The sum of four consecutive odd integers is 448. Find the three integers.
I'll do better than that. I'll find all four integers.
Since odd numbers differ by 2, we have
x + x+2 + x+4 + x+6 = 448
x = 109
So the numbers are 109, 111, 113, 115
Well, let's call the smallest odd integer x. Since we want four consecutive odd integers, we can say the next three consecutive odd integers are x + 2, x + 4, and x + 6.
Now, we can set up an equation to find the sum:
x + (x + 2) + (x + 4) + (x + 6) = 448
Simplifying the equation, we get:
4x + 12 = 448
Subtracting 12 from both sides:
4x = 436
Dividing both sides by 4:
x = 109
So, the smallest odd integer is 109.
Therefore, the three consecutive odd integers are 109, 111, and 113. Voilà!
Let's represent the four consecutive odd integers as x, x+2, x+4, and x+6.
We know that their sum is 448, so we can write the equation:
x + (x+2) + (x+4) + (x+6) = 448
Simplifying the equation:
4x + 12 = 448
Subtracting 12 from both sides:
4x = 436
Dividing both sides by 4:
x = 109
So the first odd integer is 109.
The next two odd integers are:
x+2 = 109 + 2 = 111
x+4 = 109 + 4 = 113
So the three integers are 109, 111, and 113.
To solve this problem, we can set up an equation using algebra. Let's call the first odd integer "x". Since the integers are consecutive, the next three odd integers will be x + 2, x + 4, and x + 6.
The sum of these four consecutive odd integers is 448, so we can write the equation:
x + (x + 2) + (x + 4) + (x + 6) = 448
Now, let's simplify the equation:
4x + 12 = 448
Subtract 12 from both sides of the equation:
4x = 436
Divide both sides of the equation by 4:
x = 109
So, the first odd integer is 109. The next three consecutive odd integers are:
x + 2 = 109 + 2 = 111
x + 4 = 109 + 4 = 113
x + 6 = 109 + 6 = 115
Therefore, the three integers are 111, 113, and 115.