We consider the sequence $u_{n}$ as follows: $u_{0}=2010^{2010}$, and $u_{n+1}=u_{n}+7$ if $u_{n}$ is odd, and $\frac{u_{n}}{2}$ if it is even. What is the smallest integer that this sequence will reach?
First, it is evident that the $u_{n}$ are always strictly positive. Next, let's look at the possible minima: if $u_{n}$ is even, then $u_{n+1}=\frac{u_{n}}{2}$ is strictly less than $u_{n}$. If $u_{n}$ is odd and greater than 7, then $u_{n+2}=\frac{u_{n}+7}{2}$ is strictly less than $u_{n}$. It remains to consider 1, 3, 5, and 7.

- if $u_{n}=7$, then the subsequent terms are $14,7,14,7, \ldots$ and the sequence loops
- if $u_{n}=1$, then the subsequent terms are $8,4,2,1,8,4, \ldots$
- if $u_{n}=3$, then the subsequent terms are $10,5,12,6,3,10, \ldots$, the same for $u_{n}=5$

Therefore, there are three possible minima: 1, 3, or 7. It is easy to verify that $7 \mid u_{n+1}$ if and only if $7 \mid u_{n}$. Since $2010^{2010}$ is not divisible by 7, the sequence never passes through 7. We try to use the same idea to separate the last two cases: we separate the residues modulo 7 into $A=\{1,2,4\}$ and $B=\{3,5,6\}$. I leave it to you to verify that $u_{n+1} \in A$ if and only if $u_{n} \in A \bmod 7$, and similarly for $B$. It remains to determine which of the two $2010^{2010}$ belongs to. The integer 2010 is divisible by 6, so by Fermat's Little Theorem, $2010^{2010} \equiv 1 \pmod{7}$, and the minimum value of the sequence is 1.