The mean (average) of 5 consecutive integers is 9 . What is the smallest of these 5 integers?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Solution 1

Since the mean of five consecutive integers is 9 , then the middle of these five integers is 9 .

Therefore, the integers are $7,8,9,10,11$, and so the smallest of the five integers is 7 .

## Solution 2

Suppose that $x$ is the smallest of the five consecutive integers.

Then the integers are $x, x+1, x+2, x+3$, and $x+4$.

The mean of these integers is $\frac{x+(x+1)+(x+2)+(x+3)+(x+4)}{5}=\frac{5 x+10}{5}=x+2$.

Since the mean is 9 , then $x+2=9$ or $x=7$.

Thus, the smallest of the five integers is 7 .

ANSWER: (D)