Find four integers such that when added three at a time, they give $6, 7, 8$, and 9.
Let $a, b, c$ and $d$ be the numbers sought. The given numbers are

$$
a+b+c, a+b+d, a+c+d \quad \text { and } \quad b+c+d
$$

$\log o$

$6+7+8+9=(a+b+c)+(a+b+d)+(a+c+d)+(b+c+d)=3(a+b+c+d)$.

Thus,

$$
a+b+c+d=\frac{30}{3}=10
$$

Note that each number is equal to the difference between the sum of the four numbers and the sum of the other three. For example: $c=(a+b+c+d)-(a+b+d)$. Therefore, the numbers sought are

$$
10-6=4 \quad, \quad 10-7=3 \quad, \quad 10-8=2 \quad \text { and } \quad 10-9=1
$$