We are asked to find four consecutive integers whose sum is 206.

1. Problem Description

We are asked to find four consecutive integers whose sum is

2

0

6.

2. Solution Steps

Let the four consecutive integers be $n, n+1, n+2, n+3$.

Their sum is $n + (n+1) + (n+2) + (n+3)$.

We are given that this sum is

2

0

6. Therefore, we have the equation:

$n + (n+1) + (n+2) + (n+3) = 206$

Combining like terms, we have:

$4n + 6 = 206$

Subtracting 6 from both sides gives:

$4n = 206 - 6$

$4n = 200$

Dividing both sides by 4, we find:

$n = \frac{200}{4}$

$n = 50$

The four consecutive integers are then:

$n = 50$

$n+1 = 50+1 = 51$

$n+2 = 50+2 = 52$

$n+3 = 50+3 = 53$

The integers are 50, 51, 52,

5

3. Let's check if their sum is 206:

$50 + 51 + 52 + 53 = 206$

The solution is correct.

3. Final Answer

50, 51, 52, 53