integers, let's assume the first even integer as 'x'.
Since the integers are consecutive, the next two even integers would be 'x + 2' and 'x + 4'.
According to the problem, the sum of the smallest (x) and largest (x + 4) is 36. We can express this in an equation:
x + (x + 4) = 36
Simplifying the equation:
2x + 4 = 36
Subtracting 4 from both sides:
2x = 36 - 4
2x = 32
Dividing both sides by 2:
x = 32/2
x = 16
So, the first even integer is 16.
The next two consecutive even integers are 16 + 2 = 18 and 16 + 4 = 20.
Therefore, the three consecutive even integers such that the sum of the smallest and largest is 36 are 16, 18, and 20.