What is the arithmetic mean of all three-digit integers that leave a remainder of 0 when divided by 19?

The correct option is 551.

To find the arithmetic mean of all three-digit integers that leave a remainder of 0 when divided by 19 (i.e., three-digit numbers divisible by 19), we can identify the first and last terms of this sequence and apply the properties of an arithmetic progression.

Three-digit integers range from 100 to 999.

First, find the smallest three-digit integer divisible by 19:
Dividing 100 by 19 gives approximately 5.26. The next whole number multiplier is 6.

$19\times 6=114$

Next, find the largest three-digit integer divisible by 19:
Dividing 999 by 19 gives approximately 52.58. The largest whole number multiplier is 52.

$19\times 52=988$

The three-digit multiples of 19 form an arithmetic progression (A.P.) where:
- First term ($a$) = 114
- Last term ($l$) = 988
- Common difference ($d$) = 19

For any sequence forming an arithmetic progression, the arithmetic mean (average) of all terms is simply equal to the average of the first term and the last term:

$\text{Arithmetic Mean}=\frac{\text{First term}+\text{Last term}}{2}$

Substituting the values of the first term and the last term into the formula:

$\text{Arithmetic Mean}=\frac{114+988}{2}$

$\text{Arithmetic Mean}=\frac{1102}{2}=551$

We can also verify this by calculating the total number of terms ($n$):

$l=a+(n-1)d$

$988=114+(n-1)\times 19$

$874=(n-1)\times 19$

$n-1=46\Rightarrow n=47$

The sum of all 47 terms is:

$\text{Sum}=\frac{n}{2}\times (a+l)=\frac{47}{2}\times (114+988)=47\times 551$

Dividing the sum by the total number of terms gives the mean:

$\text{Arithmetic Mean}=\frac{47\times 551}{47}=551$

Thus, the arithmetic mean of all three-digit integers divisible by 19 is 551.