When four positive integers are divided by $11$, the remainders are $2,$ $4,$ $6,$ and $8,$ respectively.

When the sum of the four integers is divided by $11$, what is the remainder?
We can call the four integers in this problem $a,$ $b,$ $c$, and $d$. Then we have \begin{align*}
a &\equiv 2\pmod{11}, \\
b &\equiv 4\pmod{11}, \\
c &\equiv 6\pmod{11}, \\
d &\equiv 8\pmod{11}.
\end{align*}Adding these congruences, we have \begin{align*}
a+b+c+d &\equiv 2+4+6+8 \\
&\equiv 20\pmod{11}.
\end{align*}Therefore, $a+b+c+d$ has the same remainder as $20$ upon division by $11$. This remainder is $\boxed{9}$.
9
Level 1
Number Theory
Problem: When four positive integers are divided by $11$, the remainders are $2,$ $4,$ $6,$ and $8,$ respectively.

When the sum of the four integers is divided by $11$, what is the remainder?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the remainder when the sum of the four integers is divided by $11$, we first denote the four integers as $a$, $b$, $c$, and $d$. According to the problem, we have the following congruences:

1. $a \equiv 2 \pmod{11}$
2. $b \equiv 4 \pmod{11}$
3. $c \equiv 6 \pmod{11}$
4. $d \equiv 8 \pmod{11}$

Next, we calculate the sum of these integers:

$a + b + c + d \equiv 2 + 4 + 6 + 8 \pmod{11}$.

Now, we perform the addition:

$2 + 4 = 6$  
$6 + 6 = 12$  
$12 + 8 = 20$.

Thus, we have:

$a + b + c + d \equiv 20 \pmod{11}$.

Next, we need to find the remainder of $20$ when divided by $11$. We perform the division:

$20 \div 11 = 1$ with a remainder of $9$.

Therefore, we conclude that:

$20 \equiv 9 \pmod{11}$.

This means that when the sum of the four integers is divided by $11$, the remainder is $9$. 

The final answer is $9$.<|END_OF_LATENT|> The final answer is $9$.