Find the largest integer $n$ that satisfies
$$\sum_{k=1}^{n} \left \lfloor{\frac 12 +\log_{10}k}\right \rfloor \leq 1000.$$
Using wolfram.alpha, I was able to approximate a few values of $k$ and $n$. Then again, I could have solved this (painstakingly) by keeping on substituting values of $k$ and $n$. However, I want to know a way to solve this problem without the use of a calculator. Help is much appreciated.
I think you'd need to calculate $\sqrt{10}$ by hand, which is $\doteq 3.16227766$.
From this, we can see that the summand increases in value when $k=4, 32,$ and $317$, and that should be enough to find $n$.
For $k<4$, the summand is zero. For $4 \leq k < 32$, the summand is $1$. For For $32 \leq k < 317$, the summand is $2$. For all the values of $k \geq 317$ that we care about, it's $3$.
Up to $k=316$, the sum is $1 \cdot 28 + 2 \cdot 285 = 598$. This means we need to add no more than $(1000-598)/3 = 134$ terms, which brings us to $n = 316+134=450$.