If the mean (average) of five consecutive integers is 21 , the smallest of the five integers is
(A) 17
(B) 21
(C) 1
(D) 18
(E) 19
Solution 1

The mean of 5 consecutive integers is equal to the number in the middle.

Since the numbers have a mean of 21 , if we were to distribute the quantities equally, we would have 21, 21, 21, 21, and 21 .

Since the numbers are consecutive, the second number is 1 less than the 21 in the middle, while the fourth number is 1 more than the 21 in the middle.

Similarly, the first number is 2 less than the 21 in the middle, while the fifth number is 2 more than the 21 in the middle.

Thus, the numbers are $21-2,21-1,21,21+1,21+2$.

The smallest of 5 consecutive integers having a mean of 21 , is 19 .

## Solution 2

Since 21 is the mean of five consecutive integers, the smallest of these five integers must be less than 21.

Suppose that the smallest number is 20 .

The mean of $20,21,22,23$, and 24 is $\frac{20+21+22+23+24}{5}=22$.

This mean of 22 is greater than the required mean of 21 ; thus, the smallest of the 5 consecutive integers must be less than 20 .

Suppose that the smallest number is 19 .

The mean of $19,20,21,22$, and 23 , is $\frac{19+20+21+22+23}{5}=21$, as required.

Thus, the smallest of the 5 consecutive integers is 19 .

ANSWER: (E)