Find all values of $r$ such that $\lfloor r \rfloor + r = 12.2$.
First, we note that $r$ must be positive, since otherwise $\lfloor r \rfloor + r$ is nonpositive. Next, because $\lfloor r \rfloor$ is an integer and $\lfloor r \rfloor + r=12.2$, the decimal part of $r$ must be $0.2$.  Therefore, $r=n+0.2$ for some integer $n$, so that $\lfloor r\rfloor =n$ and $\lfloor r \rfloor + r = 2n+0.2 =12.2$. Therefore, $n=6$, and the only value of $r$ that satisfies the equation is $\boxed{r=6.2}$.
6.2
Level 3
Algebra
Problem: Find all values of $r$ such that $\lfloor r \rfloor + r = 12.2$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation $\lfloor r \rfloor + r = 12.2$, we start by letting $\lfloor r \rfloor = n$, where $n$ is an integer. This means that $r$ can be expressed as $r = n + f$, where $f$ is the fractional part of $r$ and satisfies $0 \leq f < 1$. 

Substituting this into the equation gives us:

$n + (n + f) = 12.2$

This simplifies to:

$2n + f = 12.2$

Rearranging this, we find:

$f = 12.2 - 2n$

Since $f$ must be a valid fractional part, it must satisfy the condition $0 \leq f < 1$. Therefore, we have two inequalities to solve:

1. $12.2 - 2n \geq 0$
2. $12.2 - 2n < 1$

From the first inequality:

$12.2 \geq 2n \implies n \leq 6.1$

Since $n$ is an integer, this means $n \leq 6$.

From the second inequality:

$12.2 - 2n < 1 \implies 12.2 - 1 < 2n \implies 11.2 < 2n \implies n > 5.6$

Again, since $n$ is an integer, this means $n \geq 6$.

Combining these results, we find that $n$ must equal 6. 

Substituting $n = 6$ back into the expression for $f$:

$f = 12.2 - 2(6) = 12.2 - 12 = 0.2$

Thus, we can express $r$ as:

$r = n + f = 6 + 0.2 = 6.2$

Therefore, the only value of $r$ that satisfies the original equation is $6.2$.<|END_OF_LATENT|> The final answer is $6.2$.